Title: Chapter 12
Course: quiz duplicate error

Q: Name the edges of the solid.

[Image]
- [Image], [Image], [Image], [Image], [Image], [Image], [Image], [Image], [Image], [Image], [Image], and [Image] (The edges of a solid are the line segments formed by the intersection of the faces. This figure has 12 such segments: 4 on top, 4 in the middle and 4 on the bottom.)
- [Image], [Image], [Image], and [Image] (The edges of a solid are the line segments formed by the intersection of the faces. This figure has 12 such segments: 4 on top, 4 in the middle and 4 on the bottom.)
- [Image], [Image], [Image], and [Image] (The edges of a solid are the line segments formed by the intersection of the faces. This figure has 12 such segments: 4 on top, 4 in the middle and 4 on the bottom.)
- [Image], [Image], [Image], and [Image] (The edges of a solid are the line segments formed by the intersection of the faces. This figure has 12 such segments: 4 on top, 4 in the middle and 4 on the bottom.)
A: [Image], [Image], [Image], [Image], [Image], [Image], [Image], [Image], [Image], [Image], [Image], and [Image]

Q: Identify the solid.

[Image]
- cylinder (A solid with a circular base and a vertex is called a cone.)
- pyramid (A solid with a circular base and a vertex is called a cone.)
- cone (A solid with a circular base and a vertex is called a cone.)
- prism (A solid with a circular base and a vertex is called a cone.)
A: cone

Q: Draw the corner view of a figure given each orthogonal drawing.

top |  | left |  | front |  | right
view |  | view |  | view |  | view
- [Image] (The top view indicates three stacks of blocks at varying heights. The left view shows that the left side is four blocks high and the right side is one block high. The front view shows varying heights on the left column and three blocks high on the right. The right view shows a mirror view of the left with the exception that some heights vary.)
- [Image] (The top view indicates three stacks of blocks at varying heights. The left view shows that the left side is four blocks high and the right side is one block high. The front view shows varying heights on the left column and three blocks high on the right. The right view shows a mirror view of the left with the exception that some heights vary.)
- [Image] (The top view indicates three stacks of blocks at varying heights. The left view shows that the left side is four blocks high and the right side is one block high. The front view shows varying heights on the left column and three blocks high on the right. The right view shows a mirror view of the left with the exception that some heights vary.)
- [Image] (The top view indicates three stacks of blocks at varying heights. The left view shows that the left side is four blocks high and the right side is one block high. The front view shows varying heights on the left column and three blocks high on the right. The right view shows a mirror view of the left with the exception that some heights vary.)
A: [Image]

Q: Name the edges of the solid.

[Image]
- [Image], [Image], and [Image] (The edges of a solid are the line segments formed by the intersection of the faces. This figure has six such segments.)
- [Image], [Image], and [Image] (The edges of a solid are the line segments formed by the intersection of the faces. This figure has six such segments.)
- [Image], [Image], [Image] , [Image], and [Image] (The edges of a solid are the line segments formed by the intersection of the faces. This figure has six such segments.)
- [Image], [Image], [Image] , [Image], [Image], and [Image] (The edges of a solid are the line segments formed by the intersection of the faces. This figure has six such segments.)
A: [Image], [Image], [Image] , [Image], [Image], and [Image]

Q: Name the faces of the solid.

[Image]
- DABC, DABD, and BCFED (A face of a solid is a flat surface. This solid contains six such surfaces: five lateral triangles and one pentagonal face on the base.)
- DABC, DABD, DACF, DAFE, DAED, and BCFED (A face of a solid is a flat surface. This solid contains six such surfaces: five lateral triangles and one pentagonal face on the base.)
- DABC, DABD, DACF, DAFE, and DAED (A face of a solid is a flat surface. This solid contains six such surfaces: five lateral triangles and one pentagonal face on the base.)
- DABC and DABD (A face of a solid is a flat surface. This solid contains six such surfaces: five lateral triangles and one pentagonal face on the base.)
A: DABC, DABD, DACF, DAFE, DAED, and BCFED

Q: Name the bases of the solid.
[Image]
- [Image] and [Image] (This solid is a cylinder. The bases of a cylinder are circles. In this cylinder, the circular bases are centered at points A and B, thus the correct answer is [Image] and [Image].)
- [Image] and [Image] (This solid is a cylinder. The bases of a cylinder are circles. In this cylinder, the circular bases are centered at points A and B, thus the correct answer is [Image] and [Image].)
- [Image] and [Image] (This solid is a cylinder. The bases of a cylinder are circles. In this cylinder, the circular bases are centered at points A and B, thus the correct answer is [Image] and [Image].)
- [Image] and [Image] (This solid is a cylinder. The bases of a cylinder are circles. In this cylinder, the circular bases are centered at points A and B, thus the correct answer is [Image] and [Image].)
A: [Image] and [Image]

Q: Draw the corner view of a figure given each orthogonal drawing.

top |  | left |  | front |  | right
view |  | view |  | view |  | view
- [Image] (The top view indicates two rows and columns of varying heights. The left view shows that the left side is four blocks high and the right side is one block high. The front view shows that the left back column is four blocks high and the front left column is two blocks high. The right column is one block high. The right view shows a mirror view of the left.)
- [Image] (The top view indicates two rows and columns of varying heights. The left view shows that the left side is four blocks high and the right side is one block high. The front view shows that the left back column is four blocks high and the front left column is two blocks high. The right column is one block high. The right view shows a mirror view of the left.)
- [Image] (The top view indicates two rows and columns of varying heights. The left view shows that the left side is four blocks high and the right side is one block high. The front view shows that the left back column is four blocks high and the front left column is two blocks high. The right column is one block high. The right view shows a mirror view of the left.)
- [Image] (The top view indicates two rows and columns of varying heights. The left view shows that the left side is four blocks high and the right side is one block high. The front view shows that the left back column is four blocks high and the front left column is two blocks high. The right column is one block high. The right view shows a mirror view of the left.)
A: [Image]

Q: Sketch each solid using isometric dot paper.

rectangular prism 5 units high, 5 units long, and 4 units wide
- [Image] (Starting at the front corner, the correct solid should be drawn 5 units down, 5 units to the left, and 4 units to the right.)
- [Image] (Starting at the front corner, the correct solid should be drawn 5 units down, 5 units to the left, and 4 units to the right.)
- [Image] (Starting at the front corner, the correct solid should be drawn 5 units down, 5 units to the left, and 4 units to the right.)
- [Image] (Starting at the front corner, the correct solid should be drawn 5 units down, 5 units to the left, and 4 units to the right.)
A: [Image]

Q: Draw the corner view of a figure given each orthogonal drawing.

