Question

solve the question in attachment.

Answer 1

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Answer 2

Given vertices are A(-5,7), B(-4,-5), C(-1,-6), D(4,5).

Area of Quadrilateral ABCD = Area of Δ ABD + ΔBCD.  

Finding Area of ΔABD:

We know that Area of ΔABC = (1/2)[x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)]

Here,

(i) (x1,y1) = (-5,7)

(ii) (x2,y2) = (-4,-5)

(iii) (x3,y3) = (4,5).

Now,

Area of ΔABC = (1/2)[(-5)(-5 - 5) + (-4)(5 - 7) + 4(7 + 5)]

= > (1/2)[50 + 8 + 48]

= > 106/2

= > 53.  

Finding Area of ΔBCD:

We know that Area of triangle ΔBCD = (1/2)[x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)

Here,

= > (x1,y1) = (-4,-5)

= > (x2,y2) = (-1,-6)

= > (x3,y3) = (4,5).

Now,

Area of ΔBCD = (1/2)[(-4)(-6 - 5) + (-1)(5 + 5) + (4)(-5 + 6)]

= > (1/2)[44 - 10 + 4]

= > (1/2)[38]

= > 38/2

= > 19.  

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Therefore,

= > Area of Quadrilateral ABCD = Area of ΔABD + Area of ΔBCD

= > 53 + 19

= > 72 sq.units.

Hope this helps!