Title: Chapter 11
Course: quiz duplicate error

Q: Find the perimeter and area of the parallelogram. Round to the nearest tenth if necessary.
[Image]
- 88 mm; 415.7 [Image] (The perimeter is the sum of all four sides. So add the two sides given and then double the result. This accounts for the other two sides which are the same as those given. For example, if the two given sides are 8 and 12, the perimeter is 2 ´ (8 + 12) = 40. The area is base ´ height. The base is given. The height can be found by recognizing the 30°-60°-90° triangle at the left. The height is the long leg which is found by taking the hypotenuse (the given slanted side) and dividing by 2, then multiplying by the square root of three. For example, if the slanted side is 8, the height is 8 ¸ 2 ´ [Image]. Multiply that height times the given base and you have the area.)
- 44 mm; 346.4 [Image] (The perimeter is the sum of all four sides. So add the two sides given and then double the result. This accounts for the other two sides which are the same as those given. For example, if the two given sides are 8 and 12, the perimeter is 2 ´ (8 + 12) = 40. The area is base ´ height. The base is given. The height can be found by recognizing the 30°-60°-90° triangle at the left. The height is the long leg which is found by taking the hypotenuse (the given slanted side) and dividing by 2, then multiplying by the square root of three. For example, if the slanted side is 8, the height is 8 ¸ 2 ´ [Image]. Multiply that height times the given base and you have the area.)
- 44 mm; 415.7 [Image] (The perimeter is the sum of all four sides. So add the two sides given and then double the result. This accounts for the other two sides which are the same as those given. For example, if the two given sides are 8 and 12, the perimeter is 2 ´ (8 + 12) = 40. The area is base ´ height. The base is given. The height can be found by recognizing the 30°-60°-90° triangle at the left. The height is the long leg which is found by taking the hypotenuse (the given slanted side) and dividing by 2, then multiplying by the square root of three. For example, if the slanted side is 8, the height is 8 ¸ 2 ´ [Image]. Multiply that height times the given base and you have the area.)
- 88 mm; 346.4 [Image] (The perimeter is the sum of all four sides. So add the two sides given and then double the result. This accounts for the other two sides which are the same as those given. For example, if the two given sides are 8 and 12, the perimeter is 2 ´ (8 + 12) = 40. The area is base ´ height. The base is given. The height can be found by recognizing the 30°-60°-90° triangle at the left. The height is the long leg which is found by taking the hypotenuse (the given slanted side) and dividing by 2, then multiplying by the square root of three. For example, if the slanted side is 8, the height is 8 ¸ 2 ´ [Image]. Multiply that height times the given base and you have the area.)
A: 88 mm; 415.7 [Image]

Q: Find the perimeter and area of the parallelogram. Round to the nearest tenth if necessary.
[Image]
- 16 mm; 15.4 [Image] (The perimeter is the sum of all four sides. So add the two sides given and then double the result. This accounts for the other two sides which are the same as those given. For example, if the two given sides are 4 and 6, the perimeter is 2 ´ (4 + 6) = 20. The area is base ´ height. Since this parallelogram has a right angle, that makes it a rectangle as well. So simple multiply that height times the given base and you have the area. For example, if the base is 6 and the height is 4, the area is 6 ´ 4 = 24.)
- 8 mm; 18.5 [Image] (The perimeter is the sum of all four sides. So add the two sides given and then double the result. This accounts for the other two sides which are the same as those given. For example, if the two given sides are 4 and 6, the perimeter is 2 ´ (4 + 6) = 20. The area is base ´ height. Since this parallelogram has a right angle, that makes it a rectangle as well. So simple multiply that height times the given base and you have the area. For example, if the base is 6 and the height is 4, the area is 6 ´ 4 = 24.)
- 16 mm; 18.5 [Image] (The perimeter is the sum of all four sides. So add the two sides given and then double the result. This accounts for the other two sides which are the same as those given. For example, if the two given sides are 4 and 6, the perimeter is 2 ´ (4 + 6) = 20. The area is base ´ height. Since this parallelogram has a right angle, that makes it a rectangle as well. So simple multiply that height times the given base and you have the area. For example, if the base is 6 and the height is 4, the area is 6 ´ 4 = 24.)
- 8 mm; 15.4 [Image] (The perimeter is the sum of all four sides. So add the two sides given and then double the result. This accounts for the other two sides which are the same as those given. For example, if the two given sides are 4 and 6, the perimeter is 2 ´ (4 + 6) = 20. The area is base ´ height. Since this parallelogram has a right angle, that makes it a rectangle as well. So simple multiply that height times the given base and you have the area. For example, if the base is 6 and the height is 4, the area is 6 ´ 4 = 24.)
A: 16 mm; 15.4 [Image]

Q: Given the coordinates of the vertices of a quadrilateral, determine whether it is a square, a rectangle or a parallelogram. Then find the area of the quadrilateral.

A(0, 0), B(3, 0), C(2, 4), D(5, 4)
- square, 15 units^2 (The slopes of the opposite sides are equal, making it a parallelogram. It is not a rectangle because the slopes of the consecutive sides are not negative reciprocals of each other. The measure of the base is 3 and the height is 4, so the area is 3 ´ 4 = 12.)
- parallelogram, 12 units^2 (The slopes of the opposite sides are equal, making it a parallelogram. It is not a rectangle because the slopes of the consecutive sides are not negative reciprocals of each other. The measure of the base is 3 and the height is 4, so the area is 3 ´ 4 = 12.)
- rectangle, 10 units^2 (The slopes of the opposite sides are equal, making it a parallelogram. It is not a rectangle because the slopes of the consecutive sides are not negative reciprocals of each other. The measure of the base is 3 and the height is 4, so the area is 3 ´ 4 = 12.)
- parallelogram, 16 units^2 (The slopes of the opposite sides are equal, making it a parallelogram. It is not a rectangle because the slopes of the consecutive sides are not negative reciprocals of each other. The measure of the base is 3 and the height is 4, so the area is 3 ´ 4 = 12.)
A: parallelogram, 12 units^2

Q: Given the coordinates of the vertices of a quadrilateral, determine whether it is a square, a rectangle or a parallelogram. Then find the area of the quadrilateral.

E(0, –2), F(3, –2), G(–2, 3), H(1, 3)
- parallelogram, 15 units^2 (The slopes of the opposite sides are equal, making it a parallelogram. It is not a rectangle because the slopes of the consecutive sides are not negative reciprocals of each other. The measure of the base is 3 and the height is 5, so the area is 3 ´ 5 = 15.)
- parallelogram, 12 units^2 (The slopes of the opposite sides are equal, making it a parallelogram. It is not a rectangle because the slopes of the consecutive sides are not negative reciprocals of each other. The measure of the base is 3 and the height is 5, so the area is 3 ´ 5 = 15.)
- square, 10 units^2 (The slopes of the opposite sides are equal, making it a parallelogram. It is not a rectangle because the slopes of the consecutive sides are not negative reciprocals of each other. The measure of the base is 3 and the height is 5, so the area is 3 ´ 5 = 15.)
- rectangle, 18 units^2 (The slopes of the opposite sides are equal, making it a parallelogram. It is not a rectangle because the slopes of the consecutive sides are not negative reciprocals of each other. The measure of the base is 3 and the height is 5, so the area is 3 ´ 5 = 15.)
A: parallelogram, 15 units^2

Q: Given the coordinates of the vertices of a quadrilateral, determine whether it is a square, a rectangle or a parallelogram. Then find the area of the quadrilateral.

J(–2, –2), K(4, –2), L(–2, 2), M(4, 2)
- parallelogram, 30 units^2 (This quadrilateral is a rectangle because the slopes of its consecutive sides are negative reciprocals of each other, making the sides perpendicular and forming right angles. The measure of the base is 6 and the height is 4, so the area is 6 ´ 4 = 24.)
- rectangle, 30 units^2 (This quadrilateral is a rectangle because the slopes of its consecutive sides are negative reciprocals of each other, making the sides perpendicular and forming right angles. The measure of the base is 6 and the height is 4, so the area is 6 ´ 4 = 24.)
- parallelogram, 24 units^2 (This quadrilateral is a rectangle because the slopes of its consecutive sides are negative reciprocals of each other, making the sides perpendicular and forming right angles. The measure of the base is 6 and the height is 4, so the area is 6 ´ 4 = 24.)
- rectangle, 24 units^2 (This quadrilateral is a rectangle because the slopes of its consecutive sides are negative reciprocals of each other, making the sides perpendicular and forming right angles. The measure of the base is 6 and the height is 4, so the area is 6 ´ 4 = 24.)
A: rectangle, 24 units^2

Q: Given the coordinates of the vertices of a quadrilateral, determine whether it is a square, a rectangle or a parallelogram. Then find the area of the quadrilateral.

