se 170 d region in each of the given figures. 10 m (i (ii) 3 m 6m-> - 12 m 14 m 4 m - 18 m
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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
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