Question

The sides of a quadrangular field, taken in order are 26 m, 27 m, 7 m are 24 m respectively. The angle contained by the last two sides is a right angle. Find its area.

Answer 1

We assume quadrilateral ABCD be the quadrangular field having sides AB, BC, CD, DA and.

We take a diagonal AC, where AC divides quadrilateral ABCD into two triangles ΔABC and ΔADC

We will find the area of two triangles ΔABC and ΔADC separately and add them to find the area of the quadrangular ABCD.

In triangle ΔADC, we have

AD = 24 m; DC = 7 m

We use Pythagoras theorem to find side AC,

AC² = AD² + DC²

Area of right angled triangle ΔADC, say A₁ is given by

Where, Base = DA = 24 m; Height = DC = 7 m

Area of triangle ΔABC, say A₂ having sides a, b, c and s as semi-perimeter is given by

Where, a = AC = 25 m; b = AB = 26 m; c = BC = 27 m

Area of quadrilateral ABCD, say A

A = Area of triangle ΔADC + Area of triangle ΔABC