N(–4, –4), O(3, –4), P(–3, 1), Q(4, 1)
- parallelogram, 35 units^2 (The slopes of the opposite sides are equal, making it a parallelogram. It is not a rectangle because the slopes of the consecutive sides are not negative reciprocals of each other. The measure of the base is 7 and the height is 5, so the area is 7 ´ 5 = 35.)
- square, 28 units^2 (The slopes of the opposite sides are equal, making it a parallelogram. It is not a rectangle because the slopes of the consecutive sides are not negative reciprocals of each other. The measure of the base is 7 and the height is 5, so the area is 7 ´ 5 = 35.)
- rectangle, 30 units^2 (The slopes of the opposite sides are equal, making it a parallelogram. It is not a rectangle because the slopes of the consecutive sides are not negative reciprocals of each other. The measure of the base is 7 and the height is 5, so the area is 7 ´ 5 = 35.)
- parallelogram, 40 units^2 (The slopes of the opposite sides are equal, making it a parallelogram. It is not a rectangle because the slopes of the consecutive sides are not negative reciprocals of each other. The measure of the base is 7 and the height is 5, so the area is 7 ´ 5 = 35.)
A: parallelogram, 35 units^2

Q: Given the coordinates of the vertices of a quadrilateral, determine whether it is a square, a rectangle or a parallelogram. Then find the area of the quadrilateral.

R(–5, 5), S(1, 5), T(–5, –1), V(1, –1)
- rectangle, 36 units^2 (This quadrilateral is a square because all sides are the same length and the slopes of its consecutive sides are negative reciprocals of each other, making the sides perpendicular and forming right angles. The measure of each side is 6, so the area is [Image] = 36.)
- square, 36 units^2 (This quadrilateral is a square because all sides are the same length and the slopes of its consecutive sides are negative reciprocals of each other, making the sides perpendicular and forming right angles. The measure of each side is 6, so the area is [Image] = 36.)
- square, 30 units^2 (This quadrilateral is a square because all sides are the same length and the slopes of its consecutive sides are negative reciprocals of each other, making the sides perpendicular and forming right angles. The measure of each side is 6, so the area is [Image] = 36.)
- rectangle, 30 units^2 (This quadrilateral is a square because all sides are the same length and the slopes of its consecutive sides are negative reciprocals of each other, making the sides perpendicular and forming right angles. The measure of each side is 6, so the area is [Image] = 36.)
A: square, 36 units^2

Q: Given the coordinates of the vertices of a quadrilateral, determine whether it is a square, a rectangle or a parallelogram. Then find the area of the quadrilateral.

W(–5, 4), X(2, 4), Y(–2, –4), Z(5, –4)
- square, 63 units^2 (The slopes of the opposite sides are equal, making it a parallelogram. It is not a rectangle because the slopes of the consecutive sides are not negative reciprocals of each other. The measure of the base is 7 and the height is 8, so the area is 7 ´ 8 = 56.)
- rectangle, 48 units^2 (The slopes of the opposite sides are equal, making it a parallelogram. It is not a rectangle because the slopes of the consecutive sides are not negative reciprocals of each other. The measure of the base is 7 and the height is 8, so the area is 7 ´ 8 = 56.)
- parallelogram, 56 units^2 (The slopes of the opposite sides are equal, making it a parallelogram. It is not a rectangle because the slopes of the consecutive sides are not negative reciprocals of each other. The measure of the base is 7 and the height is 8, so the area is 7 ´ 8 = 56.)
- parallelogram, 49 units^2 (The slopes of the opposite sides are equal, making it a parallelogram. It is not a rectangle because the slopes of the consecutive sides are not negative reciprocals of each other. The measure of the base is 7 and the height is 8, so the area is 7 ´ 8 = 56.)
A: parallelogram, 56 units^2

Q: Given the coordinates of the vertices of a quadrilateral, determine whether it is a square, a rectangle or a parallelogram. Then find the area of the quadrilateral.

A(–1, 3), B(3, 1), C(1, 7), D(5, 5)
- square, 20 units^2 (This quadrilateral is a square because all sides are the same length and the slopes of its consecutive sides are negative reciprocals of each other, making the sides perpendicular and forming right angles. The measure of each side is [Image], so the area is [Image] = 20.)
- square, 24 units^2 (This quadrilateral is a square because all sides are the same length and the slopes of its consecutive sides are negative reciprocals of each other, making the sides perpendicular and forming right angles. The measure of each side is [Image], so the area is [Image] = 20.)
- parallelogram, 20 units^2 (This quadrilateral is a square because all sides are the same length and the slopes of its consecutive sides are negative reciprocals of each other, making the sides perpendicular and forming right angles. The measure of each side is [Image], so the area is [Image] = 20.)
- parallelogram, 24 units^2 (This quadrilateral is a square because all sides are the same length and the slopes of its consecutive sides are negative reciprocals of each other, making the sides perpendicular and forming right angles. The measure of each side is [Image], so the area is [Image] = 20.)
A: square, 20 units^2

Q: Given the coordinates of the vertices of a quadrilateral, determine whether it is a square, a rectangle or a parallelogram. Then find the area of the quadrilateral.

E(–7, –1), F(2, –1), G(–4, –8), H(5, –8)
- rectangle, 56 units^2 (The slopes of the opposite sides are equal, making it a parallelogram. It is not a rectangle because the slopes of the consecutive sides are not negative reciprocals of each other. The measure of the base is 9 and the height is 7, so the area is 9 ´ 7 = 63.)
- rectangle, 70 units^2 (The slopes of the opposite sides are equal, making it a parallelogram. It is not a rectangle because the slopes of the consecutive sides are not negative reciprocals of each other. The measure of the base is 9 and the height is 7, so the area is 9 ´ 7 = 63.)
- parallelogram, 54 units^2 (The slopes of the opposite sides are equal, making it a parallelogram. It is not a rectangle because the slopes of the consecutive sides are not negative reciprocals of each other. The measure of the base is 9 and the height is 7, so the area is 9 ´ 7 = 63.)
- parallelogram, 63 units^2 (The slopes of the opposite sides are equal, making it a parallelogram. It is not a rectangle because the slopes of the consecutive sides are not negative reciprocals of each other. The measure of the base is 9 and the height is 7, so the area is 9 ´ 7 = 63.)
A: parallelogram, 63 units^2

Q: Given the coordinates of the vertices of a quadrilateral, determine whether it is a square, a rectangle or a parallelogram. Then find the area of the quadrilateral.

J(–6, 4), K(–1, 4), L(–6, –4), M(–1, –4)
- parallelogram, 40 units^2 (This quadrilateral is a rectangle because the slopes of its consecutive sides are negative reciprocals of each other, making the sides perpendicular and forming right angles. The measure of the base is 5 and the height is 8, so the area is 5 ´ 8 = 40.)
- rectangle, 40 units^2 (This quadrilateral is a rectangle because the slopes of its consecutive sides are negative reciprocals of each other, making the sides perpendicular and forming right angles. The measure of the base is 5 and the height is 8, so the area is 5 ´ 8 = 40.)
- parallelogram, 32 units^2 (This quadrilateral is a rectangle because the slopes of its consecutive sides are negative reciprocals of each other, making the sides perpendicular and forming right angles. The measure of the base is 5 and the height is 8, so the area is 5 ´ 8 = 40.)
- rectangle, 32 units^2 (This quadrilateral is a rectangle because the slopes of its consecutive sides are negative reciprocals of each other, making the sides perpendicular and forming right angles. The measure of the base is 5 and the height is 8, so the area is 5 ´ 8 = 40.)
A: rectangle, 40 units^2

Q: Given the coordinates of the vertices of a quadrilateral, determine whether it is a square, a rectangle or a parallelogram. Then find the area of the quadrilateral.

