Question

*The length of the shorter altitude and the shorter side of a parallelogram are 9cm and 82 cm. The length of a longer diagonal is 15 cm. What is the area of this parallelogram?

Answer 1

Refer to the image attached.

Given: Altitude AC = 9 cm, Diagonal AD = 15 cm, side AB = 82 cm.

To find: Area of parallelogram

Solution:

Since, area of parallelogram = base $\times$ height

= $BD \times AC$

We have to determine the base BD.

Consider the triangle ABC,

by Pythagoras theorem,

$(AB)^2 = (BC)^2 + (AC)^2$

$(82)^2 = (BC)^2 + (9)^2$

$6724-81= (BC)^2$

BC = 81.5 cm

Now, Consider the triangle ACD,

by Pythagoras theorem,

$(AD)^2 = (AC)^2 + (CD)^2$

$(15)^2 = (9)^2 + (CD)^2$

$225-81= (CD)^2$

CD = 12 cm

Now, base BD = BC + CD

= 81.5+12

= 93.5 cm

Area of parallelogram = BD $\times AC$

= 93.5 x 9

= 841.5 square centimeters.

Therefore, the area of parallelogram is 841.5 square centimeters.