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Find the area of quadrilateral ABCD whose vertices are A(3, -1

Question

Find the area of quadrilateral ABCD whose vertices are $A (3, - 1), B (9, - 5), C (14, 0)$ and $D (9, 19)$ .

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Find the area of quadrilateral ABCD whose vertices are

A (3, - 1), B (9, - 5), C (14, 0)

and

D (9, 19)

9 mins ago

Discuss this question LIVE

9 mins ago

Text solutionVerified

Vertices of quadrilateral ABCD are

A (3, - 1), B (9, - 5), C (14, 0)

and

D (9, 19)

Construction: Join diagonal AC.

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 (- 5 - 0) + 9 (0 + 1) + 14 (- 1 + 5)]

= \frac{1}{2} [- 15 + 9 + 14 (4)]

= \frac{1}{2} [- 15 + 9 + 56]

= \frac{1}{2} \times 50

= 25

sq. units.

Area of triangle ADC:

= \frac{1}{2} [3 (0 - 19) + 14 (19 + 1) + 9 (- 1 + 0)]

= \frac{1}{2} [3 (- 19) + 14 \times 20 + 9 \times (- 1)]

= \frac{1}{2} [- 57 + 280 - 9]

= \frac{1}{2} \times 214

= 107

sq. units.

Now, Area of quadrilateral ABCD

=

Area of triangle ABC

+

ARea of triangle ADC

= 25 + 107

132

sq.units.

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Question Text	Find the area of quadrilateral ABCD whose vertices are $A (3, - 1), B (9, - 5), C (14, 0)$ and $D (9, 19)$ .
Answer Type	Text solution:1
Upvotes	150