An isosceles trapezoid has parallel base lengths of 10 and 26, and an altitude of 15. A li… — GRE General Test

An isosceles trapezoid has parallel base lengths of $10$ and $26$ , and an altitude of $15$ . A line segment connects the midpoints of the two non-parallel legs, dividing the figure into two smaller trapezoids. What is the area of the larger of these two smaller trapezoids?

Answer: 165

Answer

165

The midsegment of a trapezoid connects the midpoints of the non-parallel legs, and its length is the average of the two parallel bases:

\frac{10 + 26}{2} = 18

. Because the line connects midpoints, it also bisects the altitude, making the height of each smaller trapezoid equal to

\frac{15}{2} = 7.5

. The larger of the two resulting trapezoids has bases of lengths

18

and

26

. Using the trapezoid area formula

\text{Area} = \frac{b_1 + b_2}{2} \times h

, we obtain

\frac{18 + 26}{2} \times 7.5 = 22 \times 7.5 = 165

Step-by-Step Solution

Calculate the length of the midsegment connecting the midpoints of the non-parallel legs.

The midsegment length is

18

The midsegment of a trapezoid is parallel to the bases and its length equals the average of the two base lengths:

\frac{10 + 26}{2} = 18

Determine the height of the smaller subtrapezoid.

The height of the subtrapezoid is

7.5

The segment connecting the midpoints of the legs bisects the overall altitude of

15

, giving a height of

\frac{15}{2} = 7.5

for each subtrapezoid.

Compute the area of the larger subtrapezoid.

The area is

165

The larger subtrapezoid is bounded by the midsegment (length

18

) and the bottom base (length

26

). Applying the trapezoid area formula yields

\text{Area} = \frac{18 + 26}{2} \times 7.5 = 22 \times 7.5 = 165

Key Concept

Trapezoid Midsegment Theorem and Subdivided Area Calculation

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