Question

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

Answer 1

Assume

$EC=DF=x$

In ADE

$AD^2=AE^2+ED^2$

$25^2=15^2+ED^2$

$ED=20$

from figure

$AB=EF=20-x$

$CD=20+x$

formula

$A=\frac{1}{2} (AB+CD)*(AE)$

we get

$A=\frac{1}{2} ((20-x)+(20+x))*(15)$

$A=300$

area  is 300 square cm

Answer 2

The area of a trapezoid that has the length of a diagonal of 25cm and the length of an altitude of 15cm is 300 cm sq.

What is the area of the trapezoid?

The area of a trapezoid can be calculated if the height of the parallel side and distance are given.

The area of a trapezoid = 1/2 (sum of parallel side) (height)

Consider a trapezoid ABCD that has the length of a diagonal is 25cm and the length of an altitude is 15cm.

Lets EC = DF =x

In triangle ADE

$\rm AD^2 = AE^2 + ED^2$

$\rm 25^2 = 15^2 + ED^2\\\rm ED = 20$

AB = EF = 20 - x

CD = 20 + x

The area of a trapezoid = 1/2 (sum of parallel side) (height)

                                        = 1/2 (  20 - x+ 20 + x)(15)

                                        = 300

Thus, The area of a trapezoid given figure is 300 cm sq.

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