Question

look at the measures shown in the adjacent figure and find the area of Quadrilateral PQRS

Answer 1

Area of quadrilateral PQRS :-

= area of ∆PSR + area of ∆PRQ

Area of ∆PRS :

angle S = 90° ,

${{PR}^{2}={PS}^{2}{SR}^{2}}$

$= \sqrt{ {36}^{2} + {15}^{2} } \\ \\ = \sqrt{1521} \\ \\ = 39\: cm$

area of ∆PRS :)

.

$= \frac{1}{2} \times base \times height \\ \\ = \frac{1}{2} \times 36 \times 15 \\ \\ = 270 {cm}^{2}$

Area of ∆PQR :

For ∆PQR,

$s = \frac{56 + 25 + 39}{2} = 60 \: cm$

Area of ∆PQR :)

$= \sqrt{s(s - a)(s - b)(s - c)} \\ \\ = \sqrt{60 \times 4 \times 35 \times 21} \\ \\ = \sqrt{176400} \\ \\ = 420 { \: cm}^{2}$

Area of quadrilateral PQRS :-

$= 420 + 270 \\ \\ = 690 { \: cm}^{2}$