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.

 

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)

adding (1) and(2), we get,

area quad. ABCD = (29+19)

  = 48 sq cm....ans... :)

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

now arADC

=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.............

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

so theres the answer.

cheeers...........

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