Question

bro plz fast important by using heron's formula

Answer 1

The area of the figure=Area of ∆ABC+ Area of ∆ACD

Area of∆ABC=√s(s-a)(s-b)(s-c). [where s= semiperimeter of ∆ABC]

Perimeter of∆ABC=AB+BC+CA
= (360+200+240)m
=800m
S=400m

Now,

By heron's formula

Ar.∆ABC=√s(s-a)(s-b)(s-c).
√400(400-360)(400-200)(400-240)
√400*40*200*160
√512,000,000
22627 m ^2(appx.)

Perimeter of∆ACD=AD+AC+CD
=400+320+240
=960m

s=480m

Area of∆ACD=√s(s-a)(s-b)(s-c). [where s= semiperimeter of ∆ACD]
=√480(480-400)(480-320)(480-240)
√480*80*160*240
√1,47,45,60,00
38400m^2

Area of the figure=22627+38400
=61027 m^2