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Solid Geometry

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

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

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

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 ACT, 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).

Circles and Conics questions on the ACT are straightforward and formula driven. The following chapters will list all the formulas you need to know to answer the circle questions on the ACT Math section.

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