Question

In an isosceles trapezoid the length of a diagonal is 25 cm and the length of an altitude is 15 cm. Find the area of the trapezoid.
Lots of points

Answer 1

Answer:

The area of the trapezoid is $525\ cm^{2}$

Step-by-step explanation:

we know that

The area of a isosceles trapezoid is equal to the area of two isosceles right triangles plus the area of a rectangle

step 1

Find the area of the  isosceles right triangle

Remember that

In a isosceles right triangle the height is equal to the base of the triangle

we have

$h=15\ cm$

so

$b=15\ cm$

The area is equal to

$A=\frac{1}{2}(b)(h)$

substitute the values

$A=\frac{1}{2}(15)(15)=112.5\ cm^{2}$

step 2

Find the area of the rectangle

The area of the rectangle is equal to

$A=LW$

we have

$W=15\ cm$ -----> is the height of the trapezoid

$d=25\ cm$  -----> the diagonal of the rectangle

Applying the Pythagoras Theorem

$25^{2}=L^{2}+15^{2}\\L^{2}=25^{2}-15^{2} \\ L^{2} =400\\L=20\ cm$

The area of the rectangle is

$A=(20)(15)=300\ cm^{2}$

step 3

Find the area of the trapezoid

$A=2(112.5\ cm^{2})+300\ cm^{2}=525\ cm^{2}$

Answer 2

Answer: The answer is 300 cm^2 I think.