Find the area of the quadrilateral whose vertices, taken in order, are (- 4, - 2), (- 3, - 5), (3, - 2) and (2, 3)
Solution:
Let ABC be any triangle whose vertices are A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃).
Then, Area of a triangle is given by 1/2 [x₁ (y₂ - y₃) + x₂ (y₃- y₁) + x₃ (y₁ - y₂)] ------ (1)
Let the vertices of the quadrilateral be A (- 4, - 2), B (- 3, - 5), C (3, - 2), and D (2, 3)
Join AC to form two triangles ∆ABC and ∆ACD.
We know that, 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 ΔABC = 1/2 [(- 4){(- 5) - (- 2)} + (- 3){(- 2) - (- 2)} + 3{(- 2) - (- 5)}]
= 1/2 (12 + 0 + 9)
= 21/2 square units
By substituting the values of vertices, A, C, D in the Equation (1),
Area of ΔACD = 1/2 [(- 4){(- 2) - 3} + 3{(3) - (- 2)} + 2{(- 2) - (- 2)}]
= 1/2 (20 + 15 + 0)
= 35/2 square units
Area of ABCD = Area of ΔABC + Area of ΔACD
= (21/2 + 35/2) square units
= 28 square units
☛ Check: NCERT Solutions for Class 10 Maths Chapter 7
Video Solution:
Find the area of the quadrilateral whose vertices, taken in order, are (- 4, - 2), (- 3, - 5), (3, - 2) and (2, 3)
NCERT Class 10 Maths Solutions - Chapter 7 Exercise 7.3 Question 4
Summary:
The area of the quadrilateral whose vertices, taken in order, are (- 4, - 2), (- 3, - 5), (3, - 2) and (2, 3) is 28 square units.
☛ Related Questions:
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