top |  | left |  |  | front |  |  | right
view |  | view |  |  | view |  |  | view
- [Image] (The top view indicates three rows and two columns of varying heights. The left view shows a descending pattern from three blocks high to one block high. The front view shows two columns, at most three blocks high. The right view shows a mirror view of the left.)
- [Image] (The top view indicates three rows and two columns of varying heights. The left view shows a descending pattern from three blocks high to one block high. The front view shows two columns, at most three blocks high. The right view shows a mirror view of the left.)
- [Image] (The top view indicates three rows and two columns of varying heights. The left view shows a descending pattern from three blocks high to one block high. The front view shows two columns, at most three blocks high. The right view shows a mirror view of the left.)
- [Image] (The top view indicates three rows and two columns of varying heights. The left view shows a descending pattern from three blocks high to one block high. The front view shows two columns, at most three blocks high. The right view shows a mirror view of the left.)
A: [Image]

Q: Draw the corner view of a figure given each orthogonal drawing.

top |  |  | left |  |  | front |  |  |  | right
view |  |  | view |  |  | view |  |  |  | view
- [Image] (The top view indicates eight columns with one a different height from the others. The left view shows three by three arrangement. The front view shows that the same thing as the left. The right view shows a mirror view of the left but with numerous depth and height changes.)
- [Image] (The top view indicates eight columns with one a different height from the others. The left view shows three by three arrangement. The front view shows that the same thing as the left. The right view shows a mirror view of the left but with numerous depth and height changes.)
- [Image] (The top view indicates eight columns with one a different height from the others. The left view shows three by three arrangement. The front view shows that the same thing as the left. The right view shows a mirror view of the left but with numerous depth and height changes.)
- [Image] (The top view indicates eight columns with one a different height from the others. The left view shows three by three arrangement. The front view shows that the same thing as the left. The right view shows a mirror view of the left but with numerous depth and height changes.)
A: [Image]

Q: Name the bases of the prism.
[Image]
- DUWY and DVXZ (The bases of a prism are congruent, parallel polygons. Here those polygons are the right triangles UWY and VXZ.)
- UVZY and WXZY (The bases of a prism are congruent, parallel polygons. Here those polygons are the right triangles UWY and VXZ.)
- UVZY and UVXW (The bases of a prism are congruent, parallel polygons. Here those polygons are the right triangles UWY and VXZ.)
- UVXW and WXZY (The bases of a prism are congruent, parallel polygons. Here those polygons are the right triangles UWY and VXZ.)
A: DUWY and DVXZ

Q: Draw the corner view of a figure given each orthogonal drawing.

top |  |  | left |  |  | front |  | right
view |  |  | view |  |  | view |  | view
- [Image] (The top view indicates two rows and three columns of varying heights. The left view shows that the left and right sides are both two blocks high. The front view shows three columns, one that is two blocks high and two that are one block high. The right view shows a mirror view of the left, but with varying heights.)
- [Image] (The top view indicates two rows and three columns of varying heights. The left view shows that the left and right sides are both two blocks high. The front view shows three columns, one that is two blocks high and two that are one block high. The right view shows a mirror view of the left, but with varying heights.)
- [Image] (The top view indicates two rows and three columns of varying heights. The left view shows that the left and right sides are both two blocks high. The front view shows three columns, one that is two blocks high and two that are one block high. The right view shows a mirror view of the left, but with varying heights.)
- [Image] (The top view indicates two rows and three columns of varying heights. The left view shows that the left and right sides are both two blocks high. The front view shows three columns, one that is two blocks high and two that are one block high. The right view shows a mirror view of the left, but with varying heights.)
A: [Image]

Q: Name the faces of the solid.

[Image]
- DWUY, DXVZ, UVZY, UVXW, and WXZY (A face of a solid is a flat surface. This solid contains five such surfaces: three lateral rectangles and two triangular bases.)
- DWUY, DXVZ, UVZY, and UVXW (A face of a solid is a flat surface. This solid contains five such surfaces: three lateral rectangles and two triangular bases.)
- UVZY, UVXW, and WXZY (A face of a solid is a flat surface. This solid contains five such surfaces: three lateral rectangles and two triangular bases.)
- DWUY and DXVZ (A face of a solid is a flat surface. This solid contains five such surfaces: three lateral rectangles and two triangular bases.)
A: DWUY, DXVZ, UVZY, UVXW, and WXZY

Q: Name the vertices of the solid.

[Image]
- P, Q, R, and S (All points that are the result of intersections of the edges of a solid are vertices. This pyramid has five such points: P, Q, R, S and T.)
- P, Q, R, S and T (All points that are the result of intersections of the edges of a solid are vertices. This pyramid has five such points: P, Q, R, S and T.)
- T (All points that are the result of intersections of the edges of a solid are vertices. This pyramid has five such points: P, Q, R, S and T.)
- P, S, and T (All points that are the result of intersections of the edges of a solid are vertices. This pyramid has five such points: P, Q, R, S and T.)
A: P, Q, R, S and T

Q: Identify the solid.

[Image]
- cone (The set of all points in space that are a given distance from a given point is called a sphere. It is the three-dimensional equivalent of a circle.)
- sphere (The set of all points in space that are a given distance from a given point is called a sphere. It is the three-dimensional equivalent of a circle.)
- cylinder (The set of all points in space that are a given distance from a given point is called a sphere. It is the three-dimensional equivalent of a circle.)
- prism (The set of all points in space that are a given distance from a given point is called a sphere. It is the three-dimensional equivalent of a circle.)
A: sphere

Q: Name the vertices of the solid.

[Image]
- A, C, I, G, E, and K (In a solid, all points that represent intersections of edges are vertices. Therefore all of the points shown are vertices.)
- B, D, J, H, F, and L (In a solid, all points that represent intersections of edges are vertices. Therefore all of the points shown are vertices.)
- A, B, C, and D (In a solid, all points that represent intersections of edges are vertices. Therefore all of the points shown are vertices.)
- A, B, C, D, E, F, G, H, I, J, K, and L (In a solid, all points that represent intersections of edges are vertices. Therefore all of the points shown are vertices.)
A: A, B, C, D, E, F, G, H, I, J, K, and L

Q: Find the surface area of each prism. Round to the nearest tenth if necessary while doing your calculations as well as in your final answer.

[Image]
- 146 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 139 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 132 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 130 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
A: 146 units^2

Q: Sketch each solid using isometric dot paper.

cube 4 units on each edge
- [Image] (Starting at the front corner, the correct solid should be drawn 4 units down, 4 units to the left, and 4 units to the right.)
- [Image] (Starting at the front corner, the correct solid should be drawn 4 units down, 4 units to the left, and 4 units to the right.)
- [Image] (Starting at the front corner, the correct solid should be drawn 4 units down, 4 units to the left, and 4 units to the right.)
- [Image] (Starting at the front corner, the correct solid should be drawn 4 units down, 4 units to the left, and 4 units to the right.)
A: [Image]

Q: Find the surface area of each prism. Round to the nearest tenth if necessary while doing your calculations as well as in your final answer.

[Image]
- 248.4 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 256.8 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 293.9 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 339.4 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
A: 339.4 units^2

Q: For each solid, draw a net and find the surface area. Round to the nearest tenth if necessary.

