Question

of the figures given below: 40 m 3 m 2 m - 40 m
​

Answer 1

Area of shaded region=arΔABC−arΔDBCarΔABC=s(s−a)(s−b)(s−c)s=2a+b+c=2122+22+120=132m∴arΔABC=132(132−122)(132−22)(132−120)=132×10×110×12=11×12×10×10×11×12=10×11×12=1320m2arΔBDC=s(s−a)(s−−b)(s−c)s

Answer 2

Answer:

Area of shaded region=arΔABC−arΔDBC

arΔABC=

s(s−a)(s−b)(s−c)

s=

2

a+b+c

=

2

122+22+120

=132m

∴arΔABC=

132(132−122)(132−22)(132−120)

=

132×10×110×12

=

11×12×10×10×11×12

=10×11×12=1320m

2

arΔBDC=

s(s−a)(s−−b)(s−c)

s=

2

a+b+c

=

2

36+22+24

=36m

∴arΔBDC=

36(36−26)(36−22)(36−24)

=

36×10×14×12

=

12×3×2×5×2×7×12

=2×12

105

=24×10.24=245.76m

2

∴ area of shaded region=1320−245.76=1074.24m

2

≈1074m

2