You have studied in Class IX (Chapter 9, Example 3), 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)
Solution:
Let ABC be any triangle whose vertices are A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃).
Area of a triangle = 1/2 [x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)]
Let the vertices of the triangle be A (4, - 6), B (3, - 2), and C (5, 2).
Let M be the mid-point of side BC of ∆ABC.
Therefore, AM is the median in ∆ABC.
According to the mid point formula,
O(x, y) = [(x₁ + x₂) / 2, (y₁ + y₂) / 2]
Since M is the mid point of the points joining C(5, 2) and B(3, -2),
Coordinates of point M = [(3 + 5) / 2, (- 2 + 2) / 2] = (4, 0)
Area of a triangle = 1/2 [x₁ (y₂ - y₃) + x₂ (y₃- y₁) + x₃-(y₁ - y₂)]
Considering ΔABM with A (4, - 6), B (3, - 2) and M(4, 0), by substituting the values of vertices, A, B, M in the formula.
Area of ΔABM = 1/2 [(4){(- 2) - (0)} + (3){(0) - (- 6)} + (4){(- 6) - (- 2)}] square units
= 1/2 (- 8 + 18 - 16) square units
= -3 square units
Considering ΔAMC with A (4, - 6), M(4, 0) and C (5, 2), by substituting the values of vertices, A, M, C in the formula.
Area of ΔAMC = 1/2 [(4){(0) - (2)} + (4){2 - (- 6)} + (5){(- 6) - (0)}] square units
= 1/2 (- 8 + 32 - 30) square units
= - 3 square units
However, the area cannot be negative. Therefore, the area of ∆AMC and ∆ABM is 3 square units.
Hence, clearly, median AM has divided ΔABC in two triangles of equal areas.
☛ Check: NCERT Solutions for Class 10 Maths Chapter 7
Video Solution:
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).
NCERT Class 10 Maths Solutions Chapter 7 Exercise 7.3 Question 5
Summary:
You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Therefore for ∆ABC whose vertices are A (4, - 6), B (3, - 2) and C (5, 2) we see that, median AM has divided ΔABC in two triangles of equal areas.
☛ 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 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.
- Find the area of the quadrilateral whose vertices, taken in order, are (- 4, - 2), (- 3, - 5), (3, - 2) and (2, 3).