[Image]
- [Image] 48 units^2 (The net consists of 4 congruent rectangles with area 10 units^2 and 2 congruent squares with area 4 units^2. So the total surface area is [Image] units^2.)
- [Image] 40 units^2 (The net consists of 4 congruent rectangles with area 10 units^2 and 2 congruent squares with area 4 units^2. So the total surface area is [Image] units^2.)
- [Image] 40 units^2 (The net consists of 4 congruent rectangles with area 10 units^2 and 2 congruent squares with area 4 units^2. So the total surface area is [Image] units^2.)
- [Image] 48 units^2 (The net consists of 4 congruent rectangles with area 10 units^2 and 2 congruent squares with area 4 units^2. So the total surface area is [Image] units^2.)
A: [Image] 48 units^2

Q: Find the surface area of each prism. Round to the nearest tenth if necessary while doing your calculations as well as in your final answer.

[Image]
- 340 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 301 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 262 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 241 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
A: 340 units^2

Q: Find the lateral area of each prism. Round to the nearest tenth if necessary.

[Image]
- 146.4 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 150.5 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 165.4 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 180.5 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
A: 150.5 units^2

Q: Find the lateral area of each prism. Round to the nearest tenth if necessary.

[Image]
- 64 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 256 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 320 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 384 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
A: 256 units^2

Q: For each solid, draw a net and find the surface area. Round to the nearest tenth if necessary.

[Image]
- [Image] 37.5 units^2 (This net consists of 6 congruent squares each having area 6.25 units^2. So the total surface area is [Image] units^2.)
- [Image] 30 units^2 (This net consists of 6 congruent squares each having area 6.25 units^2. So the total surface area is [Image] units^2.)
- [Image] 30 units^2 (This net consists of 6 congruent squares each having area 6.25 units^2. So the total surface area is [Image] units^2.)
- [Image] 37.5 units^2 (This net consists of 6 congruent squares each having area 6.25 units^2. So the total surface area is [Image] units^2.)
A: [Image] 37.5 units^2

Q: Find the lateral area of a cylinder with the given measurements. Use 3.14 for pi and round your answer to the nearest tenth.

height: 6 yards
diameter: 10 yards
- 96.2 [Image] (The lateral surface of a cylinder is a rectangle with the circumference and height of the cylinder as its dimensions. The formula for the circumference is 2 ´ p ´ r, so the lateral area is found by the formula 2 ´ p ´ r ´ h.)
- 345.4 [Image] (The lateral surface of a cylinder is a rectangle with the circumference and height of the cylinder as its dimensions. The formula for the circumference is 2 ´ p ´ r, so the lateral area is found by the formula 2 ´ p ´ r ´ h.)
- 193.0 [Image] (The lateral surface of a cylinder is a rectangle with the circumference and height of the cylinder as its dimensions. The formula for the circumference is 2 ´ p ´ r, so the lateral area is found by the formula 2 ´ p ´ r ´ h.)
- 188.4 [Image] (The lateral surface of a cylinder is a rectangle with the circumference and height of the cylinder as its dimensions. The formula for the circumference is 2 ´ p ´ r, so the lateral area is found by the formula 2 ´ p ´ r ´ h.)
A: 188.4 [Image]

Q: Find the lateral area of each prism. Round to the nearest tenth if necessary.

[Image]
- 84 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 105.2 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 108.5 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 126.4 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
A: 84 units^2

Q: Find the lateral area of each prism. Round to the nearest tenth if necessary.

The dimension labeled 10 is the height of the prism.
[Image]
- 120 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 160 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 180 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 220 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
A: 180 units^2

Q: Find the lateral area of a cylinder with the given measurements. Use 3.14 for pi and round your answer to the nearest tenth.

height: 18 inches
radius: 9 inches
- 957.3 [Image] (The lateral surface of a cylinder is a rectangle with the circumference and height of the cylinder as its dimensions. The formula for the circumference is 2 ´ p ´ r, so the lateral area is found by the formula 2 ´ p ´ r ´ h.)
- 1017.4 [Image] (The lateral surface of a cylinder is a rectangle with the circumference and height of the cylinder as its dimensions. The formula for the circumference is 2 ´ p ´ r, so the lateral area is found by the formula 2 ´ p ´ r ´ h.)
- 955.9 [Image] (The lateral surface of a cylinder is a rectangle with the circumference and height of the cylinder as its dimensions. The formula for the circumference is 2 ´ p ´ r, so the lateral area is found by the formula 2 ´ p ´ r ´ h.)
- 1526.0 [Image] (The lateral surface of a cylinder is a rectangle with the circumference and height of the cylinder as its dimensions. The formula for the circumference is 2 ´ p ´ r, so the lateral area is found by the formula 2 ´ p ´ r ´ h.)
A: 1017.4 [Image]

Q: Find the surface area of each prism. Round to the nearest tenth if necessary while doing your calculations as well as in your final answer.

[Image]
- 460.3 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 516.3 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 524.3 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 560.3 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
A: 516.3 units^2

Q: For each solid, draw a net and find the surface area. Round to the nearest tenth if necessary.

[Image]
- [Image] 28 units^2 (The net consists of a square with area 9 units^2 and 4 congruent triangles that are 6 units^2 each. So the total surface area is [Image] units^2.)
- [Image] 28 units^2 (The net consists of a square with area 9 units^2 and 4 congruent triangles that are 6 units^2 each. So the total surface area is [Image] units^2.)
- [Image] 33 units^2 (The net consists of a square with area 9 units^2 and 4 congruent triangles that are 6 units^2 each. So the total surface area is [Image] units^2.)
- [Image] 33 units^2 (The net consists of a square with area 9 units^2 and 4 congruent triangles that are 6 units^2 each. So the total surface area is [Image] units^2.)
A: [Image] 33 units^2

Q: Find the lateral area of each prism. Round to the nearest tenth if necessary.

[Image]
- 240 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. What is the perimeter of the base?)
- 258 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. What is the perimeter of the base?)
- 275.8 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. What is the perimeter of the base?)
- 292 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. What is the perimeter of the base?)
A: 240 units^2

Q: Find the lateral area of each prism. Round to the nearest tenth if necessary.

The dimension labeled 11 is the height of the prism.
[Image]
- 299 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 312 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 264 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 334 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
A: 264 units^2

Q: Find the lateral area of each prism. Round to the nearest tenth if necessary.

[Image]
- 456 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 225 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 184 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 429 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
A: 184 units^2

Q: Find the lateral area of each prism. Round to the nearest tenth if necessary.

[Image]
- 162 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 186 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 210 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 224 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
A: 162 units^2

Q: For each solid, draw a net and find the surface area. Round to the nearest tenth if necessary.