N(0, 5), O(5, 0), P(–5, 0), Q(0, –5)
- square, 25 units^2 (This quadrilateral is a square because all sides are the same length and the slopes of its consecutive sides are negative reciprocals of each other, making the sides perpendicular and forming right angles. The measure of each side is [Image], so the area is [Image] = 50.)
- rectangle, 25 units^2 (This quadrilateral is a square because all sides are the same length and the slopes of its consecutive sides are negative reciprocals of each other, making the sides perpendicular and forming right angles. The measure of each side is [Image], so the area is [Image] = 50.)
- square, 50 units^2 (This quadrilateral is a square because all sides are the same length and the slopes of its consecutive sides are negative reciprocals of each other, making the sides perpendicular and forming right angles. The measure of each side is [Image], so the area is [Image] = 50.)
- rectangle, 50 units^2 (This quadrilateral is a square because all sides are the same length and the slopes of its consecutive sides are negative reciprocals of each other, making the sides perpendicular and forming right angles. The measure of each side is [Image], so the area is [Image] = 50.)
A: square, 50 units^2

Q: Find the area of the figure. Round to the nearest tenth if necessary.
[Image]
- 13.1 [Image] (The area of a triangle is found by the formula [Image]. For example, if the base of the triangle is 6 and the height is 4, the area is [Image] ´ (6 ´ 4) = 12.)
- 12 [Image] (The area of a triangle is found by the formula [Image]. For example, if the base of the triangle is 6 and the height is 4, the area is [Image] ´ (6 ´ 4) = 12.)
- 23.9 [Image] (The area of a triangle is found by the formula [Image]. For example, if the base of the triangle is 6 and the height is 4, the area is [Image] ´ (6 ´ 4) = 12.)
- 12.8 [Image] (The area of a triangle is found by the formula [Image]. For example, if the base of the triangle is 6 and the height is 4, the area is [Image] ´ (6 ´ 4) = 12.)
A: 12 [Image]

Q: Find the area of the figure. Round to the nearest tenth if necessary.
[Image]
- 39 [Image] (The area of a triangle is found by the formula [Image]. For example, if the base of the triangle is 6 and the height is 4, the area is [Image] ´ (6 ´ 4) = 12.)
- 36.4 [Image] (The area of a triangle is found by the formula [Image]. For example, if the base of the triangle is 6 and the height is 4, the area is [Image] ´ (6 ´ 4) = 12.)
- 72.8 [Image] (The area of a triangle is found by the formula [Image]. For example, if the base of the triangle is 6 and the height is 4, the area is [Image] ´ (6 ´ 4) = 12.)
- 33.9 [Image] (The area of a triangle is found by the formula [Image]. For example, if the base of the triangle is 6 and the height is 4, the area is [Image] ´ (6 ´ 4) = 12.)
A: 36.4 [Image]

Q: Find the area of the figure. Round to the nearest tenth if necessary.
[Image]
- 160 [Image] (The area of a trapezoid is found by the formula [Image]. For example, if the bases of the trapezoid are 8 and 10 and the height is 6, the area is [Image] ´ 6 ´ (8 + 10) = 54.)
- 360 [Image] (The area of a trapezoid is found by the formula [Image]. For example, if the bases of the trapezoid are 8 and 10 and the height is 6, the area is [Image] ´ 6 ´ (8 + 10) = 54.)
- 180 [Image] (The area of a trapezoid is found by the formula [Image]. For example, if the bases of the trapezoid are 8 and 10 and the height is 6, the area is [Image] ´ 6 ´ (8 + 10) = 54.)
- 130 [Image] (The area of a trapezoid is found by the formula [Image]. For example, if the bases of the trapezoid are 8 and 10 and the height is 6, the area is [Image] ´ 6 ´ (8 + 10) = 54.)
A: 180 [Image]

Q: Find the area of the figure. Round to the nearest tenth if necessary.
[Image]
- 194.2 [Image] (The area of a trapezoid is found by the formula [Image]. For example, if the bases of the trapezoid are 8 and 10 and the height is 6, the area is [Image] ´ 6 ´ (8 + 10) = 54.)
- 88.7 [Image] (The area of a trapezoid is found by the formula [Image]. For example, if the bases of the trapezoid are 8 and 10 and the height is 6, the area is [Image] ´ 6 ´ (8 + 10) = 54.)
- 99.3 [Image] (The area of a trapezoid is found by the formula [Image]. For example, if the bases of the trapezoid are 8 and 10 and the height is 6, the area is [Image] ´ 6 ´ (8 + 10) = 54.)
- 97.1 [Image] (The area of a trapezoid is found by the formula [Image]. For example, if the bases of the trapezoid are 8 and 10 and the height is 6, the area is [Image] ´ 6 ´ (8 + 10) = 54.)
A: 97.1 [Image]

Q: Find the area of the figure. Round to the nearest tenth if necessary.
[Image]
- 66 [Image] (The area of a rhombus is found by the formula [Image]. For example, if the diagonals of the rhombus are 20 and 30, the area is [Image] ´ 20 ´ 30 = 300.)
- 1064 [Image] (The area of a rhombus is found by the formula [Image]. For example, if the diagonals of the rhombus are 20 and 30, the area is [Image] ´ 20 ´ 30 = 300.)
- 532 [Image] (The area of a rhombus is found by the formula [Image]. For example, if the diagonals of the rhombus are 20 and 30, the area is [Image] ´ 20 ´ 30 = 300.)
- 266 [Image] (The area of a rhombus is found by the formula [Image]. For example, if the diagonals of the rhombus are 20 and 30, the area is [Image] ´ 20 ´ 30 = 300.)
A: 532 [Image]

Q: Find the area of the figure. Round to the nearest tenth if necessary.
[Image]
- 672 [Image] (The area of a rhombus is found by the formula [Image]. For example, if the diagonals of the rhombus are 20 and 30, the area is [Image] ´ 20 ´ 30 = 300.)
- 74 [Image] (The area of a rhombus is found by the formula [Image]. For example, if the diagonals of the rhombus are 20 and 30, the area is [Image] ´ 20 ´ 30 = 300.)
- 336 [Image] (The area of a rhombus is found by the formula [Image]. For example, if the diagonals of the rhombus are 20 and 30, the area is [Image] ´ 20 ´ 30 = 300.)
- 1344 [Image] (The area of a rhombus is found by the formula [Image]. For example, if the diagonals of the rhombus are 20 and 30, the area is [Image] ´ 20 ´ 30 = 300.)
A: 672 [Image]

Q: Find the area of a regular hexagon with a perimeter of 90 centimeters. Round to the nearest tenth if necessary.
- 195 units^2 (The formula for the area of a regular polygon is [Image], where P is the perimeter and a is the length of the apothem. Some relationships you will need in these types of problems are as follows: Perimeter = number of sides ´ length of side. Apothem = length of side ¸ (2 ´ tan(angle)). Angle = 360 ¸ (2 ´ number of sides).)
- 585 units^2 (The formula for the area of a regular polygon is [Image], where P is the perimeter and a is the length of the apothem. Some relationships you will need in these types of problems are as follows: Perimeter = number of sides ´ length of side. Apothem = length of side ¸ (2 ´ tan(angle)). Angle = 360 ¸ (2 ´ number of sides).)
- 1170 units^2 (The formula for the area of a regular polygon is [Image], where P is the perimeter and a is the length of the apothem. Some relationships you will need in these types of problems are as follows: Perimeter = number of sides ´ length of side. Apothem = length of side ¸ (2 ´ tan(angle)). Angle = 360 ¸ (2 ´ number of sides).)
- 97.5 units^2 (The formula for the area of a regular polygon is [Image], where P is the perimeter and a is the length of the apothem. Some relationships you will need in these types of problems are as follows: Perimeter = number of sides ´ length of side. Apothem = length of side ¸ (2 ´ tan(angle)). Angle = 360 ¸ (2 ´ number of sides).)
A: 585 units^2

