Prove that the area of triangle whose vertices are (5 ,2),(-9,-3),&(-3,-5) is four times the area of the triangle formed by joining the mid points of the sides
Answers
Formula:
If A (x₁, y₁), B (x₂, y₂), C (x₃, y₃) be the vertices of a triangle, then the area of triangle ABC is
= 1/2 * {x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)} square units
For the same triangle, the mid-points of the sides are given by
( (x₁ + x₂)/2 , (y₁ + y₂)/2 )
( (x₂ + x₃)/2 , (y₂ + y₃)/2 )
( (x₃ + x₁)/2 , (y₃ + y₁)/2 )
Solution:
Step1.
The coordinates of the vertices a triangle is given (5, 2), (- 9, -3) and (- 3, - 5)
Thus, its area is A₁
= 1/2 * [5 (- 3 + 5) - 9 (- 5 - 2) - 3 (2 + 3)]
= 1/2 * [5 (2) - 9 (- 7) - 3 (5)]
= 1/2 * (10 + 63 - 15)
= 1/2 * 58
= 29 square units
Step 2.
The coordinates of the mid-points of the triangle are (- 2, -1/2), (- 6, - 4) and (1, - 3/2)
If a triangle is formed with these mid-points, then the area of the triangle be A₂
= 1/2 * [- 2 (- 4 + 3/2) - 6 (- 3/2 + 1/2) + 1 (- 1/2 + 4)]
= 1/2 * [(- 2) (- 5/2) - 6 (- 1) + (1) (7/2)]
= 1/2 * (5 + 6 + 7/2)
= 1/2 * 29/2
= 29/4 square units
∴ A₁ = 4 * A₂
Hence, proved.
Answer:
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