Question

plzZzzxxxxxxxxzzzzzzxzzzzzxgive me answer​

Answer 1

Given

$= >$

In a given figure,

DH=12cm

GK=20cm

A(Traingle DEF) =300cm^2

To find

(1) EF

(2)A(triangleGEF)

(3)A(⬛DEFG)

Explanation

A(triangle DEF) = 1/2x b X h

= 1/2 X b X (DH)

( 300) = 1/2 X b X (12)

[ 300 X 2]/12 = b = 50

[↪EF=b=50cm•]

Now,

A(triangleGEF)= 1/2x b X h

= 1/2 X (EF)x(GK)

=1/2 X (50)x (20)

= 500

[ A(triangleGEF)=500cm^2•]

Now,

A(⬛ DEFG) =

A(triangleDEF)+A(triangleGEF)

= 500+300

= 800cm^2

[ A(⬛ DEFG) =800cm^2]