Question

P is a point interior to rectangle ABCD and such that PA=3 inches, PD=4 inches and PC=5 inches. Then PB in inches equals to

• A. $2\sqrt{3}$
• B. $3\sqrt{2}$
• C. $3\sqrt{3}$
• D. $4\sqrt{2}$

Answer 1

The correct option is B $3\sqrt{2}$

Let there be a point $P$ inside the rectangle.

Given: $PA=3$inches, $PB=4$inches, and $PC=5$inches.

The relation between $PA,PB,PC,PD$ is given as:

${PA}^2+{PC}^2={PB}^2+{PD}^2$

${PD}^2={PA}^2+{PC}^2-{PB}^2=3^2+5^2-4^2=9+25-16=18$

$\therefore PD=\sqrt{9\times 2}=3\sqrt{2}$inches