Question

solve it and send the image

Answer 1

Let we consider for ∆ABC
draw a perpendicular bisector AO on BC
whose sides are 6 cm

Take ∆ AOC & ∆ AOB
apply the rule of congruency of RHS
so OB =OC

Now in ∆ AOC

AO²+OC²=AC²
AO²=AC²-OC²=6²-3²=>36-9
AO²=>27
AO=>5.19 cm

area of ∆ ABC=1/2×6×5.19
=3×5.19
=15.57

area of ∆ABC= area of imaginary ∆EDF
15.57=9y
y=15.57÷9
y=1.73=rarely 2

By herons formula in image

Answer 2

Hello dear your answer is in up
and yes I think that you will not understand my S










Thanks