he sides of a quadrangular field, taken in order are 26 m, 27 m, 7 m, are 24 m respectively.

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

Let consider a quadrilateral ABCD

In ∆ADC;

AC = \(\sqrt{(24)^2+7^2}\) = 25 cm

In ∆ABC

AB = a = 26 cm, BC = b = 27 cm, AC = c = 25 cm

Let a, b and c are the sides of triangle and s is

the semi-perimeter, then its area is given by:

A = \(\sqrt{s(s-a)(s-b)(s-c)}\)where s = \(\frac{a+b+c}2\)

s = \(\frac{a+b+c}2\) = \(\frac{26+27+25}2\) = 39

A₁ = \(\sqrt{39(39-26)(39-27)(39-25)}\)

A₁ = \(\sqrt{39\times13\times12\times14}\) = 291.85 cm²

In ∆ADC;

DA = a = 24 cm, CD = b = 7cm, AC = c = 25 cm

Let a, b and c are the sides of triangle and s is

the semi-perimeter, then its area is given by:

A = \(\sqrt{s(s-a)(s-b)(s-c)}\)where s = \(\frac{a+b+c}2\)[Heron’s Formula]

s = \(\frac{a+b+c}2\) = \(\frac{24+7+25}2\) = 28

A₂ = \(\sqrt{28(28-24)(28-7)(28-25)}\)

A₂ = \(\sqrt{28\times4\times21\times3}\) = 84 cm²

Therefore area of quadrilateral ABCD = A₁ + A₂ = 291.85+84 = 375.85 cm²