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?
Option 1: $4\sqrt5$
Option 2: $4\sqrt4$
Option 3: $6\sqrt3$
Option 4: $4\sqrt6$
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Correct Answer: $4\sqrt5$
Solution :
Given: ABCD is a square whose side is 4 cm. P is a point on the side AD.
Let P be the midpoint of AD and draw a line from P parallel to DC which divides BC at Q into two equal parts.
Applying Pythagoras theorem in $\triangle$PQC,
CP = $\sqrt{4^2+2^2}$
⇒ CP = $2\sqrt5$ cm
Similarly, we will get,
BP = $2\sqrt5$ cm
So, the minimum value of (BP + CP) is $(2\sqrt5+2\sqrt5)=4\sqrt5$ cm
Hence, the correct answer is $4\sqrt5$.
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