Question

In the given figure, ABCD is a square whose side is 4 cm. P is a point on the side AD. What is the minimum value (in cm) of BP + CP ?

A) 4√5 B) 4√4 C) 6√3 D) 4√6

Answer 1

Answer:

Given :-

• ABCD is a square whose side is 4 cm
• P is a point on the side AD

To Find :-

• What is the minimum value (in cm) of BP + CP ?

Answer

• A) 4√5

Answer 2

Answer:

Minimum possible length of BP and Cp together is 4√5 = 8.9 cm

Step-by-step explanation:

Given: Side of Square ABCD = 4 cm

          Pont P on Side AD

To find: PB + CP

Case 1: when P is mid point of AD

Draw PX ⊥ BC

now, in Δ PBX

by Pythagoras theorem,

PB² = PX² + BX²

PB² = 4² + 2² ( PX = AB = 4 cm and BX =  = 2 cm )

PB² = 16 +4

PB² = 20

PB = √20

PB = 2√5 cm

Also, CP = 2√5 cm

⇒ BP + CP = 2√5 + 2√5 = 4√5 cm = 8.9 cm (approx.)

Case 2: when Point P is 1 cm away from point A

Draw PY ⊥ BC

now, in Δ PBY

by Pythagoras theorem,

PB² = PY² + BY²

PB² = 4² + 1² ( PY = AB = 4 cm and BY = AP = 1 cm )

PB² = 16 + 1

PB² = 17

PB = √17

PB = √17 cm

in Δ PCY

by Pythagoras theorem,

PC² = PY² + CY²

PC² = 4² + 3² ( PY = AB = 4 cm and CY = CB - BY = 4 - 1 = 3 cm )

PC² = 16 + 9

PC² = 25

PC = √25

PC = 5 cm

⇒ BP + CP = √17 + 5 = 5√17 cm = 9.1 cm (approx.)

Therefore, Minimum possible length of BP and Cp together is 4√5 = 8.9 cm