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Ex 7.3 , 4 Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3). Let the vertices of quadrilateral be A (−4, −2), B (−3, −5) C (3, −2), D (2, 3) Joining AC, There are 2 triangles formed ABC & ACD Hence, Area of quadrilateral ABCD = Area of ∆ ABC + Area of ∆ ADC Finding area ∆ ABC Area of triangle ABC = 1/2 [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] Here x1 = −4 , y1 = −2 x2 = −3 , y2 = −5 x3 = 3 , y3 = −2 Putting values Area of triangle ABC = 1/2 [ −4(−5 –(−2)) + (−3)(−2 – (−2)) + 3(−2 – (−5))] = 1/2 [ −4(−5 + 2) – 3(−2 + 2) + 3(−2 + 5)] = 1/2 [ −4(−3) – 3(0) + 3(3)] = 1/2 [12 – 0 + 9] = 1/2 [21] square units Finding area ∆ ADC Area of triangle ADC = 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ] Here x1 = −4 , y1 = −2 x2 = 2 , y2 = 3 x3 = 3 , y3 = −2 Putting values Area of triangle ADC = 1/2 [ −4(3 –(−2)) + 2(−2 – (−2)) + 3(−2 – 3) ] = 1/2 [ −4(3 + 2) – 3(−2 + 2) + 3(−2 − 3)] = 1/2 [ −4(5) – 3(0) + 3(−5)] = 1/2 [ −20 – 0 − 15] = 1/2 [ −35] But area cannot be negative, So, Area of triangle ADC = 1/2 [ 35] square units Hence, Area of quadrilateral ABCD = Area of ∆ ABC + Area of ∆ ADC = 21/2 + 35/2 = 56/2 = 28 square units

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.