Find the area of the triangle whose vertices are:
(i) (2, 3), (- 1, 0), (2, - 4) (ii) (- 5, - 1), (3, - 5), (5, 2)
Solution:
Let ABC be any triangle whose vertices are A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃).
The area of a triangle = 1/2 [x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)]
(i) Let A(x₁, y₁) = (2, 3), B(x₂, y₂) = (- 1 , 0), and C(x₃, y₃) = (2, - 4)
Area of a triangle is given by 1/2 [x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)]
By substituting the values of vertices, A, B, C in the formula.
Area of the given triangle = 1/2 [2(0 - (-4)) + (-1)((- 4) - 3) + 2(3 - 0)]
= 1/2 (8 + 7 + 6)
= 21/2 square units
(ii) Let A(x₁, y₁) = (- 5, - 1), B(x₂, y₂) = (3, - 5) and C(x₃, y₃) = (5, 2)
Area of a triangle = 1/2 [x₁ (y₂ - y₃) + x₂ (y₃- y₁) + x₃ (y₁ - y₂)]
By substituting the values of vertices, A, B, C in the formula.
Area of the given triangle = 1/2 [(- 5)((- 5) - 2) + 3(2 - ( -1)) + 5(- 1 - ( - 5))]
= 1/2 (35 + 9 + 20)
= 32 square units
☛ Check: NCERT Solutions for Class 10 Maths Chapter 7
Video Solution:
Find the area of the triangle whose vertices are: (i) (2, 3), (- 1, 0), (2, - 4) (ii) (- 5, - 1), (3, - 5), (5, 2)
NCERT Class 10 Maths Solutions Chapter 7 Exercise 7.3 Question 1
Summary:
The area of the triangle whose vertices are (2, 3), (- 1, 0), (2, - 4) is 21/2 square units and (- 5, - 1), (3, - 5), (5, 2) is 32 square units
☛ Related Questions:
- 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).
- 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)
visual curriculum