Question

If A(–5, 7), B(–4, –5), C(–1, –6) and D(4, 5) are the vertices of a quadrilateral, find the

area of the quadrilateral ABCD.

Answer 1

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Answer 2

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By joining B to D, we will get two triangles ABD and BCD

Now,

$\tt the \: area \: of \: \triangle \: BCD = \dfrac{1}{2} [-4(-6-5)-1(5+5)+4(-5+6)]\\ \tt = \dfrac{1}{2} (44 - 10 + 4) = 19 \: sq \: units$

Also,

$\tt the \: area \: of \: \triangle \: ABD = \dfrac{1}{2}[ - 5( - 5 - 5) + ( - 4)(5 - 7) + 4(7 + 5)] \\ \\ \tt= \frac{1}{2} (50 + 8 + 48) \\ \\ \tt = \dfrac{106}{2} = 53 \: sq. \: units$

So, the area of quadrilateral ABCD = 53 + 19 = 72 sq. units.

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Note : To find the area of a polygon, we divide it into triangular regions, which have no common area, and add the areas of these regions.

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