Question

ABCD is a rectangle and point P is such that PA = 3 cm, PC = 3$\surd$2 cm and PD = 5 cm, find the value of PB.

Answer 1

Given $PA=3cm$

$PC=3\sqrt{2}cm$

$PD=5cm$

$AC=PA+PC$

$AC=PA+PC$

$=3+3\sqrt{2}$

$=3(1+\sqrt{3})$

$=3(1+1.414)$

$=3\left(2.414\right)$

$AC=7.242$

We know that in rectangle

$AC=BD$

$AC=PD+PB$

$7.242=5+PB$

$PB=7.242-5$

$PB=2.242$

$PB=2.242cm$