Question

6) Find the area of the quadrilateral ABCD whose vertices are:
A(4,-2), B (-3,-5), C (3,-2) and D (2,3).​

Answer 1

➫QUESTION:-

Find the area of the quadrilateral ABCD whose vertices are: A(4,-2), B (-3,-5), C (3,-2) and D (2,3).

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SOLUTION:-

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•In quadrilateral ABCD join A and C.

Then area of quadrilateral ABCD

$⠀⠀⠀⠀ = \sf \: ar( \triangle ABC)+ar(\triangle ACD)$

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➲Area of $\sf \triangle ABC$

=$\bf \dfrac{1}{2}×\{(-4)×(-5+2)-3(-2+2)+3(-2+5)\}$

=$\bf \dfrac{1}{2}×\{(-4×(-3)-3×0+3×3\}$

=$\bf \dfrac{1}{2}×\{12-0+9\}$

=$\bf \dfrac{21}{2}sq.units$

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➲Area of $\sf \triangle ACD$

=$\bf \dfrac{1}{2}×\{(-4)×(-2-3)+3(3+2)+2(-2+2)\}$

=$\bf \dfrac{1}{2}×\{(-4)×(-5)+3×5+2×0\}$

=$\bf \dfrac{1}{2}×\{20+15+0\}$

=$\bf \dfrac{35}{2}sq.unit$

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Area of quadrilateral ABCD= =$\sf \bigg(\dfrac{21}{2}+\dfrac{35}{2}\bigg)$ sq.unit=28 sq.unit

Answer 2

Step-by-step explanation:

28 Square unit...............