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SAT Math : Trapezoids

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Example Question #1 : Trapezoids

A trapezoid has a base of length 4, another base of length s, and a height of length s. A square has sides of length s. What is the value of s such that the area of the trapezoid and the area of the square are equal?

Possible Answers:

$2\sqrt{2}$

$\sqrt{2}$

Correct answer:

Explanation:

In general, the formula for the area of a trapezoid is (1/2)(a + b)(h), where a and b are the lengths of the bases, and h is the length of the height. Thus, we can write the area for the trapezoid given in the problem as follows:

area of trapezoid = (1/2)(4 + s)(s)

Similarly, the area of a square with sides of length a is given by a². Thus, the area of the square given in the problem is s².

We now can set the area of the trapezoid equal to the area of the square and solve for s.

(1/2)(4 + s)(s) = s²

Multiply both sides by 2 to eliminate the 1/2.

(4 + s)(s) = 2s²

Distribute the s on the left.

4s + s² = 2s²

Subtract s² from both sides.

4s = s²

Because s must be a positive number, we can divide both sides by s.

4 = s

This means the value of s must be 4.

The answer is 4.

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Example Question #1 : Quadrilaterals

Find the area of a trapezoid given bases of length 1 and 2 and height of 2.

Possible Answers:

$\frac{3}{2}$

Correct answer:

Explanation:

To solve, simply use the formula for the area of a trapezoid. Thus,

$A=\frac{1}{2}(b1+b2)h=\frac{1}{2}(1+2)(2)=3$

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Example Question #2 : Trapezoids

Square 1

The above figure shows Square ABCD . is the midpoint of $\overline{AB}$ ; is the midpoint of $\overline{CD}$ ; is the midpoint of $\overline{CN}$ . Construct $\overline{MO}$ .

If Square ABCD has area , what is the area of Quadrilateral AMOD ?

Possible Answers:

$\frac{1}{2} X$

$\frac{5}{8} X$

$\frac{\sqrt{17}}{4} X$

$\frac{\sqrt{17}}{2} X$

$\frac{3}{4} X$

Correct answer:

$\frac{5}{8} X$

Explanation:

Let be the common sidelength of the square. The area of the square is $X = t^{2}$ .

Construct segment $\overline{MN}$ . This divides the square into Rectangle AMND and a right triangle. The dimensions of Rectangle AMND are

AD = t

and

$ND = \frac{1}{2} CD = \frac{1}{2} t$ .

The area of Rectangle AMND s the product of these dimensions:

$A_{1} = AD \cdot ND = t \cdot \frac{1}{2} t = \frac{1}{2} t ^{2}$

The lengths of the legs of Right $\bigtriangleup MNO$ are

MN = AD = t

and

$NO = \frac{1}{2} CN = \frac{1}{2} \cdot \frac{1}{2} CD = \frac{1}{2} \cdot \frac{1}{2} t = \frac{1}{4} t$

The area of this right triangle is half the product of these lengths, or

$A_{2} = \frac{1}{2} \cdot MN \cdot NO = \frac{1}{2} \cdot t \cdot \frac{1}{4} t = \frac{1}{8} t ^{2}$

This is seen below:

Square 2

The sum of these areas is the area of Quadrilateral AMOD

$A = A_{1}+ A_{2}= t^{2}+\frac{1}{8}t ^{2} = \frac{5}{8} t ^{2}$ .

Substituting for $t^{2}$ , the area is $\frac{5}{8} X$ .

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Example Question #2 : Quadrilaterals

Square 1

The above figure shows Square ABCD . is the midpoint of $\overline{AB}$ ; is the midpoint of $\overline{CD}$ ; is the midpoint of $\overline{CN}$ . Construct $\overline{MO}$ .

AM = t . Which of the following expresses the length of $\overline{MO}$ in terms of ?

Possible Answers:

$\frac{\sqrt{5}}{2} t$

$\frac{\sqrt{17}}{4} t$

$\frac{\sqrt{65}}{8} t$

$\frac{\sqrt{63}}{8} t$

$\frac{\sqrt{15}}{4} t$

Correct answer:

$\frac{\sqrt{17}}{4} t$

Explanation:

Construct $\overline{MN}$ as shown in the diagram below:

Square 1

Quadrilateral AMND is a rectangle, so opposite sides are congruent. Therefore, MN = AD = t .

Since is the midpoint of $\overline{CD}$ ,

$CN = \frac{1}{2} CD = \frac{1}{2} t$

Since is the midpoint of $\overline{CN}$ ,

$ON = \frac{1}{2} CN = \frac{1}{2} \cdot \frac{1}{2} t = \frac{1}{4} t$ .

$\bigtriangleup MNO$ is a right triangle, so, by the Pythagorean Theorem,

$MO = \sqrt{(MN)^{2}+(NO)^{2} }$

Substituting:

$MO= \sqrt{t^{2}+\left (\frac{1}{4}t \right)^{2} }$

$MO= \sqrt{t^{2}+ \frac {1}{16} t ^{2} }$

$MO= \sqrt{ \frac {17}{16} t ^{2} }$

Apply the Product of Radicals and Quotient of Radicals Rules:

$MO= \sqrt{ \frac {17}{16} t ^{2} }$

$= \sqrt{ \frac {17}{16} } \cdot \sqrt{t^{2}}$

$= \sqrt{ \frac {17}{16} } t$

$= \frac { \sqrt{17}}{ \sqrt{16}} \cdot t$

$= \frac { \sqrt{17}}{4} t$

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SAT Math : Trapezoids

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #1 : Trapezoids

Example Question #1 : Quadrilaterals

Example Question #2 : Trapezoids

Example Question #2 : Quadrilaterals

All SAT Math Resources

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