Find the area of the triangle whose vertices are:

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

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

Asked by Pragya Singh | 1 year ago |  64

1 Answer

Solution :-

Area of a triangle formula = \( \dfrac{1}{2}\)× [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 = \( \dfrac{1}{2}\)[2 {0- (-4)} + (-1) {(-4) – (3)} + 2 (3 – 0)]

=\( \dfrac{1}{2}\) {8 + 7 + 6}

=\( \dfrac{21}{2}\)

So, area of triangle is \( \dfrac{21}{2}\)square units.

 

(ii) Here,

x1 = -5, x2 = 3, x3 = 5, y1 = -1, y2 = -5 and y3 = 2

Area of the triangle = \( \dfrac{1}{2}\)[-5 { (-5)- (2)} + 3(2-(-1)) + 5{-1 – (-5)}]

= \( \dfrac{1}{2}\){35 + 9 + 20} = 32

Therefore, the area of the triangle is 32 square units.

Answered by Abhisek | 1 year ago

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