Question

whoever does this will receive brainliest​

Answer 1

Answer:

Solution : 

Area of quad. ABCD = ar(ΔABC)+ar(ΔACD).ar(ΔABC)+ar(ΔACD).



For ΔABCΔABC, we have

(x1=−4,y1=8),(x2=−3,y2=−4)and(x3=0,y3=−5)(x1=-4,y1=8),(x2=-3,y2=-4)and(x3=0,y3=-5)

∴ar(ΔABC)=12|x1(y2−y3)+x2(y3−y1)+x3(y1−y2)|∴ar(ΔABC)=12|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|

=12|(−4)⋅(−4+5)+(−3)⋅(−5−8)+0⋅(8+4)|=12|(-4)⋅(-4+5)+(-3)⋅(-5-8)+0⋅(8+4)|

=12|(−4)×1+(−3)×(−13)+0|=12|(-4)×1+(-3)×(-13)+0|

=12|−4+39+0|=352=12|-4+39+0|=352 sq units.

For ΔACDΔACD, we have