Find the perimeter of the triangle with vertices (i) (0, 8), (6, 0) and origin; (ii) (9, 3), (1, −3) and origin.

Formula used:

i). (0, 8), (6, 0) and (0, 0)

Let the points be A (0, 8), B (6, 0) and C (0, 0)

Distance of AB

⇒ AB = √ ((6 – 0)² + (0 – 8)²)

⇒ AB = √ ((6)² + (–8)²)

⇒ AB = √ (36 + 64)

⇒ AB = √ 100

⇒ AB = 10

Distance of BC

⇒ BC = √ ((0 – 6)² + (0 – 0)²)

⇒ BC = √ ((–6)² + (0)²)

⇒ BC = √ (36 + 0)

⇒ BC = √ 36

⇒ BC = 6

Distance of AC

⇒ AC = √ ((0 – 0)² + (0 – 8)²)

⇒ AC = √ ((0)² + (–8)²)

⇒ AC = √ (0 + 64)

⇒ AC = √ 64

⇒ AC = 8

Perimeter of ΔABC = AB + BC + AC

= 10 + 6 + 8

= 24

ii). (9, 3), (1, –3) and (0, 0)

Let the points be A (9, 3), B (1, –3) and C (0, 0)

Distance of AB

⇒ AB = √ ((1 – 9)² + (–3 – 3)²)

⇒ AB = √ ((–8)² + (–6)²)

⇒ AB = √ (64 + 36)

⇒ AB = √ 100

⇒ AB = 10

Distance of BC

⇒ BC = √ ((0 – 1)² + (0 – (–3))²)

⇒ BC = √ ((0 – 1)² + (0 + 3)²)

⇒ BC = √ ((–1)² + (3)²)

⇒ BC = √ (1 + 9)

⇒ BC = √10

Distance of AC

⇒ AC = √ ((0 – 9)² + (0 – 3)²)

⇒ AC = √ ((–9)² + (–3)²)

⇒ AC = √ (81 + 8)

⇒ AC = √ 90

⇒ AC = 3√10

Perimeter of ΔABC = AB + BC + AC

= 10 + √ 10 + 3√ 10

= 10 + 4√10