The area of a triangle with vertices (a, b + c), (b, c + a) and (c, a + b) is
a. (a + b + c)²
b. 0
c. a + b + c
d. abc

Solution:

Consider the vertices of a triangle as

A = (x₁, y₁) = (a, b + c)

B = (x₂, y₂) = (b, c + a)

C = (x₃, y₃) = (c, a + b)

We know that

Area of triangle ABC = Δ = 1/2 [x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]

Let us substitute the values

Δ = 1/2 [a (c + a - a - b) + b (a + b - b - c) + c (b + c - c - a)]

By further simplification

Δ = 1/2 [a (c - b) + b (a - c) + c (b - a)]

Δ = 1/2 [ac - ab + ab - bc + bc - ac]

So we get

Δ = 1/2 [0]

Δ = 0

Therefore, the area of the triangle is 0.

✦ Try This: The area of a triangle with vertices (p, q + r), (q, r + p) and (r, p + q) is

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 7

NCERT Exemplar Class 10 Maths Exercise 7.1 Problem 18

The area of a triangle with vertices (a, b + c), (b, c + a) and (c, a + b) is a. (a + b + c)², b. 0, c. a + b + c, d. abc

Summary:

The area of a triangle with vertices (a, b + c), (b, c + a) and (c, a + b) is 0

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