# SAT Geometry Student-Produced Response

To successfully solve solid geometry problems on the SAT, you need to know:

- the contents of the Reference Information at the beginning of each math section
- the properties of triangles, polygons, and circles
- about congruent and similar figures
- how to find the perimeter and area of triangles, polygons, and circles
- how to find the surface area and volume of rectangular solids and cylinders

## Examples

A stack of ancient coins forms a cylinder with height \(H=10\) cm. If each coin has radius \(r=2\) cm and volume \(V=2\pi\) cm\(^3,\) how many coins are there in the stack?

Correct Answer: 20

Solution:

Tip: Volume of a cylinder: \(V=\pi r^2 h.\)

We find the height of each coin:\[\begin{align} V = \pi r^2 h &= 2 \pi\\ r^2 h &= 2\\ 2^2 h &= 2\\ 4h &= 2\\ h &= 0.5\ \text{cm}\\ \end{align}\]

If the stack is \(10\) cm tall, and each coin is \(0.5\) cm, there are \(\frac{10}{0.5} = 20\) coins.

Common Mistakes:

- Misreading the prompt and thinking that the volume of the stack of coins is \(2\pi,\) rather than the volume of each coin.
- Not knowing how to find the volume of a cylinder and forgetting about the Reference Information at the beginning of each math section.
- Finding the height of each coin, instead of the number of coins in the stack.
- When finding the number of coins in the stack, multiplying the height of the stack by the height of each coin, rather than dividing the two numbers.
If you got this problem wrong, you should review SAT Solid Geometry.

The perimeter of an equilateral triangle is equal to its area. What is the length of the side of the triangle, rounded to the nearest whole number?

Correct Answer: 7

Tip: Draw a picture.

Tip: Know the \(30^\circ-60^\circ-90^\circ\) and the \(45^\circ-45^\circ-90^\circ\) Theorems.

We draw an equilateral triangle \(\triangle ABC\) with side length \(x\) and height \(h.\) Note that in an equilateral triangle, the height is also a median.

The perimeter of the triangle is \(P=3x\) and its area is \(A = \frac{1}{2}\cdot xh.\) We are told that these are equal, but in order to find \(x\) by equating \(P\) and \(A,\) we need to express \(A\) in terms of \(x\) only.

Because \(\triangle ADC\) is a \(30^\circ - 60^\circ - 90^\circ\) triangle, its longer leg is \(\sqrt{3}\) longer than its shorter leg. So,

\(h = \sqrt{3}\cdot \frac{x}{2} = \frac{\sqrt{3}x}{2}.\)

The area of the triangle is \(A = \frac{1}{2}\cdot xh = \frac{1}{2} \cdot x \cdot \frac{\sqrt{3}x}{2} = \frac{\sqrt{3}x^2}{4}.\)

We are ready to set up an equation.

\[\begin{array}{r c l} A &=& P\\ \frac{\sqrt{3}x^2}{4} &=& 3x\\ \sqrt{3}x^2 &=& 12x\\ \sqrt{3}x^2 - 12x &=& 0\\ x(\sqrt{3}x - 12) &=& 0\\ \end{array}\]

Either \(x=0\) or \(\sqrt{3}x - 12 = 0.\) \(x\) cannot be \(0.\) So,

\[\begin{align} \sqrt{3}x - 12 &=0\\ \sqrt{3}x &=12\\ x &= \frac{12}{\sqrt{3}} = \frac{12\sqrt{3}x}{3} = 4\sqrt{3} \approx 7\\ \end{align}\]

Common Mistakes:

- Not drawing a diagram.
- Not knowing the properties of equilateral triangles.
- Not knowing the properties of \(30^\circ - 60^\circ - 90^\circ.\)
- Not knowing how to factor.
- Not approximating correctly.
If you got this problem wrong, you should review SAT Right Triangles and SAT Quadratic Functions.

## For Further Study

If you thought these examples difficult and you need to review the material, these links will help:

## SAT Tips for Geometry Student-Produced Response

- Area of a triangle with height \(h\) and base \(b\): \(A_{\triangle} = \frac{1}{2}bh.\)
- Know the \(30^\circ-60^\circ-90^\circ\) and the \(45^\circ-45^\circ-90^\circ\) Theorems.
- The perimeter of a square with side length \(s\): \(P_{\square} = 4s.\)
- Area of a square with side length \(s: A_{\square} = s^2.\)
- Area of a rectangle with length \(l\) and width \(w: A = l\cdot w.\)
- The volume of a cube with edge length \(s\): \(V = s^3.\)
- The volume of a rectangular solid with length \(l,\) width \(w,\) and height \(h: V = l\cdot w \cdot h.\)
- The surface area of a cube with edge length \(s\): \(SA = 6s^2.\)
- Volume of a cylinder with base radius \(r\) and height \(h: V = \pi r^2 h.\)
- The circumference of a circle with radius \(r\) and diameter \(d: C = 2\pi r = \pi d.\)
- Area of a circle with radius \(r: A_{\bigodot} = \pi r^2.\)
- The measure of an arc equals the measure of its central angle.
- The length of an arc with measure \(x^\circ\) is \(\frac{x}{360}\cdot 2 \pi r.\)
- The area of the sector formed by an arc measuring \(x\) and two radii is \(\frac{x}{360} \cdot \pi r^2.\)
- SAT General Tips

**Cite as:**SAT Geometry Student-Produced Response.

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