Question

calculate the area of the shaded region from the following figure

Answer 1

then area of large triangle - area of small triangle
=1248-120
=1128

Answer 2

Firstly find the side QP by using the pythagorus theorem i.e.,
$h {}^{2} = b {}^{2} + p {}^{2}$
putting the values,
${qp}^{2} = 16 {}^{2} + 12 {}^{2}$
${qp}^{2} = 144 + 256$
${qp}^{2} = 400$
$qp = \sqrt{400}$
$qp = 20$
we got the side QP =20 cm
Now we have all the sides of the triangle PQR
○Perimeter of triangle PQR is
$s = (a + b + c) \div 2$
s = (52+48+20)÷2
s = 120÷2
s = 60
○ Area of triangle PQR is
$area= \sqrt{s(s - a)(s - b)(s - c)}$
$area= \sqrt{60(60 - 52)(60 - 48)(60 - 20)}$
$area = \sqrt{60(8)(12)(40)}$
$area = \sqrt{230400}$
$area= 400{cm} {}^{2}$
area of the righ angled triangle =
$(1 \div 2) \times b \times h$
= (1÷2)×16×12
$= 96 {cm}^{2}$
Area of the shaded region = Area of triangle PQR - Area of right angled triangle
$area =( 400 - 96) {cm}^{2}$
$area = 304 {cm}^{2}$
AREA OF THE SHADED REGION IS 304 CM^2