Question

In the adjoining fig. Find the area of the shaded region, using Heron's formula
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Answer 1

Step-by-step explanation:

by Pythagorean theorem

AB =20cm

first find area of ABC=

$240 \sqrt{2} \: cm {}^{2}$

then fine area of ADB=96cm sq.

area of shaded region

=

$240 \sqrt{2 \: } \: - 96 = 294 \: cm {}^{2}$

please fine first

$\sqrt{2}$

i hope it help

Answer 2

Step-by-step explanation:

By Pythagoras theorem: (AD)^2+(BD)^2=(AV)^2

(12)^2+(16)^2=AB^2

144cm+256cm=AB^2

400cm^2=AB^2

AB =20cm

in ∆abc. ad^2+bc^2

400cm^2+2304cm^2

2704cm^2

ac= 52cm

AB^2+BC^2=AC^2

ABC is a right angled ∆

*Area of shaded region *= area of ∆abc- area of ∆ abd

1/2×bc×ca-1/2ab×bd

1/2×20×48- 1/2×12×16

480 cm^2-76 cm^2

= 384 cm ^2

therefore area of shaded region is 384 cm^2

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