Q: The coordinates of the vertices of a regular polygon are given. Find the area of the polygon to the nearest tenth.
A(0, 0), B(2, –2), C(0, –4), D(–2, –2)
- 11.3 (This figure is a square or a regular quadrilateral. Begin by graphing the points and then use the distance formula with two consecutive vertices to find the length of a side. You can then find the area by squaring the length of a side, since that is the area formula for a square. For example, if the length of one side is found to be 3.5, then the area will be 3.5 ´ 3.5 = 12.25.)
- 5.7 (This figure is a square or a regular quadrilateral. Begin by graphing the points and then use the distance formula with two consecutive vertices to find the length of a side. You can then find the area by squaring the length of a side, since that is the area formula for a square. For example, if the length of one side is found to be 3.5, then the area will be 3.5 ´ 3.5 = 12.25.)
- 2.8 (This figure is a square or a regular quadrilateral. Begin by graphing the points and then use the distance formula with two consecutive vertices to find the length of a side. You can then find the area by squaring the length of a side, since that is the area formula for a square. For example, if the length of one side is found to be 3.5, then the area will be 3.5 ´ 3.5 = 12.25.)
- 8 (This figure is a square or a regular quadrilateral. Begin by graphing the points and then use the distance formula with two consecutive vertices to find the length of a side. You can then find the area by squaring the length of a side, since that is the area formula for a square. For example, if the length of one side is found to be 3.5, then the area will be 3.5 ´ 3.5 = 12.25.)
A: 8

Q: Find the area of a circle having a circumference of [Image]. Round to the nearest tenth. Use 3.14 for pi.
- 808.9 units^2 (The area formula for a circle is [Image]. To find the radius, we use the formula for the circumference, which is [Image] or [Image]. For example, if the circumference is given as [Image], the radius will be [Image]. Thus the area is [Image].)
- 907.5 units^2 (The area formula for a circle is [Image]. To find the radius, we use the formula for the circumference, which is [Image] or [Image]. For example, if the circumference is given as [Image], the radius will be [Image]. Thus the area is [Image].)
- 1814.9 units^2 (The area formula for a circle is [Image]. To find the radius, we use the formula for the circumference, which is [Image] or [Image]. For example, if the circumference is given as [Image], the radius will be [Image]. Thus the area is [Image].)
- 903.3 units^2 (The area formula for a circle is [Image]. To find the radius, we use the formula for the circumference, which is [Image] or [Image]. For example, if the circumference is given as [Image], the radius will be [Image]. Thus the area is [Image].)
A: 907.5 units^2

Q: Find the area of a circle having a circumference of 79.3. Round to the nearest tenth. Use 3.14 for p.
- 430.2 [Image] (The area formula for a circle is [Image]. To find the radius, use the formula for the circumference, which is [Image] or [Image]. For example, if the circumference is given as 20, the radius will be [Image]. Thus the area is [Image].)
- 1001.4 [Image] (The area formula for a circle is [Image]. To find the radius, use the formula for the circumference, which is [Image] or [Image]. For example, if the circumference is given as 20, the radius will be [Image]. Thus the area is [Image].)
- 500.7 [Image] (The area formula for a circle is [Image]. To find the radius, use the formula for the circumference, which is [Image] or [Image]. For example, if the circumference is given as 20, the radius will be [Image]. Thus the area is [Image].)
- 469.9 [Image] (The area formula for a circle is [Image]. To find the radius, use the formula for the circumference, which is [Image] or [Image]. For example, if the circumference is given as 20, the radius will be [Image]. Thus the area is [Image].)
A: 500.7 [Image]

Q: Find the area of the figure. Round to the nearest tenth if necessary.

[Image]
- 764.7 units^2 (This figure consists of a rectangle and two semicircles which combine to make one circle with radius 6.5. The area of the rectangle is [Image]. The area of the circle is [Image] So the total area of the figure is [Image].)
- 366.7 units^2 (This figure consists of a rectangle and two semicircles which combine to make one circle with radius 6.5. The area of the rectangle is [Image]. The area of the circle is [Image] So the total area of the figure is [Image].)
- 249.7 units^2 (This figure consists of a rectangle and two semicircles which combine to make one circle with radius 6.5. The area of the rectangle is [Image]. The area of the circle is [Image] So the total area of the figure is [Image].)
- 300.3 units^2 (This figure consists of a rectangle and two semicircles which combine to make one circle with radius 6.5. The area of the rectangle is [Image]. The area of the circle is [Image] So the total area of the figure is [Image].)
A: 366.7 units^2

Q: Find the area of the figure. Round to the nearest tenth if necessary.

[Image]
- 96 units^2 (This figure is equivalent to a rectangle with length 12 and width 8. The triangular portion that sticks out to the right is congruent to the triangular portion cut out on the left. So removing the right portion and placing it in the space on the left results in a rectangle. The area is then [Image].)
- 48 units^2 (This figure is equivalent to a rectangle with length 12 and width 8. The triangular portion that sticks out to the right is congruent to the triangular portion cut out on the left. So removing the right portion and placing it in the space on the left results in a rectangle. The area is then [Image].)
- 72 units^2 (This figure is equivalent to a rectangle with length 12 and width 8. The triangular portion that sticks out to the right is congruent to the triangular portion cut out on the left. So removing the right portion and placing it in the space on the left results in a rectangle. The area is then [Image].)
- 108 units^2 (This figure is equivalent to a rectangle with length 12 and width 8. The triangular portion that sticks out to the right is congruent to the triangular portion cut out on the left. So removing the right portion and placing it in the space on the left results in a rectangle. The area is then [Image].)
A: 96 units^2

Q: Find the area of the figure. Round to the nearest tenth if necessary.

[Image]
- 137.1 units^2 (This figure represents a rectangle with a semicircle cut out from one end. So subtract the area of the semicircle from the area of the rectangle. The area of the rectangle is [Image]. The area of the semicircle is [Image]. So the area of the figure is [Image].)
- 61.76 units^2 (This figure represents a rectangle with a semicircle cut out from one end. So subtract the area of the semicircle from the area of the rectangle. The area of the rectangle is [Image]. The area of the semicircle is [Image]. So the area of the figure is [Image].)
- 86.9 units^2 (This figure represents a rectangle with a semicircle cut out from one end. So subtract the area of the semicircle from the area of the rectangle. The area of the rectangle is [Image]. The area of the semicircle is [Image]. So the area of the figure is [Image].)
- 162.24 units^2 (This figure represents a rectangle with a semicircle cut out from one end. So subtract the area of the semicircle from the area of the rectangle. The area of the rectangle is [Image]. The area of the semicircle is [Image]. So the area of the figure is [Image].)
A: 86.9 units^2

Q: Find the area of the figure. Round to the nearest tenth if necessary.

[Image]
- 148 units^2 (This figure represents a rectangle attached to a right triangle. The area of the rectangle is [Image]. The area of the triangle is [Image]. So the total area of the figure is [Image].)
- 140 units^2 (This figure represents a rectangle attached to a right triangle. The area of the rectangle is [Image]. The area of the triangle is [Image]. So the total area of the figure is [Image].)
- 130 units^2 (This figure represents a rectangle attached to a right triangle. The area of the rectangle is [Image]. The area of the triangle is [Image]. So the total area of the figure is [Image].)
- 122 units^2 (This figure represents a rectangle attached to a right triangle. The area of the rectangle is [Image]. The area of the triangle is [Image]. So the total area of the figure is [Image].)
A: 122 units^2

Q: Find the area of the figure. Round to the nearest tenth if necessary.

[Image]
- 1920 units^2 (This figure represents a rectangle attached to a triangle. The area of the rectangle is [Image]. The area of the triangle is [Image]. So the total area of the figure is [Image].)
- 1936 units^2 (This figure represents a rectangle attached to a triangle. The area of the rectangle is [Image]. The area of the triangle is [Image]. So the total area of the figure is [Image].)
- 2304 units^2 (This figure represents a rectangle attached to a triangle. The area of the rectangle is [Image]. The area of the triangle is [Image]. So the total area of the figure is [Image].)
- 2688 units^2 (This figure represents a rectangle attached to a triangle. The area of the rectangle is [Image]. The area of the triangle is [Image]. So the total area of the figure is [Image].)
A: 2304 units^2

Q: Find the area of the figure. Round to the nearest tenth if necessary.

