Example 11 - Chapter 7 Class 10 Coordinate Geometry (Term 1)
Last updated at Aug. 16, 2021 by Teachoo
Last updated at Aug. 16, 2021 by Teachoo
Transcript
Example 11 Find the area of a triangle whose vertices are (1, –1), (– 4, 6) and (–3, –5). Let the two points be A (1, −1) , B(−4, 6) & C(−3, −5) Area of triangle ABC = 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ] Here x1 = 1 , y1 = −1 x2 = −4 , y2 = 6 x3 = −3 , y3 = −5 Putting values Area of triangle ABC = 1/2 [ 1(6 – (−5)) + (−4)(−5 – (−1) ) + (−3)(−1 − 6) ] Area of triangle ABC = 1/2 [1(6 – (−5)) + (−4)(−5 – (−1) ) + (−3)(−1 − 6)] = 1/2 [1(6 + 5) + (−4)(−5 + 1 ) + (−3)(−7)] = 1/2 [ 1(11) + (−4)(−4 ) + (−3)(−7) ] = 1/2 [ 11 + 16 + 21 ] = 1/2 [ 48 ] = 24 square units
Area of triangle
Example 12 Deleted for CBSE Board 2022 Exams
Example 13 Important Deleted for CBSE Board 2022 Exams
Ex 7.3, 1 (i) Deleted for CBSE Board 2022 Exams
Ex 7.3, 3 Important Deleted for CBSE Board 2022 Exams
Ex 7.3, 5 Deleted for CBSE Board 2022 Exams
Area of triangle
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