[Image]
- [Image] 36 units^2 (This net consists of 3 rectangles having areas 10, 6 and 8 units^2 and 2 congruent triangles each with area 6 units^2. So the total surface area is [Image] units^2.)
- [Image] 36 units^2 (This net consists of 3 rectangles having areas 10, 6 and 8 units^2 and 2 congruent triangles each with area 6 units^2. So the total surface area is [Image] units^2.)
- [Image] 34 units^2 (This net consists of 3 rectangles having areas 10, 6 and 8 units^2 and 2 congruent triangles each with area 6 units^2. So the total surface area is [Image] units^2.)
- [Image] 34 units^2 (This net consists of 3 rectangles having areas 10, 6 and 8 units^2 and 2 congruent triangles each with area 6 units^2. So the total surface area is [Image] units^2.)
A: [Image] 36 units^2

Q: Find the surface area of each prism. Round to the nearest tenth if necessary while doing your calculations as well as in your final answer.

[Image]
- 163.8 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 192.8 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 216.8 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 245.8 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
A: 245.8 units^2

Q: Sketch each solid using isometric dot paper.

cube 6 units on each edge
- [Image] (Starting at the front corner, the correct solid should be drawn 6 units down, 6 units to the left, and 6 units to the right.)
- [Image] (Starting at the front corner, the correct solid should be drawn 6 units down, 6 units to the left, and 6 units to the right.)
- [Image] (Starting at the front corner, the correct solid should be drawn 6 units down, 6 units to the left, and 6 units to the right.)
- [Image] (Starting at the front corner, the correct solid should be drawn 6 units down, 6 units to the left, and 6 units to the right.)
A: [Image]

Q: Sketch each solid using isometric dot paper.

triangular prism 6 units high, with bases that are right triangles with legs 4 units and 5 units long
- [Image] (Starting at the front corner, the correct solid should be drawn 6 units down, 4 units to the left, and 5 units to the right.)
- [Image] (Starting at the front corner, the correct solid should be drawn 6 units down, 4 units to the left, and 5 units to the right.)
- [Image] (Starting at the front corner, the correct solid should be drawn 6 units down, 4 units to the left, and 5 units to the right.)
- [Image] (Starting at the front corner, the correct solid should be drawn 6 units down, 4 units to the left, and 5 units to the right.)
A: [Image]

Q: Find the surface area of each prism. Round to the nearest tenth if necessary while doing your calculations as well as in your final answer.

[Image]
- 136 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has a an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 140 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has a an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 152 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has a an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 164 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has a an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
A: 164 units^2

Q: Find the surface area of the cylinder. Use 3.14 for pi and round your answer to the nearest tenth.

[Image]
- 355.7 [Image] (The formula for the surface area of a cylinder is [Image] where r is the radius and h is the height of the cylinder. For example, if the height is given as 6 and the radius as 4, then the surface area is computed as 2 ´ 3.14 ´ 4 ´ 6 + 2 ´ 3.14 ´ 4 ´ 4 = 251.2 units^2.)
- 320.6 [Image] (The formula for the surface area of a cylinder is [Image] where r is the radius and h is the height of the cylinder. For example, if the height is given as 6 and the radius as 4, then the surface area is computed as 2 ´ 3.14 ´ 4 ´ 6 + 2 ´ 3.14 ´ 4 ´ 4 = 251.2 units^2.)
- 175.8 [Image] (The formula for the surface area of a cylinder is [Image] where r is the radius and h is the height of the cylinder. For example, if the height is given as 6 and the radius as 4, then the surface area is computed as 2 ´ 3.14 ´ 4 ´ 6 + 2 ´ 3.14 ´ 4 ´ 4 = 251.2 units^2.)
- 276.3 [Image] (The formula for the surface area of a cylinder is [Image] where r is the radius and h is the height of the cylinder. For example, if the height is given as 6 and the radius as 4, then the surface area is computed as 2 ´ 3.14 ´ 4 ´ 6 + 2 ´ 3.14 ´ 4 ´ 4 = 251.2 units^2.)
A: 276.3 [Image]

Q: Find the surface area of each prism. Round to the nearest tenth if necessary while doing your calculations as well as in your final answer.

[Image]
- 308 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 357 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 406 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 414 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
A: 406 units^2

Q: Find the surface area of each prism. Round to the nearest tenth if necessary while doing your calculations as well as in your final answer.

[Image]
- 586 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 552 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 456 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 360 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
A: 552 units^2

Q: Find the surface area of each prism. Round to the nearest tenth if necessary while doing your calculations as well as in your final answer.

[Image]
- 112 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has a an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 120 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has a an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 144 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has a an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 168 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has a an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
A: 168 units^2

Q: Find the lateral area of each prism. Round to the nearest tenth if necessary.

[Image]
- 126 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 132 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 120 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 108 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
A: 126 units^2

Q: Sketch each solid using isometric dot paper.

triangular prism 3 units high, with bases that are right triangles with legs 3 units and 5 units long
- [Image] (Starting at the front corner, the correct solid should be drawn 3 units down, 3 units to the left, and 5 units to the right.)
- [Image] (Starting at the front corner, the correct solid should be drawn 3 units down, 3 units to the left, and 5 units to the right.)
- [Image] (Starting at the front corner, the correct solid should be drawn 3 units down, 3 units to the left, and 5 units to the right.)
- [Image] (Starting at the front corner, the correct solid should be drawn 3 units down, 3 units to the left, and 5 units to the right.)
A: [Image]

Q: For each solid, draw a net and find the surface area. Round to the nearest tenth if necessary.

[Image]
- [Image] 183 units^2 (This net consists of 4 rectangles having areas 42, 42, 35 and 28 units^2 and 2 congruent trapezoids each with area 25 units^2. So the total surface area is [Image] units^2.)
- [Image] 197 units^2 (This net consists of 4 rectangles having areas 42, 42, 35 and 28 units^2 and 2 congruent trapezoids each with area 25 units^2. So the total surface area is [Image] units^2.)
- [Image] 197 units^2 (This net consists of 4 rectangles having areas 42, 42, 35 and 28 units^2 and 2 congruent trapezoids each with area 25 units^2. So the total surface area is [Image] units^2.)
- [Image] 183 units^2 (This net consists of 4 rectangles having areas 42, 42, 35 and 28 units^2 and 2 congruent trapezoids each with area 25 units^2. So the total surface area is [Image] units^2.)
A: [Image] 197 units^2

Q: Find the lateral area of each prism. Round to the nearest tenth if necessary.

[Image]
- 720 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 682 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 656 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
- 612 units^2 (The lateral area is the sum of the areas of the lateral faces of the prism. If a right prism has a lateral area of L square units, a height of h units, and each base has a perimeter of P units, then L = Ph. The perimeter is the sum of all of the sides of the base.)
A: 612 units^2

Q: Find the surface area of each prism. Round to the nearest tenth if necessary while doing your calculations as well as in your final answer.

[Image]
- 442 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 396 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 342 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
- 288 units^2 (The surface area is the sum of areas of the lateral faces and both bases of the prism. If a right prism has a surface area of T square units, a height of h units, and each base has an area of B square units and a perimeter of P units, then T = L + 2B. The perimeter is the sum of all of the sides of the base.)
A: 396 units^2

Q: Determine whether each statement is true or false. If false, give a counterexample.

