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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Solution

$\begin{matrix} L e t t h e v e r t i c e s o f q u a d r i l a t e r a l b e A (- 4, - 2), B (- 3, - 5) C = (3, - 2) D (2, 3) j o i n i n g A C i n t h e f i g u r e T h e r e a r e 2 t r i n a g l e s f o r m e d A B C & A C D N o w A r e a o f t r i a n g l e A B C = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [- 4 (- 5 + 2) - 3 (- 2 + 2) + 3 (- 2 + 5)] = \frac{1}{2} [12 - 0 + 9] = \frac{1}{2} [21] s q . u n i t s S i m i l a r l y A r e a o f t i r a n g l e A D C = \frac{1}{2} (- 35) = \frac{1}{2} [35] [a r e a c a n t b e n e g a t i v e] H e n c e, a r e a o f q u a d r i l a t e r a l A B C D = a r e a o f Δ A B C + a r e a o f Δ A D C = \frac{21}{2} + \frac{35}{2} = \frac{56}{2} = 28 s q . u n i t s . \end{matrix}$

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