Question

The figure shows the front side of a purse in the shape of a trapezoid.

What is the area of this trapezoid?

Responses

110 in²
110 in²

121 in²
121 in²

130 in²
130 in²

220 in²
220 in²
Isosceles trapezoid A B C D with parallel sides BC and A D. A D is 13 inches. B C is 9 inches. A dotted segment runs from B to E where E is on side A D and angle B E F is a right angle. A dotted segment runs from point C to F where F is on side A D. Angle C F E is a right angle. C F is 10 inches.

Answer 1

To find the area of the trapezoid \(ABCD\) with parallel sides \(AD\) and \(BC\), we can use the area formula for a trapezoid:

\[
\text{Area} = \frac{1}{2} \times (b_1 + b_2) \times h
\]

where \(b_1\) and \(b_2\) are the lengths of the two bases (the parallel sides) and \(h\) is the height of the trapezoid.

In this case:
- \(b_1 = AD = 13 \text{ inches}\)
- \(b_2 = BC = 9 \text{ inches}\)

To find \(h\), we can use the information given about the lengths \(BE\) and \(CF\). We know that:
- \(CF = 10 \text{ inches}\)
- Since \(BE\) is the height from point \(B\) to line \(AD\), we can deduce that \(h\) is also equal to \(10 \text{ inches}\) (because \(C\) and \(B\) are on a horizontal line above \(AD\)).

Now we can substitute the values into the area formula:

\[
\text{Area} = \frac{1}{2} \times (13 + 9) \times 10
\]

Calculating the sum of the bases:

\[
13 + 9 = 22
\]

Now substitute this value back into the area formula:

\[
\text{Area} = \frac{1}{2} \times 22 \times 10 = 11 \times 10 = 110 \text{ in}^2
\]

Thus, the area of the trapezoid is

\[
\boxed{110 \text{ in}^2}
\]