Question

In the adjoining figure, ∠PQR = 90°, PR = 10 cm, QR = 6 cm and SR = 9 cm. Find PS.

Answer 1

Given:

From the figure:

∠PQR = 90°

PR = 10 cm

QR = 6 cm

SR = 9 cm

Step-wise calculation:

1. Place Q at (0, 0). 

Since QR is horizontal with a length of 6 to the left. 

R = (–6, 0)

SR = 9 

So, S = (–6 – 9, 0)

= (–15, 0)

2. Let P = (0, y) 

P is vertically above Q because ∠PQR = 90°.

Use PR = 10:

PR² = (0 – (–6))² + (y – 0)² 

= 6² + y²

= 36 + y²

= 10²

= 100

⇒ y² = 64

⇒ y = 8

So P = (0, 8).

3. Compute PS: 

PS² = (0 – (–15))² + (8 – 0)²

= 15² + 8²

= 225 + 64

= 289

⇒ `PS = sqrt(289)`

⇒ PS = 17 cm