# SAT Polygons

To successfully solve problems about polygons on the SAT, you need to know:

- the Properties of Parallelograms
- ways to prove that quadrilaterals are parallelograms
- properties of special quadrilaterals: rectangle, square, and rhombus
- the sum of the interior angles of a convex polygon
- how to find the perimeter and area of squares, rectangles, parallelograms and composite figures

## Quadrilaterals

In the figure above, the area of triangle $AED$ is 30. What is the area of parallelogram $ABCD?$

(A) $\ \ 120$

(B) $\ \ 150$

(C) $\ \ 180$

(D) $\ \ 240$

(C) $\ \ 260$

Correct Answer: D

Solution:

Tip: $A_{parallelogram} = bh,$ where $b$ is its base, and $h$ is its height.

Tip: Area of a triangle with height $h$ and base $b$: $A_{\triangle} = \frac{1}{2}bh.$

To find the area of the parallelogram, we must know the lengths of its base and height. The length of its base is $AB = AE + EB = 5+ 15 = 20,$ and we can find the length of its height, $DE,$ by using the information about $\triangle AED.$$\begin{aligned} \frac{1}{2}\cdot AE \cdot DE &= A_{\triangle AED} = 30\\ 5 \cdot DE &= 60\\ DE &= 12\\ \end{aligned}$

$A_{ABCD} = AB \cdot DE = 20 \cdot 12 = 240$

Incorrect Choices:

(A)

Tip: $A_{parallelogram} = bh,$ where $b$ is its base, and $h$ is its height.

If you use $A_{\triangle} = \frac{1}{2}bh$ to find the area of the parallelogram, you will get this wrong answer.

(B)

If you estimate that about 5 triangles with the same area as $\triangle AED$ would fit into the parallelogram, you may get this wrong answer. Notice that the diagram is not drawn to scale.

(C)

Tip: Read diagrams carefully.

If you think the base of the parallelogram is 15, you will get this wrong answer. The base is $\overline{AB}$ and $AB = AE + EB = 5+ 15 = 20.$

(E)

Tip: $A_{parallelogram} = bh,$ where $b$ is its base, and $h$ is its height.

If you find the area of the parallelogram by multiplying $DA$ and $AB,$ you will get this wrong answer. $DA$ is not the parallelogram's height.

## Other Polygons

If the shapes above are regular polygons, what is the value of $x-y?$

(A) $\ \ 38$

(B) $\ \ 45$

(C) $\ \ 60$

(D) $\ \ 75$

(E) $\ \ 165$

Correct Answer: D

Solution:

Tip: The sum of the measures of the exterior angles, one per vertex, of any convex polygon is $360^\circ.$

If we count one exterior angle per vertex, then the equilateral triangle has 3 congruent exterior angles and the regular octagon has 8 congruent exterior angles. So,$x = \frac{360}{3} = 120 \quad \text{and} \quad y = \frac{360}{8} = 45 \quad \text{and}$

$x-y = 120-45 = 75.$

Incorrect Choices:

(A)

If you count two exterior angles per vertex, you will get this wrong answer.

(B)

Tip: Read the entire question carefully.

This is the measure of $\angle y,$ not $x-y.$

(C)

Tip: Read the entire question carefully.

This is the measure of an interior angle in the triangle.

(E)

Tip: Read the entire question carefully.

If you find $x+y$ instead of $x-y$ you will get this wrong answer.

## Review

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

## SAT Tips for Polygons

- Know the Properties of Parallelograms.
- $A_{parallelogram} = bh,$ where $b$ is the length of the base, and $h$ is the height.
- The perimeter of a square with side length $s$: $P_{\square} = 4s.$
- The perimeter of a rectangle with length $l$ and width $w$: $P_{\text{rectangle}} = 2(l+w).$
- Area of a triangle with height $h$ and base $b$: $A_{\triangle} = \frac{1}{2}bh.$
- Area of a square with side length $s: A_{\square} = s^2.$
- The sum of the measures of the interior angles of a convex polygon with $n$ sides is $180(n-2).$
- The sum of the measures of the exterior angles, one per vertex, of any convex polygon is $360^\circ.$
- SAT General Tips