# Find the area of quadrilateral ABCD whose vertices are A(-5,7), B(-4,-5), C(-1,-6) and D(4,5).What is answer of this question? I solved this question but my answer is coming different from answer given in sample Q-P  .. So I got confused ... In sample Q-P its given that area = 72 .but i am not getting this value at all. hoping for guide lines from you.

@Anandbhashyakarla: has correctly provided the solution.

For finding the area of quadrilateral ABCD, join AC. Area of two obtained triangles can be found out by using the formula .

Area of triangle ABC = square unit

Area of triangle ADC = square unit

∴ Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ADC = square unit = 72 square unit

• 14

Firstly, join BD. then, area of quad. ABCD = area of tri. ABD + area tri. BCD.

then, area of tri. ABD = 1/2 {-5(-5-5) -4(5-7) 4(-5+5)}

= 1/2{ 50+ 8+0)

= 29.....(1)

area of tri. BCD = 1/2{ -4(-6-5) -1(5+5) +4(-5+6)}

= 1/2{44-10 +4)

= 19.....(2)

= 48 sq cm....ans... :)

• 2

firstly join AC.

ar of ABC

=1/2(-5(-5+6)-4(-6-7)-1(7+5)

=1/2(-5(1)-4(-13)-1(12)

=1/2(-5+52-12)

=1/2(35)

=32/2

=1/2(-5(5+6)+4(-6-7)-1(7-5)

=1/2-5(11)+4(-13)-1(2)

=1/2(-55-52-2)

=1/2(-109)

=-109/2

adding these 2 areas we get -109/2+35/2

=74/2

=37sq cm.............

• 6

hey sorry in the area of ADC =-109/2,  minus is negated so -109/2 will be +109/2.

so 109/2

109/2+35/2=144/2=72