[Image]
- 240 units^2 (The figure represents two congruent trapezoids attached at one of their bases. The area formula for a trapezoid is [Image]. So for each trapezoid, the area is [Image], and the combined area of the two trapezoids is 480.)
- 480 units^2 (The figure represents two congruent trapezoids attached at one of their bases. The area formula for a trapezoid is [Image]. So for each trapezoid, the area is [Image], and the combined area of the two trapezoids is 480.)
- 960 units^2 (The figure represents two congruent trapezoids attached at one of their bases. The area formula for a trapezoid is [Image]. So for each trapezoid, the area is [Image], and the combined area of the two trapezoids is 480.)
- 1920 units^2 (The figure represents two congruent trapezoids attached at one of their bases. The area formula for a trapezoid is [Image]. So for each trapezoid, the area is [Image], and the combined area of the two trapezoids is 480.)
A: 480 units^2

Q: Find the area of the figure. Round to the nearest tenth if necessary.

[Image]
- 541.9 units^2 (This figure represents a rectangle with two identical semicircles cut out on each side. The semicircles combine to make one circle with a diameter of 24 and a radius of 12. The area of the rectangle is [Image]. The area of the circle is [Image], using 3.14 for [Image]. So the area of the figure is [Image].)
- 624 units^2 (This figure represents a rectangle with two identical semicircles cut out on each side. The semicircles combine to make one circle with a diameter of 24 and a radius of 12. The area of the rectangle is [Image]. The area of the circle is [Image], using 3.14 for [Image]. So the area of the figure is [Image].)
- 192 units^2 (This figure represents a rectangle with two identical semicircles cut out on each side. The semicircles combine to make one circle with a diameter of 24 and a radius of 12. The area of the rectangle is [Image]. The area of the circle is [Image], using 3.14 for [Image]. So the area of the figure is [Image].)
- 315.8 units^2 (This figure represents a rectangle with two identical semicircles cut out on each side. The semicircles combine to make one circle with a diameter of 24 and a radius of 12. The area of the rectangle is [Image]. The area of the circle is [Image], using 3.14 for [Image]. So the area of the figure is [Image].)
A: 315.8 units^2

Q: Find the area of the figure. Round to the nearest tenth if necessary.

[Image]
- 67.7 units^2 (This figure represents a rectangle attached to an isosceles triangle. The area of the rectangle is [Image]. The area of the triangle is [Image]. So the total area of the figure is [Image] when rounded to the nearest tenth.)
- 132.48 units^2 (This figure represents a rectangle attached to an isosceles triangle. The area of the rectangle is [Image]. The area of the triangle is [Image]. So the total area of the figure is [Image] when rounded to the nearest tenth.)
- 69.1 units^2 (This figure represents a rectangle attached to an isosceles triangle. The area of the rectangle is [Image]. The area of the triangle is [Image]. So the total area of the figure is [Image] when rounded to the nearest tenth.)
- 66.2 units^2 (This figure represents a rectangle attached to an isosceles triangle. The area of the rectangle is [Image]. The area of the triangle is [Image]. So the total area of the figure is [Image] when rounded to the nearest tenth.)
A: 66.2 units^2

Q: Find the area of the figure. Round to the nearest tenth if necessary.

[Image]
- 160 units^2 (This figure represents a rectangle with dimensions 10 by 16 attached to a semicircle with radius 5. The area of the rectangle is [Image]. The area of the semicircle is [Image]. So the area of the entire figure is [Image].)
- 167.9 units^2 (This figure represents a rectangle with dimensions 10 by 16 attached to a semicircle with radius 5. The area of the rectangle is [Image]. The area of the semicircle is [Image]. So the area of the entire figure is [Image].)
- 199.3 units^2 (This figure represents a rectangle with dimensions 10 by 16 attached to a semicircle with radius 5. The area of the rectangle is [Image]. The area of the semicircle is [Image]. So the area of the entire figure is [Image].)
- 238.5 units^2 (This figure represents a rectangle with dimensions 10 by 16 attached to a semicircle with radius 5. The area of the rectangle is [Image]. The area of the semicircle is [Image]. So the area of the entire figure is [Image].)
A: 199.3 units^2

Q: Find the area of the figure. Round to the nearest tenth if necessary.

[Image]
- 121.4 units^2 (This figure represents a rectangle with an equilateral triangle cut out of the right side. The area of the rectangle is [Image]. The area of the equilateral triangle is [Image], since the altitude of the equilateral triangle is [Image]. So the area of the figure is [Image].)
- 116.3 units^2 (This figure represents a rectangle with an equilateral triangle cut out of the right side. The area of the rectangle is [Image]. The area of the equilateral triangle is [Image], since the altitude of the equilateral triangle is [Image]. So the area of the figure is [Image].)
- 98.7 units^2 (This figure represents a rectangle with an equilateral triangle cut out of the right side. The area of the rectangle is [Image]. The area of the equilateral triangle is [Image], since the altitude of the equilateral triangle is [Image]. So the area of the figure is [Image].)
- 88.6 units^2 (This figure represents a rectangle with an equilateral triangle cut out of the right side. The area of the rectangle is [Image]. The area of the equilateral triangle is [Image], since the altitude of the equilateral triangle is [Image]. So the area of the figure is [Image].)
A: 116.3 units^2

Q: Find the area of the figure. Round to the nearest tenth if necessary.

[Image]
- 42 units^2 (The figure consists of two isosceles triangles with the same base. The area of the top triangle is [Image]. The area of the bottom triangle is [Image]. So the total area of the figure is [Image].)
- 21 units^2 (The figure consists of two isosceles triangles with the same base. The area of the top triangle is [Image]. The area of the bottom triangle is [Image]. So the total area of the figure is [Image].)
- 33 units^2 (The figure consists of two isosceles triangles with the same base. The area of the top triangle is [Image]. The area of the bottom triangle is [Image]. So the total area of the figure is [Image].)
- 30 units^2 (The figure consists of two isosceles triangles with the same base. The area of the top triangle is [Image]. The area of the bottom triangle is [Image]. So the total area of the figure is [Image].)
A: 21 units^2

Q: Find the area of the figure. Round to the nearest tenth if necessary.

[Image]
- 41.5 units^2 (The figure consists of a triangle on top of a trapezoid. The area of the triangle is [Image]. The area of the trapezoid is [Image]. So the total area of the figure is [Image].)
- 50.5 units^2 (The figure consists of a triangle on top of a trapezoid. The area of the triangle is [Image]. The area of the trapezoid is [Image]. So the total area of the figure is [Image].)
- 74 units^2 (The figure consists of a triangle on top of a trapezoid. The area of the triangle is [Image]. The area of the trapezoid is [Image]. So the total area of the figure is [Image].)
- 56.5 units^2 (The figure consists of a triangle on top of a trapezoid. The area of the triangle is [Image]. The area of the trapezoid is [Image]. So the total area of the figure is [Image].)
A: 41.5 units^2

Q: Find the area of the figure. Round to the nearest tenth if necessary.

[Image]
- 149.0 units^2 (The figure consists of a 6-by-6 square and two identical semicircles that make one full circle with radius 3. The area of the square is [Image]. The area of the combined semicircles is [Image]. So the total area of the figure is [Image].)
- 58.3 units^2 (The figure consists of a 6-by-6 square and two identical semicircles that make one full circle with radius 3. The area of the square is [Image]. The area of the combined semicircles is [Image]. So the total area of the figure is [Image].)
- 71.7 units^2 (The figure consists of a 6-by-6 square and two identical semicircles that make one full circle with radius 3. The area of the square is [Image]. The area of the combined semicircles is [Image]. So the total area of the figure is [Image].)
- 64.3 units^2 (The figure consists of a 6-by-6 square and two identical semicircles that make one full circle with radius 3. The area of the square is [Image]. The area of the combined semicircles is [Image]. So the total area of the figure is [Image].)
A: 64.3 units^2

Q: Find the area of the figure. Round to the nearest tenth if necessary.

[Image]
- 51 units^2 (The figure consists of two congruent trapezoids sharing a common base. The area of each trapezoid is [Image]. So the total area of the figure is [Image].)
- 48 units^2 (The figure consists of two congruent trapezoids sharing a common base. The area of each trapezoid is [Image]. So the total area of the figure is [Image].)
- 102 units^2 (The figure consists of two congruent trapezoids sharing a common base. The area of each trapezoid is [Image]. So the total area of the figure is [Image].)
- 96 units^2 (The figure consists of two congruent trapezoids sharing a common base. The area of each trapezoid is [Image]. So the total area of the figure is [Image].)
A: 48 units^2

Q: Find the area of the figure. Round to the nearest tenth if necessary.

