Question

Find the area of the following polygon​

Answer 1

Given Find the area of the quadrilateral ABCD in which AB=9cm , BC=40cm , CD=28cm, DA =15cm and ∆ABC = 90°

Now ABCD is a quadrilateral.

Triangle ABC = 90

In right angle triangle ABC

AC^2 = AB^2 + BC^2

= 9^2 + 40^2

= 81 + 1600

= 1681

So AC = 41 cm

Now area of triangle = 1/2 x b x h

= 1/2 x 9 x 40 = 180 sq cm

Now to find the other side of triangle using heron's formula we get

s = a + b + c / 2

s = 15 + 41 + 28 / 2

s = 42 cm

Now area of triangle ACD

= √s (s - a)(s - b)(s - c)

= √42(42 - 15)(42 - 41)(42 - 28)

= √42 x 27 x 1 x 14

= 126 sq cm

So Area of quadrilateral will be 126 + 180 = 306 sq cm