ABCD is a rectangle. P is a point on the side AB as shown in the given figure. If DP = 13, CP = 10 and BP = 6, then what is the value of AP?

ABCD is a rectangle. P is a point on the side AB as shown in the given figure. If DP = 13, CP = 10 and BP = 6, then what is the value of AP?

Answer: Option A

Solution

Answer: Option A

Solution:

$$\begin{array}{rl}& \text{In }\Delta PCB\\ & BC=\sqrt{{10}^2-6^2}=8\text{cm}\\ & BC=AD=8\text{cm}\\ & \text{In }\Delta ADP\\ & AP=\sqrt{{13}^2-8^2}\\ & AP=\sqrt{169-64}=\sqrt{105}\end{array}$$