*\(V_{cube}=l\: w\: h=s^{3}\)*

*\(Surface\: Area=6s^{2}\)*

*\(Lateral\: Surface\: Area=4s^{2}\)*

A.

Solution

1 gallon = 32 \(ft^{2}\)

1 wall = 2 gallons

Therefore, 1 wall = 2(32 \(ft^{2}\)) = 64 \(ft^{2}\)

Now, since there are four walls, the total square feet required is 4 x 64 = 256 \(ft^{2}\)

that can be packed into a cube shaped box whose interiors measures 6 inches on an edge?

A.

Solution

You can use the formula for the volume of a cube: \(s^{3}\rightarrow 6^{3}=216\)

Now find the volume of one of the smaller rectangular solids: LWH \(\Rightarrow (3)(2)(1)=6\)

And divide the larger rectangular solid by the smaller to find out how many of the smaller rectangular solids can fit inside the larger: \(\frac{216}{6}=36\)

The cuboid is often called a rectangular solid or a rectangular and triangular.

*Volume = l w hSurface Area = 2lw + 2lh + 2wh = 2(lw + lh + wh)Lateral Surface Area = 2lh + 2wh = 2(lh + wh)*

of water, what are the dimensions of the tank?

A.

Solution: Since the tank is 75% full and the volume is 3375 /(cm^{3}\) when 75% full, the entire volume of the prism is = 4500 /(cm^{3}\)

Additionally, we are given that:

\(w=3l\rightarrow l=\frac{w}{3}\)

\(w=2h\rightarrow h=\frac{w}{2}\)

Therefore, V = l w h

\(4500=\frac{w}{3}\) x w x \(\frac{w}{2}=\frac{w^{3}}{6}\)

\(w^{3}=2700\)

w = 30

\(l=\frac{w}{3}=10\)

\(h=\frac{w}{2}=15\)

a = height of the triangular base

c = side length of triangle

h = height of the prism

\(Surface\: Area=2(A_{1})+A_{11}+A_{12}+A_{13}\)

\(A_{1}\) = Area of the triangle faces

\(A_{11},A_{12}\: and\: A_{13} =\) the areas of each of the rectangular faces

Diane is creating a triangular prism and wants to know how many jelly beans can fill the object. The triangular base has a side length of 4 in and a height of 3 in. The height of the prism is 10 in and 20 jellybeans fill 25 /(in^{3}\) . Approximately how many jellybeans will fill the prism?

A.

Solution:

\(Volume=\frac{1}{2}\: a.c.h\: =\frac{1}{2}\: 4.3.10\: =60\: in^{3}\)

Knowing the volume of the prism, we can proceed to solve for how many jellybeans will fill it.

We know that 20 jelly beans fill 25 /(in^{3}\) of space, so we can set up a proportion to solve for how many jelly beans fill 60 /(in^{3}\) of space: \(\frac{20\: jelly\: beans}{25\: in^{3}}\) x \(\frac{x\: jelly\: beans}{60\: in^{3}}\)

x = 48 jellybeans

Approximately 48 jellybeans would fit in the prism described, so

*\(Volume=\frac{1}{3}Area(Base).H\)*

Area = Area of the Base

H = height of the pyramid

A.

Solution:

b = 2

h = 1.732

H = 4

\(Volume =\frac{1}{6}\cdot 2\cdot 1.732\cdot 4=2.309\: cm^{3}\)

To find the total surface area of a cylinder, add the area of the circular top and bottom, as well as the area of the rectangle that wraps around the outside.

*\(Area_{cylinder}=2\: Circles+1\: Rectangle=2(\pi r^{2})+2\pi rh\)*

The volume of a cylinder measures how much “stuff” it can hold inside. In order to find the volume of a cylinder, use the following formula: *\(Volume_{cylinder}=\pi r^{2}h\)*

A.

Solution:

Substituting values, we get the answer to be 15 π/\(in^{2}\).

*\(Volume=\frac{4}{3}\pi r^{3}\)**Surface Area = 4πr\(^{2}\)*

Jenny gave Melissa a basketball for her birthday and she wrapped it in decorative paper. If the basketball has a diameter of 12 inches how much paper does Jenny use?

A.

Solution

\(Surface\: Area=4\pi r^{2}=4\pi 6^{2}=144\pi\)

C is the correct answer.

Suppose there is a sphere inscribed in a cube-like the one shown above on the left. The cube has a side length of 8 cm. If the sphere is filled completely with water, what is the volume of the cube that is not filled with water?

A.

Solution:

\(V_{cube}=S^{3}=8^{3}=512\)

\(V_{sphere}=\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi 4^{3}=85.33\pi\)

Take the difference between the cube’s volume and the sphere’s volume to find your answer: \(V_{cube}-V_{sphere}=512-85.33\pi =249.93\: cm^{3}\)

The correct answer is A.

- You can find this diagonal by breaking up the figure into two flat triangles and using the Pythagorean Theorem for both.
- First, find the length of the diagonal (hypotenuse) of the base of the solid using the Pythagorean Theorem

\(c^{2}=l^{2}+w^{2}\) - Next, use that length as one of the smaller sides of a new triangle with the diagonal of the rectangular solid as the new hypotenuse:

\(d^{2}=c^{2}+h^{2}\)

The formulae is *\(Diagonal\: of\: a\: Cuboid=\sqrt{(Length^{2}+Width^{2}+Height^{2})}\)*

On the SAT, at times you’ll be asked about removing a slice or a cross-section from a 3D figure. Students often lose it when they see such a question on the test and rarely even attempt to solve it. However, these kinds of questions are pretty straightforward once you realize how cross-sections work.

- If we cut a cube like below, we end up with a square (the shaded Z-D shape).
- If we cut a cube in a diagonal direction like below, we end up with a rectangle (as the shaded 2-D shape).
- If we slice a cylinder horizontally, the resulting Z-D cross section is a circle (the shaded region).
- If we slice a cylinder vertically, theresulting 2-D cross section is a rectangle (the shaded region).