1. Find the area of the triangle whose vertices are:

(i) (2, 3), (-1, 0), (2, -4)

(ii) (-5, -1), (3, -5), (5, 2)

Question

1. Find the area of the triangle whose vertices are: (i) (2, 3), (-1, 0), (2, -4) (ii) (-5, -1), (3, -5), (5, 2)

Pawan Prajapati · Answer

Area of a triangle formula = 1/2 × [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] (i) Here, x1 = 2, x2 = -1, x3 = 2, y1 = 3, y2 = 0 and y3 = -4 Substitute all the values in the above formula, we get Area of triangle = 1/2 [2 {0- (-4)} + (-1) {(-4) – (3)} + 2 (3 – 0)] = 1/2 {8 + 7 + 6} = 21/2 So, area of triangle is 21/2 square units. (ii) Here, x1 = -5, x2 = 3, x3 = 5, y1 = -1, y2 = -5 and y3 = 2 Area of the triangle = 1/2 [-5 { (-5)- (2)} + 3(2-(-1)) + 5{-1 – (-5)}] = 1/2{35 + 9 + 20} = 32 Therefore, the area of the triangle is 32 square units.

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1. Find the area of the triangle whose vertices are: (i) (2, 3), (-1, 0), (2, -4) (ii) (-5, -1), (3, -5), (5, 2)

1. Find the area of the triangle whose vertices are:

(i) (2, 3), (-1, 0), (2, -4)

(ii) (-5, -1), (3, -5), (5, 2)

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