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Find the area of quadrilateral ABCD whose vertices are A(-3, -

Question

Find the area of quadrilateral ABCD whose vertices are $A (- 3, - 1), B (- 2, - 4), C (4, - 1)$ and $D (3, 4)$ .

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Find the area of quadrilateral ABCD whose vertices are

A (- 3, - 1), B (- 2, - 4), C (4, - 1)

and

D (3, 4)

6 mins ago

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6 mins ago

Text solutionVerified

Given: ABCD is a quadrilateral whose vertices are

A (- 3, - 1), B (- 2, - 4), C (4, - 1)

and

D (3, 4)

By joining AC, we get two triangles ABC and ADC

We know that:

Area of

Δ A BC = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]

Area of triangle ABC:

= \frac{1}{2} [- 3 (- 4 + 1) + (- 2) (- 1 + 1) + 4 (- 1 + 4)]

= \frac{1}{2} [- 3 (- 5) + (- 2) \times 0 + 4 \times 3]

= \frac{1}{2} [15 - 0 + 12] = \frac{1}{2} \times 27 = \frac{27}{2}

sq. units

Area of triangle ADC.

= \frac{1}{2} [- 3 (- 4) + 4 (4 + 1) + (3) (- 1 + 1)]

= \frac{1}{2} [- 3 (- 3) + 4 \times 5 + 3 \times (0)]

= \frac{1}{2} [9 + 20 - 0] = \frac{1}{2} \times 29 = \frac{29}{2}

sq. units.

Now, Area of quadrilateral PQRS

=

Area of triangle ABC

+

Area of triangle ADC

= 27/2 + 29/2

= 28

sq. units.

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Question Text	Find the area of quadrilateral ABCD whose vertices are $A (- 3, - 1), B (- 2, - 4), C (4, - 1)$ and $D (3, 4)$ .
Answer Type	Text solution:1
Upvotes	150