If ABCD is a rectangle and P is a point inside it such that AP = 33, BP = 16, DP = 63. Find CP.
Answers
Given : ABCD is a rectangle and P is a point inside it such that AP = 33, BP = 16, DP = 63.
To Find : CP
Solution:
Draw 4 perpendiculars from point P on each side of rectangle
PE ⊥ AB = X
PF ⊥ BC = Y
PH ⊥ CD = Z
PG ⊥ AD = W
AP² = X² + W² = 33²
BP² = X² + Y² = 16²
CP² = Y² + Z²
DP² = W² + Z² = 63²
BP² + DP² - AP² = X² + Y² + W² + Z² - ( X² + W² )
=> BP² + DP² - AP² = Y² + Z²
=> BP² + DP² - AP² = CP²
=> 16² + 63² - 33² = CP²
=> 56² = CP²
=> CP = 56
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