# Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, - 1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle

**Solution:**

Let ABC be any triangle whose vertices are A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}).

Let A(x_{1}, y_{1}) = (0, - 1), B(x_{2}, y_{2}) = (2 , 1) and C(x_{3}, y_{3}) = (0, 3)

Area of a triangle = 1/2 (x_{1} (y_{2} - y_{3}) + x_{2} (y_{3}- y_{1}) + x_{3} (y_{1} - y_{2})) ...Equation(1)

By substituting the values of vertices, A, B, C in (1).

Let P, Q, R be the mid-points of the sides of this triangle.

Coordinates of P, Q, and R are given by

P = [(0 + 0)/2, (3 - 1)/2] = (0, 1)

Q = [(0 + 2)/2, (- 1 + 1)/2] = (1, 0)

R = [(2 + 0)/2, (1 + 3)/2] = (1, 2)

By substituting the values of Points P, Q, R in the formula, we get

Area of ΔPQR = 1 ((2 - 1) +1(1 - 0) + 0(0 - 2))

= 1/2 (1 + 1) square units

= 1 square units

By substituting the values of points A, B, C in the formula, we get

Area if ΔABC = 1/2 [0(1 - 3) + 2{3 - (- 1)} + 0(- 1 - 1)]

= 1/2 × 8 square units

= 4 square units

Therefore, the ratio of the area of ∆PQR to the area of the triangle ΔABC is 1 : 4

**Video Solution:**

## Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, - 1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle

### NCERT Class 10 Maths Solutions - Chapter 7 Exercise 7.3 Question 3

Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, - 1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle

The area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, - 1), (2, 1) and (0, 3). Then the ratio of this area to the area of the given triangle is 1 : 4