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:
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