The intersection of a plane and a sphere will always be a circle.
- True (A plane may intersect a sphere in a circle or at a single point.)
- False, the intersection can be a point or a circle. (A plane may intersect a sphere in a circle or at a single point.)
- False, the intersection can only be a point. (A plane may intersect a sphere in a circle or at a single point.)
- False, the intersection can only be a line segment. (A plane may intersect a sphere in a circle or at a single point.)
A: False, the intersection can be a point or a circle.

Q: Find the lateral area of the cone. Use 3.14 for p. Round to the nearest tenth if necessary.
[Image]
- 35.0 [Image] (The formula for the lateral area of a cone is p [Image]r [Image] where r is the radius of the circular base and [Image] is the slant height of the cone. For example, if the slant height is given as 10 and the radius as 6, then the lateral area is then computed as [Image] units^2.)
- 109.9 [Image] (The formula for the lateral area of a cone is p [Image]r [Image] where r is the radius of the circular base and [Image] is the slant height of the cone. For example, if the slant height is given as 10 and the radius as 6, then the lateral area is then computed as [Image] units^2.)
- 188.4 [Image] (The formula for the lateral area of a cone is p [Image]r [Image] where r is the radius of the circular base and [Image] is the slant height of the cone. For example, if the slant height is given as 10 and the radius as 6, then the lateral area is then computed as [Image] units^2.)
- 183.2 [Image] (The formula for the lateral area of a cone is p [Image]r [Image] where r is the radius of the circular base and [Image] is the slant height of the cone. For example, if the slant height is given as 10 and the radius as 6, then the lateral area is then computed as [Image] units^2.)
A: 109.9 [Image]

Q: Find the surface area of the regular pyramid. Round to the nearest tenth if necessary.

[Image]
- 385.0 [Image] (The formula for the surface area of a regular pyramid is [Image] [Image] + B where P is the perimeter of the base, [Image] is the slant height and B is the area of the base of the pyramid. To find the perimeter of the base, multiply the length of one side by 4. To find the area of the base, square the length of the side. For example, if the slant height is given as 10 and the side length as 8, then the surface area is then computed as [Image] units^2.)
- 264.0 [Image] (The formula for the surface area of a regular pyramid is [Image] [Image] + B where P is the perimeter of the base, [Image] is the slant height and B is the area of the base of the pyramid. To find the perimeter of the base, multiply the length of one side by 4. To find the area of the base, square the length of the side. For example, if the slant height is given as 10 and the side length as 8, then the surface area is then computed as [Image] units^2.)
- 528.0 [Image] (The formula for the surface area of a regular pyramid is [Image] [Image] + B where P is the perimeter of the base, [Image] is the slant height and B is the area of the base of the pyramid. To find the perimeter of the base, multiply the length of one side by 4. To find the area of the base, square the length of the side. For example, if the slant height is given as 10 and the side length as 8, then the surface area is then computed as [Image] units^2.)
- 1452.0 [Image] (The formula for the surface area of a regular pyramid is [Image] [Image] + B where P is the perimeter of the base, [Image] is the slant height and B is the area of the base of the pyramid. To find the perimeter of the base, multiply the length of one side by 4. To find the area of the base, square the length of the side. For example, if the slant height is given as 10 and the side length as 8, then the surface area is then computed as [Image] units^2.)
A: 385.0 [Image]

Sketch each solid using isometric dot paper.

rectangular prism 4 units high, 6 units long, and 5 units wide

For each solid, draw a net and find the surface area. Round to the nearest tenth if necessary.

[Image]

Sketch each solid using isometric dot paper.

rectangular prism 2 units high, 5 units long, and 3 units wide

Draw the corner view of a figure given each orthogonal drawing.

top |  |  | left |  |  | front |  |  | right
view |  |  | view |  |  | view |  |  | view

Q: Find the lateral area of a cone if the height is 12 meters and the slant height is 16 meters. Use 3.14 for p. Round to the nearest tenth if necessary.
- 1875.6 meters^2 (The formula for the lateral area of a cone is p [Image]r [Image] where r is the radius of the circular base and [Image] is the slant height of the cone. In this problem, you must use the Pythagorean Theorem to find the radius (r) from the given height (h) and slant height ( [Image]). For example, if the slant height is given as 10 and the height as 8, then the radius is [Image]. So the lateral area is then computed as [Image] units^2.)
- 169.3 meters^2 (The formula for the lateral area of a cone is p [Image]r [Image] where r is the radius of the circular base and [Image] is the slant height of the cone. In this problem, you must use the Pythagorean Theorem to find the radius (r) from the given height (h) and slant height ( [Image]). For example, if the slant height is given as 10 and the height as 8, then the radius is [Image]. So the lateral area is then computed as [Image] units^2.)
- 531.7 meters^2 (The formula for the lateral area of a cone is p [Image]r [Image] where r is the radius of the circular base and [Image] is the slant height of the cone. In this problem, you must use the Pythagorean Theorem to find the radius (r) from the given height (h) and slant height ( [Image]). For example, if the slant height is given as 10 and the height as 8, then the radius is [Image]. So the lateral area is then computed as [Image] units^2.)
- 883.4 meters^2 (The formula for the lateral area of a cone is p [Image]r [Image] where r is the radius of the circular base and [Image] is the slant height of the cone. In this problem, you must use the Pythagorean Theorem to find the radius (r) from the given height (h) and slant height ( [Image]). For example, if the slant height is given as 10 and the height as 8, then the radius is [Image]. So the lateral area is then computed as [Image] units^2.)
A: 531.7 meters^2

Q: Determine whether each statement is true or false. If false, give a counterexample.

A chord that contains the center of the sphere is a diameter of the sphere.
- True (A chord that intersects a sphere through its center is a diameter by definition. So the statement is true.)
- False, a chord cannot contain the center of the sphere. (A chord that intersects a sphere through its center is a diameter by definition. So the statement is true.)
- False, a chord containing the center is a radius, not a diameter. (A chord that intersects a sphere through its center is a diameter by definition. So the statement is true.)
- False, a sphere does not have a diameter. (A chord that intersects a sphere through its center is a diameter by definition. So the statement is true.)
A: True

Q: Each side of the square base of a regular pyramid is 6.9 feet. The sides rise at an angle of 52°. Find the lateral area of the pyramid. Round to the nearest tenth.
- 84.2 feet^2 (The formula for the lateral area of a regular pyramid is [Image] [Image] where P is the perimeter of the base and [Image] is the slant height of a face of the pyramid. For example, if the slant height is given as 5 and the perimeter as 24, then the lateral area is computed as [Image] units^2. To find the perimeter, multiply the length of one side by 4. To find the slant height, use the formula [Image] where side is the length of the side of the base and a is the angle which the sides rise by as given in the question.)
- 77.3 feet^2 (The formula for the lateral area of a regular pyramid is [Image] [Image] where P is the perimeter of the base and [Image] is the slant height of a face of the pyramid. For example, if the slant height is given as 5 and the perimeter as 24, then the lateral area is computed as [Image] units^2. To find the perimeter, multiply the length of one side by 4. To find the slant height, use the formula [Image] where side is the length of the side of the base and a is the angle which the sides rise by as given in the question.)
- 70.2 feet^2 (The formula for the lateral area of a regular pyramid is [Image] [Image] where P is the perimeter of the base and [Image] is the slant height of a face of the pyramid. For example, if the slant height is given as 5 and the perimeter as 24, then the lateral area is computed as [Image] units^2. To find the perimeter, multiply the length of one side by 4. To find the slant height, use the formula [Image] where side is the length of the side of the base and a is the angle which the sides rise by as given in the question.)
- 60.9 feet^2 (The formula for the lateral area of a regular pyramid is [Image] [Image] where P is the perimeter of the base and [Image] is the slant height of a face of the pyramid. For example, if the slant height is given as 5 and the perimeter as 24, then the lateral area is computed as [Image] units^2. To find the perimeter, multiply the length of one side by 4. To find the slant height, use the formula [Image] where side is the length of the side of the base and a is the angle which the sides rise by as given in the question.)
A: 77.3 feet^2

Q: The solid below is a composite of a cube and a square pyramid. The base of the solid is the base of the cube. Find the lateral area of the solid.
[Image]
- 80 [Image] (The formula for the lateral area of a regular pyramid is [Image] [Image] where P is the perimeter of the base and [Image] is the slant height of a face of the pyramid. For example, if the slant height is given as 12 and the side as 14, then the perimeter is 56 (4 time 14, since it is a square), and the lateral area is [Image] units^2. The lateral area of the cube is perimeter times height. The perimeter in this example as 56, and the height is 14 since it is a cube. So the lateral area of the cube is [Image]. Finally, the total lateral area of the solid is the sum of the lateral area of the pyramid and the lateral area of the cube. Add the two previous answers to get [Image] units^2.)
- 660 [Image] (The formula for the lateral area of a regular pyramid is [Image] [Image] where P is the perimeter of the base and [Image] is the slant height of a face of the pyramid. For example, if the slant height is given as 12 and the side as 14, then the perimeter is 56 (4 time 14, since it is a square), and the lateral area is [Image] units^2. The lateral area of the cube is perimeter times height. The perimeter in this example as 56, and the height is 14 since it is a cube. So the lateral area of the cube is [Image]. Finally, the total lateral area of the solid is the sum of the lateral area of the pyramid and the lateral area of the cube. Add the two previous answers to get [Image] units^2.)
- 560 [Image] (The formula for the lateral area of a regular pyramid is [Image] [Image] where P is the perimeter of the base and [Image] is the slant height of a face of the pyramid. For example, if the slant height is given as 12 and the side as 14, then the perimeter is 56 (4 time 14, since it is a square), and the lateral area is [Image] units^2. The lateral area of the cube is perimeter times height. The perimeter in this example as 56, and the height is 14 since it is a cube. So the lateral area of the cube is [Image]. Finally, the total lateral area of the solid is the sum of the lateral area of the pyramid and the lateral area of the cube. Add the two previous answers to get [Image] units^2.)
- 720 [Image] (The formula for the lateral area of a regular pyramid is [Image] [Image] where P is the perimeter of the base and [Image] is the slant height of a face of the pyramid. For example, if the slant height is given as 12 and the side as 14, then the perimeter is 56 (4 time 14, since it is a square), and the lateral area is [Image] units^2. The lateral area of the cube is perimeter times height. The perimeter in this example as 56, and the height is 14 since it is a cube. So the lateral area of the cube is [Image]. Finally, the total lateral area of the solid is the sum of the lateral area of the pyramid and the lateral area of the cube. Add the two previous answers to get [Image] units^2.)
A: 560 [Image]

Q: Find the surface area of a sphere if the circumference of a great circle is 43.96 centimeters. Use 3.14 for p. Round to the nearest tenth.
- 4308.1 [Image] (The surface area of a sphere is found by the formula [Image], where r is the radius of the sphere. This problem gives the circumference, so begin by computing the radius which can be found by the formula [Image]. For example, if the circumference is given as 12.56, then the radius is [Image] and the surface area would be [Image]. If the solid is a hemisphere, then the formula for the surface area is [Image], since a hemisphere has half the surface area of the sphere [Image] plus the area of the great circle on its base [Image].)
- 196 [Image] (The surface area of a sphere is found by the formula [Image], where r is the radius of the sphere. This problem gives the circumference, so begin by computing the radius which can be found by the formula [Image]. For example, if the circumference is given as 12.56, then the radius is [Image] and the surface area would be [Image]. If the solid is a hemisphere, then the formula for the surface area is [Image], since a hemisphere has half the surface area of the sphere [Image] plus the area of the great circle on its base [Image].)
- 153.9 [Image] (The surface area of a sphere is found by the formula [Image], where r is the radius of the sphere. This problem gives the circumference, so begin by computing the radius which can be found by the formula [Image]. For example, if the circumference is given as 12.56, then the radius is [Image] and the surface area would be [Image]. If the solid is a hemisphere, then the formula for the surface area is [Image], since a hemisphere has half the surface area of the sphere [Image] plus the area of the great circle on its base [Image].)
- 615.4 [Image] (The surface area of a sphere is found by the formula [Image], where r is the radius of the sphere. This problem gives the circumference, so begin by computing the radius which can be found by the formula [Image]. For example, if the circumference is given as 12.56, then the radius is [Image] and the surface area would be [Image]. If the solid is a hemisphere, then the formula for the surface area is [Image], since a hemisphere has half the surface area of the sphere [Image] plus the area of the great circle on its base [Image].)
A: 615.4 [Image]

Draw the corner view of a figure given each orthogonal drawing.

top |  |  | left |  |  | front |  |  |  | right
view |  |  | view |  |  | view |  |  |  | view

Q: Determine whether each statement is true or false. If false, give a counterexample.

