Question

find the area of trapezoid vwxy. round your answer to the nearest tenth if necessary.

image of trapezoid vwxy with sides labeled: xy = 6.6 in, xw = 8.8 in, the segment from x to the side vw (with right angle) is 8.8 in, vw = 11 in, yv = 9.4 in

Explanation:

Step1: Recall the formula for the area of a trapezoid

The formula for the area of a trapezoid is \( A=\frac{1}{2}(b_1 + b_2)h \), where \( b_1 \) and \( b_2 \) are the lengths of the two parallel sides (bases) and \( h \) is the height (the perpendicular distance between the bases).

Step2: Identify the bases and the height

From the diagram, the two parallel sides (bases) seem to be \( YV = 9.4\) in and \( XW \)? Wait, no, looking at the sides: the two parallel sides (the ones that are the "bases") - actually, the height is the perpendicular distance, which is 11 in? Wait, no, let's re - examine. Wait, the trapezoid has two parallel sides: one is \( YX +...\) Wait, no, the two parallel sides (bases) are \( YV = 9.4\) in and \( XW\)? Wait, no, the lengths of the two bases: looking at the sides, the two parallel sides (the ones that are the "top" and "bottom") - actually, the two bases are \( 6.6 + 9.4\)? No, wait, the formula for the area of a trapezoid is the average of the two bases times the height. Wait, in the diagram, the two parallel sides (bases) are \( 6.6\) in and \( 9.4\) in? No, that doesn't seem right. Wait, no, the height is the perpendicular distance between the two bases. Wait, the height here is 11 in? Wait, no, the side with length 11 in is perpendicular to the two bases. Wait, actually, the two bases are \( 6.6\) in and \( 9.4\) in? No, wait, let's look again. The trapezoid has two parallel sides: let's see, the sides \( YV = 9.4\) in and \( XW\)? Wait, no, the correct bases: the two parallel sides (the ones that are the "bases") are \( 6.6\) in and \( 9.4\) in? Wait, no, the height is the distance between them, which is 11 in? Wait, no, the formula is \( A=\frac{(a + b)}{2}\times h\), where \( a \) and \( b \) are the lengths of the two parallel sides, and \( h \) is the height (the perpendicular distance between them).

Wait, actually, the two bases are \( 6.6\) in and \( 9.4\) in, and the height is 11 in? Wait, no, that can't be. Wait, no, the side with length 11 in is perpendicular to the two bases. Wait, let's check the lengths: the two parallel sides (bases) are \( 6.6\) in and \( 9.4\) in, and the height (the perpendicular distance between them) is 11 in? Wait, no, that would make the area \( \frac{(6.6 + 9.4)}{2}\times11\). Wait, \( 6.6+9.4 = 16\), then \( \frac{16}{2}=8\), and \( 8\times11 = 88\)? But that seems too simple. Wait, no, maybe I misidentified the bases. Wait, another way: the trapezoid can also be thought of as having bases \( 8.8\) and \( 8.8\)? No, that would be a parallelogram. Wait, no, looking at the diagram again: the sides \( YX = 6.6\) in, \( XW = 8.8\) in, \( WV = 11\) in (perpendicular), \( VY = 9.4\) in. So the two parallel sides (bases) are \( YX = 6.6\) in and \( VZ\) (where \( Z\) is a point on \( XW\))? No, wait, the correct bases are \( 6.6\) in and \( 9.4\) in, and the height is 11 in? Wait, no, the height is the length of the side that is perpendicular to the two bases, which is 11 in. And the two bases are \( 6.6\) in and \( 9.4\) in.

Wait, let's calculate: \( A=\frac{(6.6 + 9.4)}{2}\times11\)

Step3: Calculate the sum of the bases

First, add the two bases: \( 6.6+9.4 = 16\)

Step4: Calculate the average of the bases

Then, take the average: \( \frac{16}{2}=8\)

Step5: Multiply by the height

Then, multiply by the height (11 in): \( 8\times11 = 88\)

Wait, but that seems too straightforward. Wait, maybe I made a mistake in identifying the bases. Wait, another approach: the trapezoid is composed of a rectangle and a triangle? No, wait, the two parallel sides (bases) are \( 6.6\) and \( 9.4\), height is 11. Let's re - check the formula. The area of a trapezoid is \( A=\frac{1}{2}(b_1 + b_2)h\), where \( b_1\) and \( b_2\) are the lengths of the two parallel sides. If \( b_1 = 6.6\), \( b_2=9.4\), and \( h = 11\), then:

\( A=\frac{1}{2}(6.6 + 9.4)\times11=\frac{1}{2}(16)\times11 = 8\times11=88\) square inches.

Answer:

The area of trapezoid \( VWXY\) is \(\boldsymbol{88.0}\) square inches (rounded to the nearest tenth, it is still 88.0 as it is a whole number).