Question

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15. Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and ( 2 , 3)

Standard:- 10

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Answer 1

Let the vertices of the Quadrilateral are:

A(-4,-2), B(-3,-5), C(3,-2) and D(2,3).

Area of triangle ABD:

Here, 

x1 = -4, x2 = -2.

x2 = -3, y2 = -5

x3 = 2, y3 = 3.

$= \ \textgreater \ \frac{1}{2} (x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))$

$= \ \textgreater \ \frac{1}{2} (-4(-5 - 3) + (-3)(3 - (-2)) + 2(-2 - (-5))$

$= \ \textgreater \ \frac{1}{2} (-4(-8) + (-3)(5) + 2(3))$

$= \ \textgreater \ \frac{1}{2}(32 - 15 + 6)$

$= \ \textgreater \ \frac{1}{2} (23)$

$= \ \textgreater \ 11.5$


Area of Triangle BCD:

Here,

x1 = -3, y1 = -5

x2 = 3, y2 = -2

x3 = 2, y3 = 3

$= \ \textgreater \ \frac{1}{2}(x1(y2 - y3) + x2(y2 - y3) + x3(y1 - y2))$

$= \ \textgreater \ \frac{1}{2}(-3(-2 - 3) + 3(3 - (-5)) + 2(-5 - (-2))$

$= \ \textgreater \ \frac{1}{2} ( -3(-5) + 3(8) + 2(-3))$

$= \ \textgreater \ \frac{1}{2} (15 + 24 - 6)$

$= \ \textgreater \ \frac{1}{2}(33)$

$= \ \textgreater \ 16.5$



Therefore:

Area of Quadrilateral ABCD = Area of Triangle ABD + Area of triangle BCD

                                                = 11.5 + 16.5

                                               = 28 square units.



Hope this helps!