Question

In ∆ABC, P(–4, 2), Q(1, 1) and R(–2, 6) are the mid points of AB, BC and AC respectively. Then, the perimeter of ∆ABC is

Answer 1

The perimeter of the Triangle is 2√26+2√34+4√5

Given:

In ∆ABC, P(–4, 2), Q(1, 1) and R(–2, 6) are the midpoints of AB, BC, and AC respectively

To find:

The perimeter of ∆ABC is

Solution:

Given midpoints of AB, BC, and AC are P(–4, 2), Q(1, 1) and R(–2, 6)

From above points

PQ = √(1 +4)² + (1 - 2)² = √25 + 1 = √26

QR = √(-2-1)² + (6 - 1)² = √9+25 = √34

PR = √(-4+2)² + (2 - 6)² = √4 +16 = √20

From the midpoint theorem

The line segment in a triangle formed by the midpoints of any two sides of a triangle will be parallel to the third side and it will equal half of the third side.

From above observation

=> PQ = 1/2 (AC)

=> QR = 1/2 (AB)

=> PR = 1/2 (BC)

=> PQ+QR+PR = 1/2 (AC) +1/2 (AB) +1/2 (BC)

=> PQ+QR+PR = 1/2 [(AC) + (AB) + (BC) ]

=> AC + AB + BC = 2(PQ+QR+PR)

As we know the perimeter of the triangle is the sum of the sides

=> AC + AB + BC = 2(√26+√34+√20)

=> AC + AB + BC = 2(√26+√34+2√5)

=> AC + AB + BC = 2√26+2√34+4√5

The perimeter of the Triangle is 2√26+2√34+4√5

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