A great circle of a sphere separates it into two congruent halves, each called a hemisphere.
- True (That is the definition of a hemisphere, so the statement is true.)
- False, a hemisphere is not created by great circles. (That is the definition of a hemisphere, so the statement is true.)
- False, the two hemispheres will not be congruent. (That is the definition of a hemisphere, so the statement is true.)
- False, a hemisphere is not half of a sphere. (That is the definition of a hemisphere, so the statement is true.)
A: True

Q: Find the slant height of the cone with the given measurements, rounded to the nearest hundredth. Then use your result to find the surface area of the cone. Use 3.14 for p. Round the final answer to the nearest ten-thousandth.

height: 17 feet
radius: 13 feet
- 873.5480 feet^2 (The formula for the surface area of a cone is pr [Image] + pr^2 where r is the radius of the circular base and [Image] is the slant height of the cone. In this problem, you must use the Pythagorean Theorem to find the slant height ( [Image]) from the given height (h) and radius (r). For example, if the radius is given as 6 and the height as 8, then the slant height is [Image]. So the surface area is then computed as [Image] units^2.)
- 278.2000 feet^2 (The formula for the surface area of a cone is pr [Image] + pr^2 where r is the radius of the circular base and [Image] is the slant height of the cone. In this problem, you must use the Pythagorean Theorem to find the slant height ( [Image]) from the given height (h) and radius (r). For example, if the radius is given as 6 and the height as 8, then the slant height is [Image]. So the surface area is then computed as [Image] units^2.)
- 3785.3747 feet^2 (The formula for the surface area of a cone is pr [Image] + pr^2 where r is the radius of the circular base and [Image] is the slant height of the cone. In this problem, you must use the Pythagorean Theorem to find the slant height ( [Image]) from the given height (h) and radius (r). For example, if the radius is given as 6 and the height as 8, then the slant height is [Image]. So the surface area is then computed as [Image] units^2.)
- 1404.2080 feet^2 (The formula for the surface area of a cone is pr [Image] + pr^2 where r is the radius of the circular base and [Image] is the slant height of the cone. In this problem, you must use the Pythagorean Theorem to find the slant height ( [Image]) from the given height (h) and radius (r). For example, if the radius is given as 6 and the height as 8, then the slant height is [Image]. So the surface area is then computed as [Image] units^2.)
A: 1404.2080 feet^2

Q: Determine whether each statement is true or false. If false, give a counterexample.

All radii of the same sphere are congruent.
- True (The radii of a sphere, like those of a circle, are all the same length. So the statement is true.)
- False, no two radii of the sphere are congruent. (The radii of a sphere, like those of a circle, are all the same length. So the statement is true.)
- False, some radii will be congruent, but not all. (The radii of a sphere, like those of a circle, are all the same length. So the statement is true.)
- False, a segment joining two points on the sphere is a tangent. (The radii of a sphere, like those of a circle, are all the same length. So the statement is true.)
A: True

Q: Determine whether each statement is true or false. If false, give a counterexample.

A sphere has exactly one great circle.
- True (A sphere may have an infinite number of great circles.)
- False, a sphere has an infinite number of great circles. (A sphere may have an infinite number of great circles.)
- False, a sphere has exactly two great circles. (A sphere may have an infinite number of great circles.)
- False, a sphere has exactly four great circles. (A sphere may have an infinite number of great circles.)
A: False, a sphere has an infinite number of great circles.

Draw the corner view of a figure given each orthogonal drawing.

top |  | left |  | front |  | right
view |  | view |  | view |  | view

Q: Find the surface area of the sphere. Use 3.14 for p. Round to the nearest tenth.
[Image]
- 113 [Image] (The surface area of a sphere is found by the formula [Image], where r is the radius of the sphere. This diagram gives the diameter, so begin by computing the radius which is half the diameter. For example, if the diameter is given as 12, then the radius is 6 and the surface area would be [Image].)
- 339.1 [Image] (The surface area of a sphere is found by the formula [Image], where r is the radius of the sphere. This diagram gives the diameter, so begin by computing the radius which is half the diameter. For example, if the diameter is given as 12, then the radius is 6 and the surface area would be [Image].)
- 28.3 [Image] (The surface area of a sphere is found by the formula [Image], where r is the radius of the sphere. This diagram gives the diameter, so begin by computing the radius which is half the diameter. For example, if the diameter is given as 12, then the radius is 6 and the surface area would be [Image].)
- 36 [Image] (The surface area of a sphere is found by the formula [Image], where r is the radius of the sphere. This diagram gives the diameter, so begin by computing the radius which is half the diameter. For example, if the diameter is given as 12, then the radius is 6 and the surface area would be [Image].)
A: 113 [Image]

A large model of a pencil was constructed for an art show. It consists of two parts: a cone for the top and a right cylinder for the base, as shown below.
[Image]
a.      What is the surface area of the cone to the nearest square inch? Use 3.14 for p.
b.      What is the surface area of the cylinder to the nearest square inch? Use 3.14 for p.
c.      What is the surface area of the pencil once it is assembled? Round to the nearest square inch? Use 3.14 for p.

Q: Determine whether each statement is true or false. If false, give a counterexample.

A segment with endpoints that are points on the sphere is a radius.
- True (A segment joining two points on the sphere is a chord.)
- False, a segment joining two points on the sphere is a diameter. (A segment joining two points on the sphere is a chord.)
- False, a segment joining two points on the sphere is a chord. (A segment joining two points on the sphere is a chord.)
- False, a segment joining two points on the sphere is a tangent. (A segment joining two points on the sphere is a chord.)
A: False, a segment joining two points on the sphere is a chord.

Q: Determine whether each statement is true or false. If false, give a counterexample.

A tangent of a sphere is a line that intersects a sphere in exactly two points.
- True (A line that intersects a sphere in exactly one point is a tangent by definition.)
- False, a tangent cannot intersect a sphere. (A line that intersects a sphere in exactly one point is a tangent by definition.)
- False, a tangent is a line that intersects the center of a sphere. (A line that intersects a sphere in exactly one point is a tangent by definition.)
- False, a tangent of a sphere is a line that intersects a sphere in exactly one point. (A line that intersects a sphere in exactly one point is a tangent by definition.)
A: False, a tangent of a sphere is a line that intersects a sphere in exactly one point.

Q: Determine whether each statement is true or false. If false, give a counterexample.

Two different great circles will intersect in exactly one point.
- True (Two different great circles will always intersect in exactly two points.)
- False, two different great circles will not intersect at all. (Two different great circles will always intersect in exactly two points.)
- False, two great circles will intersect in exactly two points. (Two different great circles will always intersect in exactly two points.)
- False, two great circles will intersect in exactly four points. (Two different great circles will always intersect in exactly two points.)
A: False, two great circles will intersect in exactly two points.

A cylindrical pole is being used to support a large banner. The diameter of the base of the pole is 8 inches and the height is 12 feet.
a.      Draw a net of the cylinder and label the dimensions.
b.      What is the lateral area of the pole to the nearest square foot? Use 3.14 for p.
c.      What is the surface area of the pole to the nearest square foot? Use 3.14 for p.

Sketch each solid using isometric dot paper.

triangular prism 4 units high, with bases that are right triangles with legs 2 units and 5 units long

Sketch each solid using isometric dot paper.

triangular prism 5 units high, with bases that are right triangles with legs 3 units and 4 units long

Q: Determine whether each statement is true or false. If false, give a counterexample.

If a plane intersects a sphere so that it contains the center of the sphere, then that intersection will sometimes be a great circle.
- True (A plane that intersects a sphere through its center will result in a great circle by definition.)
- False, the intersection will never be a great circle. (A plane that intersects a sphere through its center will result in a great circle by definition.)
- False, the intersection will always be a great circle. (A plane that intersects a sphere through its center will result in a great circle by definition.)
- False, the intersection of a plane and a sphere cannot contain the center of the sphere. (A plane that intersects a sphere through its center will result in a great circle by definition.)
A: False, the intersection will always be a great circle.

