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# 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(x₁, y₁), B(x₂, y₂) and C(x₃, y₃).

Let A(x₁, y₁) = (0, - 1), B(x₂, y₂) = (2, 1) and C(x₃, y₃) = (0, 3)

Area of a triangle = 1/2 [x₁ (y₂ - y₃) + x₂ (y₃- y₁) + x₃ (y₁ - y₂)] ...Equation(1)

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

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

According to the mid point formula,

O(x, y) = [(x₁ + x₂) / 2, (y₁ + y₂) / 2]

Coordinates of P, Q, and R are given as follows:

Considering P to be the mid point of AC with A(0, -1) and C(0, 3) we get,

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

Similarly, Q is the mid point of AB with A(0, -1) and B(2,1) we get,

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

Similarly, Q is the mid point of CB with C(0, 3) and B(2,1) we get,

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

Now, let P(x₁, y₁) = (0, 1), Q(x₂, y₂) = (1, 0) and R(x₃, y₃) = (1, 2)

By substituting the values of Points P, Q, R in the formula 1/2 [x₁ (y₂ - y₃) + x₂ (y₃- y₁) + x₃ (y₁ - y₂)] ,we get

Area of ΔPQR = 1/2 [0(0 - 2) +1(2 - 1) + 1(1 - 0)]

= 1/2 (1 + 1) square units

= 1 square unit

Now, let A(x₁, y₁) = (0, - 1), B(x₂, y₂) = (2, 1) and C(x₃, y₃) = (0, 3)

By substituting the values of points A, B, C in the formula 1/2 [x₁ (y₂ - y₃) + x₂ (y₃- y₁) + x₃ (y₁ - y₂)] , 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.

**☛ Check: **NCERT Solutions Class 10 Maths Chapter 7

**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

**Summary:**

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) is 1 square unit. Then the ratio of this area to the area of the given triangle is 1 : 4.

**☛ Related Questions:**

- Find the area of the triangle whose vertices are:(i) (2, 3), (- 1, 0), (2, - 4)(ii) (- 5, - 1), (3, - 5), (5, 2)
- In each of the following find the value of ‘k’, for which the points are collinear.(i) (7, - 2), (5, 1), (3, k)(ii) (8, 1), (k, - 4), (2, - 5)
- Find the area of the quadrilateral whose vertices, taken in order, are (- 4, - 2), (- 3, - 5), (3, - 2) and (2, 3).
- You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ∆ABC whose vertices are A (4, - 6), B (3, - 2) and C (5, 2)

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