Question

ABCD is a rectangle. AP = 6 cm, PB = 5 cm, AQ = 8 cm, QD = 4 cm, DR = 3 cm. Find PQ, QR and PC.

Answer 1

AP = 6 cm, PB = 5 cm

AB = AP + PB

= 6 + 5

= 11 cm

AQ = 8 cm, QD = 4 cm

AD = AQ + QD

= 8 + 4

= 12 cm

DR = 3 cm

CR = CD − DR = ?

Since ABCD is a rectangle:

AB = CD = 11 cm, AD = BC = 12 cm

Thus,

CR = CD − DR

= 11 − 3

= 8 cm

1. Find PQ:

Coordinates help:

Let, A = (0, 12), B = (11, 12), D = (0, 0), C = (11, 0).

Then P is 6 cm from A along AB.

⇒ P = (6,12)

Q is 8 cm from A down AD.

⇒ Q = (0, 4)

Distance PQ:

PQ² = (6 − 0)² + (12 − 4)²

= 6² + 8²

= 36 + 64

PQ² = 100

PQ = 10 cm

2. Find QR:

Coordinates: Q = (0, 4)

R is 3 cm from D on DC.

⇒ R = (3, 0)

QR² = (3 − 0)² + (0 − 4)²

= 3² + 4²

= 9 + 16

QR² = 25

QR = 5 cm

3. Find PC:

Coordinates: P = (6, 12), C = (11, 0)

PC² = (11 − 6)² + (12 − 0)²

= 5² + 12²

= 25 + 144

PC² = 169

PC = 13 cm