Question 4 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
Let the vertices of the quadrilateral be A (- 4, - 2), B ( - 3, - 5), C (3, - 2), and D (2, 3). Join AC to form two trianglesΔABCandΔACD. Area of a triangle =12{x1(y2−y3)+x2(y3−y1)+x3(y1−y2)} Area of ΔABC =12[(−4){(−5)−(−2)}+(−3){(−2)−(−2)} +3{(−2)−(−5)}] =12(12+0+9) =212squareunits Area of ΔACD =12[(−4){(−2)−(3)}+3{(3)−(−2)} +2{(−2)−(−2)}] =12(20+15+0) =352squareunits Area of ΔABCD =AreaofΔABC+AreaofΔACD =(212+352) square units = 28 square units