SAT Math : Trapezoids

Study concepts, example questions & explanations for SAT Math

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Example Questions

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:

\(\displaystyle \sqrt{2}\)

\(\displaystyle 4\)

\(\displaystyle 2\sqrt{2}\)

\(\displaystyle 1\)

\(\displaystyle 2\)

Correct answer:

\(\displaystyle 4\)

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.

Example Question #1 : Quadrilaterals

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

Possible Answers:

\(\displaystyle 3\)

\(\displaystyle 5\)

\(\displaystyle \frac{3}{2}\)

\(\displaystyle 4\)

Correct answer:

\(\displaystyle 3\)

Explanation:

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

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

 

Example Question #2 : How To Find The Area Of A Trapezoid

Square 1

The above figure shows Square \(\displaystyle ABCD\)\(\displaystyle M\)is the midpoint of \(\displaystyle \overline{AB}\)\(\displaystyle N\) is the midpoint of \(\displaystyle \overline{CD}\)\(\displaystyle O\) is the midpoint of \(\displaystyle \overline{CN}\). Construct \(\displaystyle \overline{MO}\)

If Square \(\displaystyle ABCD\) has area \(\displaystyle X\), what is the area of Quadrilateral \(\displaystyle AMOD\)?

Possible Answers:
\(\displaystyle \frac{5}{8} X\)

\(\displaystyle \frac{3}{4} X\)

\(\displaystyle \frac{\sqrt{17}}{2} X\)

\(\displaystyle \frac{1}{2} X\)

\(\displaystyle \frac{\sqrt{17}}{4} X\)

Correct answer:
\(\displaystyle \frac{5}{8} X\)
Explanation:

Let \(\displaystyle t\) be the common sidelength of the square. The area of the square is \(\displaystyle X = t^{2}\).

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

\(\displaystyle AD = t\)

and

\(\displaystyle ND = \frac{1}{2} CD = \frac{1}{2} t\).

 

The area of Rectangle \(\displaystyle AMND\) s the product of these dimensions:

\(\displaystyle A_{1} = AD \cdot ND = t \cdot \frac{1}{2} t = \frac{1}{2} t ^{2}\)

The lengths of the legs of Right \(\displaystyle \bigtriangleup MNO\) are

\(\displaystyle MN = AD = t\)

and 

\(\displaystyle NO = \frac{1}{2} CN = \frac{1}{2} \cdot \frac{1}{2} CD = \frac{1}{2} \cdot \frac{1}{2} t = \frac{1}{4} t\)

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

\(\displaystyle A_{2} = \frac{1}{2} \cdot MN \cdot NO = \frac{1}{2} \cdot t \cdot \frac{1}{4} t = \frac{1}{8} t ^{2}\)

This is seen below:

Square 2

The sum of these areas is the area of Quadrilateral \(\displaystyle AMOD\)

\(\displaystyle A = A_{1}+ A_{2}= t^{2}+\frac{1}{8}t ^{2} = \frac{5}{8} t ^{2}\).

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

Example Question #174 : Geometry

Square 1

The above figure shows Square \(\displaystyle ABCD\)\(\displaystyle M\)is the midpoint of \(\displaystyle \overline{AB}\)\(\displaystyle N\) is the midpoint of \(\displaystyle \overline{CD}\)\(\displaystyle O\) is the midpoint of \(\displaystyle \overline{CN}\). Construct \(\displaystyle \overline{MO}\)

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

Possible Answers:

\(\displaystyle \frac{\sqrt{5}}{2} t\)

\(\displaystyle \frac{\sqrt{17}}{4} t\)

\(\displaystyle \frac{\sqrt{65}}{8} t\)

\(\displaystyle \frac{\sqrt{63}}{8} t\)

\(\displaystyle \frac{\sqrt{15}}{4} t\)

Correct answer:

\(\displaystyle \frac{\sqrt{17}}{4} t\)

Explanation:

Construct \(\displaystyle \overline{MN}\) as shown in the diagram below:

Square 1

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

Since \(\displaystyle N\) is the midpoint of \(\displaystyle \overline{CD}\),

\(\displaystyle CN = \frac{1}{2} CD = \frac{1}{2} t\)

Since \(\displaystyle O\) is the midpoint of \(\displaystyle \overline{CN}\)

\(\displaystyle ON = \frac{1}{2} CN = \frac{1}{2} \cdot \frac{1}{2} t = \frac{1}{4} t\).

\(\displaystyle \bigtriangleup MNO\) is a right triangle, so, by the Pythagorean Theorem, 

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

Substituting: 

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

\(\displaystyle MO= \sqrt{t^{2}+ \frac {1}{16} t ^{2} }\)

\(\displaystyle MO= \sqrt{ \frac {17}{16} t ^{2} }\)

Apply the Product of Radicals and Quotient of Radicals Rules:

\(\displaystyle MO= \sqrt{ \frac {17}{16} t ^{2} }\)

\(\displaystyle = \sqrt{ \frac {17}{16} } \cdot \sqrt{t^{2}}\)

\(\displaystyle = \sqrt{ \frac {17}{16} } t\)

\(\displaystyle = \frac { \sqrt{17}}{ \sqrt{16}} \cdot t\)

\(\displaystyle = \frac { \sqrt{17}}{4} t\)

 

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