Springboard, Page 272, #1

1. Springboard, Page 272, #1 This problem has an infinite number of answers. Below is just one example, but the premise is the same, no matter which numbers you use. • Choose possible numbers for two of the dimensions and substitute them into the surface area equation. • Start with small numbers because the surface area isn’t a large number. Below, we start with a length (l) of 2 feet and height (h) of 3 feet. • Solve the problem like a linear equation. • 96 = 2(2)w + 2(2)(3) + 2(3)w • 96 = 4w + 12 + 6w • 96 = 10w + 12 • -12 - 12 • 84 = 10w – Divide both sides by 10 • 8.4 = w • A possible set of dimensions for this rectangular prism is 2 x 3 x 8.4.

2. Springboard, Page 272, #2 Remember, surface area means you must find the area of each surface on the shape and add them together. Even if you don’t have a formula for the specific shape, the idea remains the same. Area of trapezoids= ½ x 4(6+12) = 2(18) = 36 Area of rectangle = 12 x 6 = 72 Area of rectangles on sides = 5 x 6 = 30 Area of square = 6 x 6 = 36 Add them all up. 36 + 36 + 72 + 30 + 30 + 36 = 240 mm2

3. Springboard, Page 272, #3 Find the area of each face and add them together. Area of each triangular base = ½(6)(4) = 12 Area of two side rectangular panels = 5 x 8 = 40 Area of bottom rectangular panel = 6 x 8 = 48 Add them all together. SA = 12 + 12 + 40 + 40 + 48 = 152 ft.2

4. Springboard, Page 272, #4 P = Perimeter of base (Add all the sides of the square base) = 14 ℓ = slant height = 6 B = Area of base (Multiply the base and height of the square base) = 12.25 SA = ½ Pℓ + B SA = ½(14)(6) + 12.25 SA = 42 + 12.25 SA = 54.25 ft2

5. Springboard, Page 272, #5 SA = 2πrh + 2πr2 SA = 2π(10)(18) + 2 π(10)2 SA = 360π + 200π SA = 560π or 1758.4 in2 r = radius, which starts at the center of the circle and stretches to any point on the circle. This problem gives you the diameter, which stretches all the way across the circle. Divide this in half to get the radius. r = 10 in.

6. Springboard, Page 272, #6 There are two different ways to solve this problem, given the information. We can use the formula for the area of a circle and use the 4/5 information given, or we can use the formula for surface area of a cone. Area of a circle = πr2 Surface Area of a cone = ½ (2πr)ℓ + B Option 1: Find the area of both circles. Area of a circle = πr2 A = π42 (Full Circle) A = π52 (Partial circle) A = 16π A = 25π A = 16π or 50.24 cm2The directions tell us that this sector is 4/5 of the area of a circle, so multiply the area by 4/5. A = 25π x 4/5 = 20π or 62.8 cm2 Add these two totals together. A = 16π+ 20π or A = 50.24 + 62.8 A = 36π cm2 or A = 113.04 cm2

7. Springboard, Page 272, #6 There are two different ways to solve this problem, given the information. We can use the formula for the area of a circle and use the 4/5 information given, or we can use the formula for surface area of a cone. Area of a circle = πr2 Surface Area of a cone = ½ (2πr)ℓ + B Surface Area of a cone = ½ (2 πr) ℓ + B R = 4cm ℓ = 5 cm B = πr2 SA = ½ (2 x 3.14 x 4) 5 + π42 SA = ½ (25.12) 5 + 16π SA = ½ (125.6) + 50.24 SA = 62.8 + 50.24 SA = 113.04 cm2

8. Springboard, Page 272, #7 As with other problems we have seen in our homework, we must look at this as two different shapes. We will find the surface area of each, then add them together. And now the pyramid SA = ½ Pℓ – In this case, we don’t need B because the base of the pyramid isn’t contributing to the surface area of the pyramid. It’s on the inside of the shape. SA = ½ Pℓ, P = 4 + 4 + 4 + 4 = 16, ℓ = 3 SA = ½(16)(3) = 24 in.2 Now that we have the surface area of the cube and pyramid, we can add them together. SA = 80 in2 + 24 in2 = 104 in2 Let’s start with the base, which is a cube. All faces of a cube have the same dimensions, so they’ll all have the same area. If you find the area of one side, you’ll know the area for each side. A = bh A = 4 x 4 A = 16 in.2 There are five faces (sides) of this cube that contribute to the surface area. The top is not on the outside and doesn’t need to be counted. SA = 16 x 5 = 80 in2

9. Springboard, Page 264, #4 What is the volume of a cone with a radius of 12 inches and a height of 11.2 inches? V = ⅓(πr2)h V = ⅓(π122)11.2 V = ⅓(π144)11.2 V = 537.6π V = 1,688.064 in.3

10. Springboard, Page 264, #5 The area of the base of a pyramid is 85 square centimeters. If its volumes is 255 cubic centimeters, find the height of the pyramid. V = ⅓Bh 255 = ⅓(85)h – Multiply both sides by 3 to get rid of ⅓. 765 = 85h – Divide both sides by 85 to get h all by itself. 9 = h The height is 9 cm.

11. Springboard, Page 264, #6 Take this problem one step at a time. We don’t have a formula to find the volume of the shape like it is, but we can take it as two separate prisms and find the volume of each. The bottom of shape is a rectangular prism. V = bwh V = 22 x 15 x 12 V = 3,960 cm3 The top of shape is a triangular prism. V = ½(bh)h V = ½(22 x 10) x 15 V = 1,650 cm3 After you have found the area of both the rectangular prism and the triangular prism, add the two volumes together. 3,960 + 1,650 = 5,610 cm3

12. Springboard, Page 264, #1 Determine the volume of the figure sketched on the dot paper. 6 4 3 Count the cubes to determine the dimensions of the prism. Volume of a rectangular prism = bwh b (base) = 3, w (width) = 4, h (height) = 6 Plugthese dimensions into the formula. • V = 3 • 4 • 6 • V = 72 cubic units or 72 units3

13. Springboard, Page 264, #2 Find the volume of the cylinder. Volume of a cylinder = πr2h • π (pi) = 3.14, r (radius) = 4 cm, h (height) = 12 cm Plugthese dimensions into the formula. • V = 3.14 • 42• 12 • V = 3.14 • 16 • 12 • V = 602.88 cm3

14. Springboard, Page 264, #3 Find the volume of the triangular prism. Volume of a triangular prism = (½bh)h • b (base of triangle) = 8 in., h (height of triangle) = 4 in, h (height of prism) = 10 in. Plugthese dimensions into the formula. • V = (½ • 8 • 4) • 10 • V = 160 cubic inches or 160 in.3