[Image]
- 57.1 units^2 (This figure consists of a semicircle on top of an isosceles triangle. The area of the semicircle is [Image]. The area of the triangle is [Image]. So the total area of the figure is [Image].)
- 66.2 units^2 (This figure consists of a semicircle on top of an isosceles triangle. The area of the semicircle is [Image]. The area of the triangle is [Image]. So the total area of the figure is [Image].)
- 47.3 units^2 (This figure consists of a semicircle on top of an isosceles triangle. The area of the semicircle is [Image]. The area of the triangle is [Image]. So the total area of the figure is [Image].)
- 41.1 units^2 (This figure consists of a semicircle on top of an isosceles triangle. The area of the semicircle is [Image]. The area of the triangle is [Image]. So the total area of the figure is [Image].)
A: 41.1 units^2

Q: Find the area of the figure. Round to the nearest tenth if necessary.

[Image]
- 36 units^2 (The semicircular region on the right is an exact match for the semicircular cut-out on the left. This makes the figure equivalent to a square that is 6-by-6. So the area is [Image].)
- 30 units^2 (The semicircular region on the right is an exact match for the semicircular cut-out on the left. This makes the figure equivalent to a square that is 6-by-6. So the area is [Image].)
- 25 units^2 (The semicircular region on the right is an exact match for the semicircular cut-out on the left. This makes the figure equivalent to a square that is 6-by-6. So the area is [Image].)
- 24 units^2 (The semicircular region on the right is an exact match for the semicircular cut-out on the left. This makes the figure equivalent to a square that is 6-by-6. So the area is [Image].)
A: 36 units^2

Q: Find the area of the figure. Round to the nearest tenth if necessary.

[Image]
- 26 units^2 (The figure consists of an isosceles triangle on top of an isosceles trapezoid. The area of the triangle is [Image]. The area of the trapezoid is [Image]. So the total area of the figure is [Image].)
- 38 units^2 (The figure consists of an isosceles triangle on top of an isosceles trapezoid. The area of the triangle is [Image]. The area of the trapezoid is [Image]. So the total area of the figure is [Image].)
- 32 units^2 (The figure consists of an isosceles triangle on top of an isosceles trapezoid. The area of the triangle is [Image]. The area of the trapezoid is [Image]. So the total area of the figure is [Image].)
- 46 units^2 (The figure consists of an isosceles triangle on top of an isosceles trapezoid. The area of the triangle is [Image]. The area of the trapezoid is [Image]. So the total area of the figure is [Image].)
A: 26 units^2

Q: Find the area of the figure. Round to the nearest tenth if necessary.

[Image]
- 42 units^2 (The figure consists of an isosceles trapezoid on top of a rectangle. The area of the trapezoid is [Image]. The area of the rectangle is [Image]. So the total area of the figure is [Image].)
- 44 units^2 (The figure consists of an isosceles trapezoid on top of a rectangle. The area of the trapezoid is [Image]. The area of the rectangle is [Image]. So the total area of the figure is [Image].)
- 46 units^2 (The figure consists of an isosceles trapezoid on top of a rectangle. The area of the trapezoid is [Image]. The area of the rectangle is [Image]. So the total area of the figure is [Image].)
- 50 units^2 (The figure consists of an isosceles trapezoid on top of a rectangle. The area of the trapezoid is [Image]. The area of the rectangle is [Image]. So the total area of the figure is [Image].)
A: 46 units^2

Q: Find the area of the figure. Round to the nearest tenth if necessary.

[Image]
- 20 units^2 (This figure consists of a right triangle on top of a parallelogram. The area of the triangle is [Image]. The area of the parallelogram is [Image]. So the area of the entire figure is [Image].)
- 26 units^2 (This figure consists of a right triangle on top of a parallelogram. The area of the triangle is [Image]. The area of the parallelogram is [Image]. So the area of the entire figure is [Image].)
- 22 units^2 (This figure consists of a right triangle on top of a parallelogram. The area of the triangle is [Image]. The area of the parallelogram is [Image]. So the area of the entire figure is [Image].)
- 34 units^2 (This figure consists of a right triangle on top of a parallelogram. The area of the triangle is [Image]. The area of the parallelogram is [Image]. So the area of the entire figure is [Image].)
A: 22 units^2

Q: Find the area of the figure. Round to the nearest tenth if necessary.

[Image]
- 44.6 units^2 (This figure consists of a rectangle attached to a semicircle. The area of the rectangle is [Image]. The area of the semicircle is [Image]. So the total area of the figure is [Image].)
- 36.3 units^2 (This figure consists of a rectangle attached to a semicircle. The area of the rectangle is [Image]. The area of the semicircle is [Image]. So the total area of the figure is [Image].)
- 38.3 units^2 (This figure consists of a rectangle attached to a semicircle. The area of the rectangle is [Image]. The area of the semicircle is [Image]. So the total area of the figure is [Image].)
- 40.5 units^2 (This figure consists of a rectangle attached to a semicircle. The area of the rectangle is [Image]. The area of the semicircle is [Image]. So the total area of the figure is [Image].)
A: 38.3 units^2

Q: Find the probability that a point chosen at random lies in the shaded region.
[Image]
- 0.58 (There are 100 boxes in the grid. So the probability is found by counting the number of shaded boxes and dividing by 100. For example, if there are 56 shaded boxes, then the probability that a point chosen at random lies in the shaded region is [Image]. Alternately, you may first want to count the number of unshaded boxes, then subtract that from 100 to get the number of shaded boxes.)
- 0.64 (There are 100 boxes in the grid. So the probability is found by counting the number of shaded boxes and dividing by 100. For example, if there are 56 shaded boxes, then the probability that a point chosen at random lies in the shaded region is [Image]. Alternately, you may first want to count the number of unshaded boxes, then subtract that from 100 to get the number of shaded boxes.)
- 0.60 (There are 100 boxes in the grid. So the probability is found by counting the number of shaded boxes and dividing by 100. For example, if there are 56 shaded boxes, then the probability that a point chosen at random lies in the shaded region is [Image]. Alternately, you may first want to count the number of unshaded boxes, then subtract that from 100 to get the number of shaded boxes.)
- 0.62 (There are 100 boxes in the grid. So the probability is found by counting the number of shaded boxes and dividing by 100. For example, if there are 56 shaded boxes, then the probability that a point chosen at random lies in the shaded region is [Image]. Alternately, you may first want to count the number of unshaded boxes, then subtract that from 100 to get the number of shaded boxes.)
A: 0.60

Q: Find the probability that a point chosen at random lies in the shaded region.
[Image]
- 0.62 (There are 100 boxes in the grid. So the probability is found by counting the number of shaded boxes and dividing by 100. For example, if there are 56 shaded boxes, then the probability that a point chosen at random lies in the shaded region is [Image]. Alternately, you may first want to count the number of unshaded boxes, then subtract that from 100 to get the number of shaded boxes.)
- 0.70 (There are 100 boxes in the grid. So the probability is found by counting the number of shaded boxes and dividing by 100. For example, if there are 56 shaded boxes, then the probability that a point chosen at random lies in the shaded region is [Image]. Alternately, you may first want to count the number of unshaded boxes, then subtract that from 100 to get the number of shaded boxes.)
- 0.72 (There are 100 boxes in the grid. So the probability is found by counting the number of shaded boxes and dividing by 100. For example, if there are 56 shaded boxes, then the probability that a point chosen at random lies in the shaded region is [Image]. Alternately, you may first want to count the number of unshaded boxes, then subtract that from 100 to get the number of shaded boxes.)
- 0.64 (There are 100 boxes in the grid. So the probability is found by counting the number of shaded boxes and dividing by 100. For example, if there are 56 shaded boxes, then the probability that a point chosen at random lies in the shaded region is [Image]. Alternately, you may first want to count the number of unshaded boxes, then subtract that from 100 to get the number of shaded boxes.)
A: 0.64

Q: Find the area of the indicated sector(s). Then find the probability of spinning the color indicated if the diameter of the spinner is 12.

