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 \(\triangle AED.\)

\[\begin{align} \frac{1}{2}\cdot AE \cdot DE &= A_{\triangle AED} = 30\\ 5 \cdot DE &= 60\\ DE &= 12\\ \end{align}\]

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

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

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