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Example 2 Important
Example 3 Important
Example 4
Example 5 Important
Example 6
Example 7 Important
Example 8
Example 9 Important
Example 10 Important
Example 11 Deleted for CBSE Board 2022 Exams
Example 12 Deleted for CBSE Board 2022 Exams
Example 13 Important Deleted for CBSE Board 2022 Exams
Example 14 Important Deleted for CBSE Board 2022 Exams
Example 15 Deleted for CBSE Board 2022 Exams You are here
Last updated at Aug. 16, 2021 by Teachoo
Example 15 If A(–5, 7), B(– 4, –5), C(–1, –6) and D(4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD. Let the vertices of quadrilateral be A(−5, 7) , B(−4, −5) C(−1, −6) , D(4, 5) 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 = −5 , y1 = 7 x2 = −4 , y2 = −5 x3 = −1 , y3 = −6 Putting values Area of triangle ABC = 1/2 [ −5(−5 –(−6)) + (−4)(−6 – 7) + (−1)(7 – (−5)) ] = 1/2 [ −5(−5 + 6) – 4(−13) + (−1)(7 + 5)] = 1/2 [ −5(1) – 4(−13) + (−1)(12)] = 1/2 [−5 + 52 − 12] = 𝟏/𝟐 [35] square units Similarly, Finding area ∆ ADC Area of triangle ADC = 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ] Here x1 = −5, y1 = 7 x2 = 4, y2 = 5 x3 = −1, y3 = −6 Area of triangle ADC = 1/2 [ −5(5 –(−6)) + 4(−6 – 7) + (−1)(7 – 5) ] = 1/2 [ −5(5 + 6) + 4(−13) + (−1)(2)] = 1/2 [ −5(11) + 4(−13) + (−1)(2)] = 1/2 [ −55 − 52 – 2] = 1/2 [ −109] But area cannot be negative, ∴ Area of triangle ADC = 𝟏/𝟐 [ 109] square units Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ADC = 1/2 [ 35 + 109] = 1/2 [ 144] = 72 square units