Question

Trapezoid QRST has two right angles. A 5-in.altitude can be drawn dividing QRDT into a rectangle and an isosceles right triangle. The longer side of the rectangle measures 9 in. What is the area of the trapezoid?

Answer 1

Answer:

57.5

Step-by-step explanation:

First you have to solve the rectangle with the formula A=B time height:

9 times 5 is 45, the area of the rectangle is 45

Next, solve the area of the triangle. The triangle is isosceles so the height and base are the same. The formula is base times height /2

Area=5 times 5 = 25. 25 divided by 2 is 12.5.

Now add the areas together (45+12.5). Your answer will be 57.5.

Answer 2

Answer:

The area of the trapezoid is 57.5 square inches

Step-by-step explanation:

we know that

The trapezoid QRST can be divided into a rectangle QRDT and an isosceles right triangle RSD

see the attached figure to better understand the problem

step 1

The area of rectangle is given by the formula

$A=((QR)(RD)$

we have

$RD=5\ in$ ----> altitude

$QR=9\ in$

substitute

$A=((9)(5)=45\ in^2$

step 2

Find the area of the isosceles right triangle

The area of triangle is given by the formula

$A=\frac{1}{2}(RD)(DS)$

we have

$RD=DS=5\ in$ ---> because is an isosceles triangle

substitute

$A=\frac{1}{2}(5)(5)=12.5\ in^2$

step 3

Adds the areas

$A=45+12.5=57.5\ in^2$