blue
[Image]
- 15.1 units^2, 0.4 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. If there is more than one sector, N will be the sum of the measures of all of the central angles of those sectors. There are two sectors that have a total measure of 144 and the radius of the circle is 6, so the area of the sectors is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which is 0.4.)
- 22.6 units^2, 0.4 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. If there is more than one sector, N will be the sum of the measures of all of the central angles of those sectors. There are two sectors that have a total measure of 144 and the radius of the circle is 6, so the area of the sectors is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which is 0.4.)
- 45.2 units^2, 0.4 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. If there is more than one sector, N will be the sum of the measures of all of the central angles of those sectors. There are two sectors that have a total measure of 144 and the radius of the circle is 6, so the area of the sectors is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which is 0.4.)
- 180.9 units^2, 0.4 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. If there is more than one sector, N will be the sum of the measures of all of the central angles of those sectors. There are two sectors that have a total measure of 144 and the radius of the circle is 6, so the area of the sectors is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which is 0.4.)
A: 45.2 units^2, 0.4

Q: Find the area of the indicated sector(s). Then find the probability of spinning the color indicated if the diameter of the spinner is 12.

pink
[Image]
- 18.8 units^2, 0.33 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. The sector measure is 60 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which is 0.17.)
- 75.4 units^2, 0.33 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. The sector measure is 60 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which is 0.17.)
- 18.8 units^2, 0.17 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. The sector measure is 60 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which is 0.17.)
- 75.4 units^2, 0.17 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. The sector measure is 60 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which is 0.17.)
A: 18.8 units^2, 0.17

Q: Find the area of the indicated sector(s). Then find the probability of spinning the color indicated if the diameter of the spinner is 12.

purple
[Image]
- 56.5 units^2, 0.25 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. The sector measure is 45 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which is 0.125 or 0.13.)
- 14.1 units^2, 0.25 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. The sector measure is 45 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which is 0.125 or 0.13.)
- 56.5 units^2, 0.13 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. The sector measure is 45 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which is 0.125 or 0.13.)
- 14.1 units^2, 0.13 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. The sector measure is 45 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which is 0.125 or 0.13.)
A: 14.1 units^2, 0.13

Q: Find the area of the indicated sector(s). Then find the probability of spinning the color indicated if the diameter of the spinner is 12.

red
[Image]
- 58.7 units^2, 0.48 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are three red sectors that have a total measure of 187 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.52.)
- 58.7 units^2, 0.52 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are three red sectors that have a total measure of 187 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.52.)
- 234.872 units^2, 0.48 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are three red sectors that have a total measure of 187 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.52.)
- 234.872 units^2, 0.52 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are three red sectors that have a total measure of 187 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.52.)
A: 58.7 units^2, 0.52

Q: Find the area of the indicated sector(s). Then find the probability of spinning the color indicated if the diameter of the spinner is 12.

green
[Image]
- 52.1 units^2, 0.46 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are three green sectors that have a total measure of 166 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.46.)
- 52.1 units^2, 0.54 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are three green sectors that have a total measure of 166 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.46.)
- 208.5 units^2, 0.54 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are three green sectors that have a total measure of 166 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.46.)
- 208.5 units^2, 0.46 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are three green sectors that have a total measure of 166 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.46.)
A: 52.1 units^2, 0.46

Q: Find the area of the indicated sector(s). Then find the probability of spinning the color indicated if the diameter of the spinner is 12.

yellow
[Image]
- 41.1 units^2, 0.39 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are three yellow sectors that have a total measure of 131 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.36.)
- 41.1 units^2, 0.36 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are three yellow sectors that have a total measure of 131 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.36.)
- 82.2 units^2, 0.36 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are three yellow sectors that have a total measure of 131 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.36.)
- 82.2 units^2, 0.39 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are three yellow sectors that have a total measure of 131 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.36.)
A: 41.1 units^2, 0.36

Q: Find the area of the indicated sector(s). Then find the probability of spinning the color indicated if the diameter of the spinner is 12.

pink
[Image]
- 37.8 units^2, 0.37 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are three pink sectors that have a total measure of 132 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.37.)
- 41.4 units^2, 0.36 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are three pink sectors that have a total measure of 132 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.37.)
- 37.8 units^2, 0.36 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are three pink sectors that have a total measure of 132 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.37.)
- 41.4 units^2, 0.37 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are three pink sectors that have a total measure of 132 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.37.)
A: 41.4 units^2, 0.37

Q: Find the area of the indicated sector(s). Then find the probability of spinning the color indicated if the diameter of the spinner is 12.

blue
[Image]
- 44.2 units^2, 0.38 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are two blue sectors that have a total measure of 138 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.38.)
- 44.2 units^2, 0.39 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are two blue sectors that have a total measure of 138 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.38.)
- 43.3 units^2, 0.38 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are two blue sectors that have a total measure of 138 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.38.)
- 43.3 units^2, 0.39 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are two blue sectors that have a total measure of 138 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.38.)
A: 43.3 units^2, 0.38

Q: Find the area of the indicated sector(s). Then find the probability of spinning the color indicated if the diameter of the spinner is 12.

white
[Image]
- 28.6 units^2, 0.28 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are two white sectors that have a total measure of 91 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.25.)
- 26.9 units^2, 0.28 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are two white sectors that have a total measure of 91 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.25.)
- 28.6 units^2, 0.25 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are two white sectors that have a total measure of 91 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.25.)
- 26.9 units^2, 0.25 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are two white sectors that have a total measure of 91 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.25.)
A: 28.6 units^2, 0.25

Q: Find the area of the indicated sector(s). Then find the probability of spinning the color indicated if the diameter of the spinner is 12.

brown
[Image]
- 64.6 units^2, 0.60 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are two brown sectors that have a total measure of 200 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.56.)
- 62.8 units^2, 0.56 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are two brown sectors that have a total measure of 200 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.56.)
- 64.6 units^2, 0.56 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are two brown sectors that have a total measure of 200 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.56.)
- 62.8 units^2, 0.60 (To find the area of a sector, use the formula [Image], where N is the measure of the central angle for that sector. There are two brown sectors that have a total measure of 200 and the radius of the circle is 6, so the area of the sector is [Image], rounding the answer to one decimal place. To find the probability of the spinning the indicated color, divide the area of the sector(s) found previously by the area of the entire circle. That would be [Image] which rounds to 0.56.)
A: 62.8 units^2, 0.56

Q: Find the area of the shaded region. Round answers to the nearest tenth. Assume all inscribed polygons are regular.

[Image]
- 22.1 units^2 (The shaded area here represents two-thirds of the area of the circle after subtracting the area of the equilateral triangle. The area of the circle is [Image]. The area of the triangle is [Image]. So the area of the shaded region is [Image].)
- 44.2 units^2 (The shaded area here represents two-thirds of the area of the circle after subtracting the area of the equilateral triangle. The area of the circle is [Image]. The area of the triangle is [Image]. So the area of the shaded region is [Image].)
- 66.3 units^2 (The shaded area here represents two-thirds of the area of the circle after subtracting the area of the equilateral triangle. The area of the circle is [Image]. The area of the triangle is [Image]. So the area of the shaded region is [Image].)
- 88.4 units^2 (The shaded area here represents two-thirds of the area of the circle after subtracting the area of the equilateral triangle. The area of the circle is [Image]. The area of the triangle is [Image]. So the area of the shaded region is [Image].)
A: 44.2 units^2

Q: Find the area of the shaded region. Round answers to the nearest tenth. Assume all inscribed polygons are regular.

[Image]
- 10.3 units^2 (The shaded area here represents two-fifths of the area of the circle after subtracting the area of the regular pentagon. The area of the circle is [Image]. The area of the pentagon is [Image]. So the area of the shaded region is [Image].)
- 9.9 units^2 (The shaded area here represents two-fifths of the area of the circle after subtracting the area of the regular pentagon. The area of the circle is [Image]. The area of the pentagon is [Image]. So the area of the shaded region is [Image].)
- 9.2 units^2 (The shaded area here represents two-fifths of the area of the circle after subtracting the area of the regular pentagon. The area of the circle is [Image]. The area of the pentagon is [Image]. So the area of the shaded region is [Image].)
- 13.8 units^2 (The shaded area here represents two-fifths of the area of the circle after subtracting the area of the regular pentagon. The area of the circle is [Image]. The area of the pentagon is [Image]. So the area of the shaded region is [Image].)
A: 9.2 units^2

Q: Find the area of the shaded region. Round answers to the nearest tenth. Assume all inscribed polygons are regular.

