Question

pls Ans asap
need it for exams

in figure d is a point on bc such that​

Answer 1

ab2=ac2+bc2

25=9+bc

bc2=25-9

bc=

$\sqrt{16}$

=4

Answer 2

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by pythagoras theorem in ΔACD

AD^2=AC^2+DC^2

4^2=3^2+DC^2

16=9+DC^2

√7=DC

IN ABCAB^2=BC^2+AC^2

5^2=3^2+BC^2

25=9+BC^2

4=BC

AREA(ACD):AREA(BCA)

AC^2:BC^2

3^2:4^2

9:16



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