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Find the area of the quadrilateral whose vertices taken in ord

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

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Find the area of the quadrilateral whose vertices taken in order, are

(- 4, - 2), (- 3, - 5), (3, - 2)

and

(2, 3)

8 mins ago

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Text solutionVerified

Let $$A(– 4, –2)$$, $$B (– 3, –5)$$, $$C (3, –2)$$ and $$D (2, 3)$$ be the vertices of the quadrilateral $$ABCD$$.

Area of a quadrilateral $$ABCD =$$ Area of $$\triangle ABC $$+ Area of $$\triangle ACD$$

By using a formula for the area of a triangle $$\displaystyle = \frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_2)+x_3(y_1-y_2)| $$

Area of $$\triangle $$ABC

=$$\dfrac{1}{2}\left[-4(-5+2)+-3(-2+2)+3(-2+5)\right]$$

$$= \dfrac{1}{2}\left[12+9\right]$$

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

Area of $$\triangle $$ACD=$$\dfrac{1}{2}\left[-4(3+2)+-2(-2+2)+3(-2-3)\right]$$

$$= \dfrac{1}{2}\left[-20-15\right]$$

$$= \dfrac{35}{2}$$sq.units

$$\therefore $$Area of quadilateral=$$\dfrac{21}{2}+\dfrac{35}{2}=\dfrac{56}{2} =28$$ sq.unit

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Question Text	Find the area of the quadrilateral whose vertices taken in order, are $(- 4, - 2), (- 3, - 5), (3, - 2)$ and $(2, 3)$ .
Answer Type	Text solution:1
Upvotes	150