[Image]
- 29.3 units^2 (The shaded area here represents three-fifths of the area of the circle after subtracting the area of the regular pentagon. The area of the circle is [Image]. The area of the pentagon is [Image]. So the area of the shaded region is [Image].)
- 19.5 units^2 (The shaded area here represents three-fifths of the area of the circle after subtracting the area of the regular pentagon. The area of the circle is [Image]. The area of the pentagon is [Image]. So the area of the shaded region is [Image].)
- 23.7 units^2 (The shaded area here represents three-fifths of the area of the circle after subtracting the area of the regular pentagon. The area of the circle is [Image]. The area of the pentagon is [Image]. So the area of the shaded region is [Image].)
- 26.1 units^2 (The shaded area here represents three-fifths of the area of the circle after subtracting the area of the regular pentagon. The area of the circle is [Image]. The area of the pentagon is [Image]. So the area of the shaded region is [Image].)
A: 29.3 units^2

Q: Find the area of the shaded region. Round answers to the nearest tenth. Assume all inscribed polygons are regular.

[Image]
- 8.5 units^2 (The shaded area here represents one-third of the area of the circle after subtracting the area of the regular hexagon. The area of the circle is [Image]. The area of the hexagon is [Image]. So the area of the shaded region is [Image].)
- 9.0 units^2 (The shaded area here represents one-third of the area of the circle after subtracting the area of the regular hexagon. The area of the circle is [Image]. The area of the hexagon is [Image]. So the area of the shaded region is [Image].)
- 4.5 units^2 (The shaded area here represents one-third of the area of the circle after subtracting the area of the regular hexagon. The area of the circle is [Image]. The area of the hexagon is [Image]. So the area of the shaded region is [Image].)
- 5.7 units^2 (The shaded area here represents one-third of the area of the circle after subtracting the area of the regular hexagon. The area of the circle is [Image]. The area of the hexagon is [Image]. So the area of the shaded region is [Image].)
A: 4.5 units^2

Q: Find the area of the shaded region. Round answers to the nearest tenth. Assume all inscribed polygons are regular.

[Image]
- 29.3 units^2 (The shaded area here represents two-thirds of the area of the circle after subtracting the area of the regular hexagon. The area of the circle is [Image]. The area of the hexagon is [Image]. So the area of the shaded region is [Image].)
- 18.7 units^2 (The shaded area here represents two-thirds of the area of the circle after subtracting the area of the regular hexagon. The area of the circle is [Image]. The area of the hexagon is [Image]. So the area of the shaded region is [Image].)
- 14.6 units^2 (The shaded area here represents two-thirds of the area of the circle after subtracting the area of the regular hexagon. The area of the circle is [Image]. The area of the hexagon is [Image]. So the area of the shaded region is [Image].)
- 12.8 units^2 (The shaded area here represents two-thirds of the area of the circle after subtracting the area of the regular hexagon. The area of the circle is [Image]. The area of the hexagon is [Image]. So the area of the shaded region is [Image].)
A: 29.3 units^2

Q: Seniors at a high school were asked what color car they drive. The results were put in a circle graph. The measure of each central angle is shown. If a senior is chosen at random from this school, find the probability of each response.

What Color Is Your Car?
[Image]

Car color is red.
- 0.38 (The probability that the senior’s car is red is found by dividing the central angle by 360. So that would be [Image] or 32%.)
- 0.36 (The probability that the senior’s car is red is found by dividing the central angle by 360. So that would be [Image] or 32%.)
- 0.32 (The probability that the senior’s car is red is found by dividing the central angle by 360. So that would be [Image] or 32%.)
- 0.28 (The probability that the senior’s car is red is found by dividing the central angle by 360. So that would be [Image] or 32%.)
A: 0.32

Q: Seniors at a high school were asked what color car they drive. The results were put in a circle graph. The measure of each central angle is shown. If a senior is chosen at random from this school, find the probability of each response.

What Color Is Your Car?
[Image]

Car color is silver.
- 0.15 (The probability that the senior’s car is silver is found by dividing the central angle by 360. So that would be [Image] or 19%.)
- 0.19 (The probability that the senior’s car is silver is found by dividing the central angle by 360. So that would be [Image] or 19%.)
- 0.18 (The probability that the senior’s car is silver is found by dividing the central angle by 360. So that would be [Image] or 19%.)
- 0.16 (The probability that the senior’s car is silver is found by dividing the central angle by 360. So that would be [Image] or 19%.)
A: 0.19

Q: Seniors at a high school were asked what color car they drive. The results were put in a circle graph. The measure of each central angle is shown. If a senior is chosen at random from this school, find the probability of each response.

What Color Is Your Car?
[Image]

Car color is white or blue.
- 0.48 (The probability that the senior’s car is white or blue is found by dividing the central angles for both colors by 360, then adding the results together. So that would be [Image] for white and [Image] for blue giving a total of 0.49.)
- 0.42 (The probability that the senior’s car is white or blue is found by dividing the central angles for both colors by 360, then adding the results together. So that would be [Image] for white and [Image] for blue giving a total of 0.49.)
- 0.46 (The probability that the senior’s car is white or blue is found by dividing the central angles for both colors by 360, then adding the results together. So that would be [Image] for white and [Image] for blue giving a total of 0.49.)
- 0.49 (The probability that the senior’s car is white or blue is found by dividing the central angles for both colors by 360, then adding the results together. So that would be [Image] for white and [Image] for blue giving a total of 0.49.)
A: 0.49

Q: Seniors at a high school were asked what color car they drive. The results were put in a circle graph. The measure of each central angle is shown. If a senior is chosen at random from this school, find the probability of each response.

What Color Is Your Car?
[Image]

Car color is not red or blue.
- 0.53 (If the car is not red or blue, then it must be white or silver. So find the individual probabilities for white and silver and add them together. The probability for silver is [Image]. The probability for white is [Image]. So the total probability is [Image].)
- 0.47 (If the car is not red or blue, then it must be white or silver. So find the individual probabilities for white and silver and add them together. The probability for silver is [Image]. The probability for white is [Image]. So the total probability is [Image].)
- 0.51 (If the car is not red or blue, then it must be white or silver. So find the individual probabilities for white and silver and add them together. The probability for silver is [Image]. The probability for white is [Image]. So the total probability is [Image].)
- 0.49 (If the car is not red or blue, then it must be white or silver. So find the individual probabilities for white and silver and add them together. The probability for silver is [Image]. The probability for white is [Image]. So the total probability is [Image].)
A: 0.53

Q: Seniors at a high school were asked what color car they drive. The results were put in a circle graph. The measure of each central angle is shown. If a senior is chosen at random from this school, find the probability of each response.

What Color Is Your Car?
[Image]

Car color is not white.
- 0.34 (The probability that the senior’s car is not white can be found by computing 1 minus the probability that it is white. The probability that it is white is [Image]. So the probability that the car is not white is [Image])
- 0.56 (The probability that the senior’s car is not white can be found by computing 1 minus the probability that it is white. The probability that it is white is [Image]. So the probability that the car is not white is [Image])
- 0.68 (The probability that the senior’s car is not white can be found by computing 1 minus the probability that it is white. The probability that it is white is [Image]. So the probability that the car is not white is [Image])
- 0.66 (The probability that the senior’s car is not white can be found by computing 1 minus the probability that it is white. The probability that it is white is [Image]. So the probability that the car is not white is [Image])
A: 0.66

The Eiffel Tower in Paris, France, is one of the most famous structures in the world. It was built in the late 1800’s and stands 1052 feet tall including its antenna. Use the diagram below to answer the following questions. Round all answers to the nearest foot.
[Image]
a.      Monique is at a point on the ground that is x feet from the base of the tower. If the angle of elevation from that point to the top of the tower is 41°, find the distance x.
b.      Pierre is at a point on the ground that is y feet from the base of the tower. If the angle of elevation from that point to the top of the tower is 32°, find the distance y.
c.      Find the distance on the ground between Monique and Pierre.

Parallelogram PQRS has vertices P(0, 0), Q(6, 0) and R(2, 5).
a.      Find the possible coordinates of S.
b.      Find the area of PQRS.