A day full of math games & activities. Find one near you.

If D(-1/2, 5/2) E(7, 3) and F(7/2, 7/2) are the midpoints of sides of ∆ ABC, find the area of the ∆ ABC

Solution:

Given, the midpoints of sides of ∆ ABC are D(-1/2, 5/2) E(7, 3) and F(7/2, 7/2)

We have to find the area of the ∆ ABC.

Let A = (x₁, y₁) B = (x₂, y₂) and C(x₃, y₃)

if-d-1-2-5-2-e-7-3-and-f-7-2-7-2-are-the-midpoints-of-sides-of-abc-find-the-area-of-the-abc

The coordinates of the mid-point of the line segment joining the points P (x₁ , y₁) and Q (x₂ , y₂) are [(x₁ + x₂)/2, (y₁ + y₂)/2]

Midpoint of AB = D

[(x₁ + x₂)/2, (y₁ + y₂)/2] = (-1/2, 5/2)

Now, (x₁ + x₂)/2 = -1/2

x₁ + x₂ = -2/2

x₁ + x₂ = -1 ----------- (1)

Also, (y₁ + y₂)/2 = 5/2

y₁ + y₂ = 5 ------------ (2)

Midpoint of BC = E

[(x₂ + x₃)/2, (y₂ + y₃)/2] = (7, 3)

Now, (x₂ + x₃)/2 = 7

x₂ + x₃ = 14 ------------------ (3)

Also, (y₂ + y₃)/2 = 3

y₂ + y₃ = 6 -------------------- (4)

Midpoint of AC = F

[(x₁ + x₃)/2, (y₁ + y₃)/2] = (7/2, 7/2)

Now, (x₁ + x₃)/2 = 7/2

x₁ + x₃ = 7 ---------------- (5)

Also, (y₁ + y₃)/2 = 7/2

y₁+y₃ = 7 ---------------- (6)

Adding (1), (3) and (5) we get,

2(x₁ + x₂ + x₃) = -1 + 14 + 7

2(x₁ + x₂ + x₃) = 20

x₁ + x₂ + x₃ = 10 --------------- (7)

Substitute (1) in (7),

-1 + x₃ = 10

x₃ = 11

Substitute (2) in (7),

x₁ + 14 = 10

x₁ = 10 - 14

x₁ = -4

Substitute (3) in (7),

x₂ + 7 = 10

x₂ = 10 - 7

x₂ = 3

Adding (2), (4) and (6) we get,

2(y₁ + y₂ + y₃) = 5 + 6 + 7

2(y₁ + y₂ + y₃) = 18

y₁ + y₂ + y₃ = 9 ------------------ (8)

Substitute (2) in (8),

5 + y₃ = 9

y₃ = 9 - 5

y₃ = 4

Substitute (4) in (8),

y₁ + 6 = 9

y₁ = 9 - 6

y₁ = 3

Substitute (6) in (8),

y₂ + 7 = 9

y₂ = 9 - 7

y₂ = 2

Therefore, the vertices of ∆ ABC are A(-4, 3) B(3, 2) and C(11, 4).

The area of a triangle with vertices A (x₁ , y₁) , B (x₂ , y₂) and C (x₃ , y₃) is

1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]

Here, (x₁ , y₁) = (-4, 3) (x₂ , y₂) = (3, 2) and (x₃ , y₃) = (11, 4)

Area of triangle ABC = 1/2[-4(2 - 4) + 3(4 - 3) + 11(3 - 2)]

= 1/2[-4(-2) + 3(1) + 11(1)]

= 1/2[8 + 3 + 11]

= 1/2[11 + 11]

= 1/2[22]

= 11 square units.

Therefore, the area of triangle ABC is 11 square units.

✦ Try This: In ∆ABC, AD is the bisector of ∠A meeting BC at D, CF ⊥ AB and E is the mid-point of AC. Then median of the triangle is

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 7

NCERT Exemplar Class 10 Maths Exercise 7.3 Problem 16

If D(-1/2, 5/2) E(7, 3) and F(7/2, 7/2) are the midpoints of sides of ∆ ABC, find the area of the ∆ ABC

Summary:

If D(-1/2, 5/2) E(7, 3) and F(7/2, 7/2) are the midpoints of sides of ∆ ABC, the area of the ∆ ABC is 11 square units

☛ Related Questions: