Question

In the Fig. 12.30, ABCD is a rectangle of length 60 m and breadth 25 m. PQ = 6 m is perpendicular to BC and MN=10 m is perpendicular to AD. Find the area of the shaded region.​

Answer 1

$\color{purple} \bf \huge Answer$

Area of shaded region = Area of rectangle - Area of △ADM & △PCB

Area of rectangle = l × b

= 60m × 25m

$\bf \: = {1500m}^{2} \\ \\ \bf Area \: \: of \: \triangle ADM = \: \frac{1}{2} \times base \: \times height$

Base = 25m

Height = 10m

$\bf \: = \frac{1}{2} \times 25m \: \times \: 10m$

$\bf \: = \frac{1}{ \cancel2} \times 25 \times \cancel{10m}$

$\bf \: = 1 \: \times \: 25m \: \times \: 5m$

$\bf \: = {125m}^{2}$

$\bf Area \: \: of \: \triangle \: PCB = \: \frac{1}{2} \times base \: \times height$

Base = 25m

Height = 6m

$\bf \: = \frac{1}{2} \times 25m \: \times \: 6m$

$\bf \: = \frac{1}{ \cancel2} \times 25 \times \cancel{6m}$

$\bf \: = 1 \: \times \: 25m \: \times \: 3m$

$\bf \: = {75m}^{2}$

Area of shaded region = Area of rectangle - Area of △ADM & △PCB

$\bf \: = {1500m}^{2} - ( {125m}^{2} + {75m}^{2})$

$\bf \: = {1500m}^{2} - {200m}^{2}$

$\bf \: = {1300m}^{2}$

Hence, area of shaded region is $\bf {1300m}^{2}$