Q: Determine whether each statement is true or false. If false, give a counterexample.

Every great circle of a sphere passes through its center.
- True (By definition, a great circle must pass through the center of a sphere. So this statement is true.)
- False, only one of a sphere’s great circles passes through its center. (By definition, a great circle must pass through the center of a sphere. So this statement is true.)
- False, none of a sphere’s great circles passes through its center. (By definition, a great circle must pass through the center of a sphere. So this statement is true.)
- False, exactly two of a sphere’s great circles passes through its center. (By definition, a great circle must pass through the center of a sphere. So this statement is true.)
A: True

Q: Find the surface area of the regular pyramid. Round to the nearest tenth if necessary.

[Image]
- 496.7 [Image] (The formula for the surface area of a regular pyramid is [Image] [Image] + B where P is the perimeter of the base, [Image] is the slant height and B is the area of the base of the pyramid. Since this is a regular (or equilateral) triangle, to find the perimeter of the base, multiply the length of one side by 3. To find the area of the triangular base, the formula is [Image]. To find the length of the slant height, use the Pythagorean Theorem as follows: [Image]. For example, if the lateral edge is given as 16 and the side as 18, then the slant height is [Image]. The surface area is [Image] units^2.)
- 5184.0 [Image] (The formula for the surface area of a regular pyramid is [Image] [Image] + B where P is the perimeter of the base, [Image] is the slant height and B is the area of the base of the pyramid. Since this is a regular (or equilateral) triangle, to find the perimeter of the base, multiply the length of one side by 3. To find the area of the triangular base, the formula is [Image]. To find the length of the slant height, use the Pythagorean Theorem as follows: [Image]. For example, if the lateral edge is given as 16 and the side as 18, then the slant height is [Image]. The surface area is [Image] units^2.)
- 864.0 [Image] (The formula for the surface area of a regular pyramid is [Image] [Image] + B where P is the perimeter of the base, [Image] is the slant height and B is the area of the base of the pyramid. Since this is a regular (or equilateral) triangle, to find the perimeter of the base, multiply the length of one side by 3. To find the area of the triangular base, the formula is [Image]. To find the length of the slant height, use the Pythagorean Theorem as follows: [Image]. For example, if the lateral edge is given as 16 and the side as 18, then the slant height is [Image]. The surface area is [Image] units^2.)
- 356.4 [Image] (The formula for the surface area of a regular pyramid is [Image] [Image] + B where P is the perimeter of the base, [Image] is the slant height and B is the area of the base of the pyramid. Since this is a regular (or equilateral) triangle, to find the perimeter of the base, multiply the length of one side by 3. To find the area of the triangular base, the formula is [Image]. To find the length of the slant height, use the Pythagorean Theorem as follows: [Image]. For example, if the lateral edge is given as 16 and the side as 18, then the slant height is [Image]. The surface area is [Image] units^2.)
A: 496.7 [Image]

Q: Find the slant height of the cone with the given measurements, rounded to the nearest hundredth. Then use your result to find the surface area of the cone. Use 3.14 for p. Round the final answer to the nearest ten-thousandth.

height: 8 yards
diameter: 8 yards
- 485.0672 yards^2 (The formula for the surface area of a cone is pr [Image] + pr^2 where r is the radius of the circular base and [Image] is the slant height of the cone. In this problem, you must first find the radius from the diameter by dividing by 2 (the radius is half the diameter). Then you must use the Pythagorean Theorem to find the slant height ( [Image]) from the given height (h) and radius (r). For example, if the diameter is given as 12, then the radius is 6 and the height is given as 8, then the slant height is [Image]. So the surface area is then computed as [Image] units^2.)
- 112.2864 yards^2 (The formula for the surface area of a cone is pr [Image] + pr^2 where r is the radius of the circular base and [Image] is the slant height of the cone. In this problem, you must first find the radius from the diameter by dividing by 2 (the radius is half the diameter). Then you must use the Pythagorean Theorem to find the slant height ( [Image]) from the given height (h) and radius (r). For example, if the diameter is given as 12, then the radius is 6 and the height is given as 8, then the slant height is [Image]. So the surface area is then computed as [Image] units^2.)
- 162.5264 yards^2 (The formula for the surface area of a cone is pr [Image] + pr^2 where r is the radius of the circular base and [Image] is the slant height of the cone. In this problem, you must first find the radius from the diameter by dividing by 2 (the radius is half the diameter). Then you must use the Pythagorean Theorem to find the slant height ( [Image]) from the given height (h) and radius (r). For example, if the diameter is given as 12, then the radius is 6 and the height is given as 8, then the slant height is [Image]. So the surface area is then computed as [Image] units^2.)
- 35.7600 yards^2 (The formula for the surface area of a cone is pr [Image] + pr^2 where r is the radius of the circular base and [Image] is the slant height of the cone. In this problem, you must first find the radius from the diameter by dividing by 2 (the radius is half the diameter). Then you must use the Pythagorean Theorem to find the slant height ( [Image]) from the given height (h) and radius (r). For example, if the diameter is given as 12, then the radius is 6 and the height is given as 8, then the slant height is [Image]. So the surface area is then computed as [Image] units^2.)
A: 162.5264 yards^2

Sketch each solid using isometric dot paper.

cube 3 units on each edge

Q: Find the surface area of the cylinder. Use 3.14 for pi and round your answer to the nearest tenth.

[Image]
- 408.8 [Image] (The formula for the surface area of a cylinder is [Image] where r is the radius and h is the height of the cylinder. For example, if the height is given as 6 and the radius as 4, then the surface area is computed as 2 ´ 3.14 ´ 4 ´ 6 + 2 ´ 3.14 ´ 4 ´ 4 = 251.2 units^2.)
- 695.0 [Image] (The formula for the surface area of a cylinder is [Image] where r is the radius and h is the height of the cylinder. For example, if the height is given as 6 and the radius as 4, then the surface area is computed as 2 ´ 3.14 ´ 4 ´ 6 + 2 ´ 3.14 ´ 4 ´ 4 = 251.2 units^2.)
- 704.7 [Image] (The formula for the surface area of a cylinder is [Image] where r is the radius and h is the height of the cylinder. For example, if the height is given as 6 and the radius as 4, then the surface area is computed as 2 ´ 3.14 ´ 4 ´ 6 + 2 ´ 3.14 ´ 4 ´ 4 = 251.2 units^2.)
- 650.2 [Image] (The formula for the surface area of a cylinder is [Image] where r is the radius and h is the height of the cylinder. For example, if the height is given as 6 and the radius as 4, then the surface area is computed as 2 ´ 3.14 ´ 4 ´ 6 + 2 ´ 3.14 ´ 4 ´ 4 = 251.2 units^2.)
A: 650.2 [Image]