A Point in Rectangle

Geometry Level 2

$ABCD$ is a rectangle and point $P$ is in the rectangle such that $PA=3\sqrt{2} \text{ cm}$ , $PB=5 \text{ cm}$ and $PC=3 \text{ cm}$ . Find the length of $PD$ .

\sqrt{3} \text{ cm}

\sqrt{5} \text{ cm}

2 \text{ cm}

\sqrt{2} \text{ cm}

2 solutions

Chew-Seong Cheong
Mar 10, 2019

Let the width of the rectangle $AB=CD=a$ , the height $DA=BC=b$ , vertex $D$ be the origin and the coordinates of $P$ be $(x,y)$ . By Pythagorean theorem , we have:

$\begin{cases} PA^2: & x^2 + (b-y)^2 & = 18 \\ PB^2: & (a-x)^2 + (b-y)^2 & = 25 \\ PC^2: & (a-x)^2 + y^2 & = 9 \end{cases}$

Since $PD^2 = x^2 + y^2 = PA^2 - PB^2 + PC^2 = 18 - 25 + 9 = 2$ , $\implies PD= \boxed{\sqrt 2}$ .

Achmad Damanhuri
Mar 10, 2019

$P$ is a point in rectangle $ABCD$ holds an equation $(AP)^2+(CP)^2=(BP)^2+(DP)^2$

@Achmad Damanhuri , for Geometry problems, we may not need to include the unit $\text{cm}$ . It does not add additional information. But once you have included in the problem it should also be included in the answer options. I have edited the problem and answer options for you.

Chew-Seong Cheong - 2 years, 3 months ago

Thank you sir 😅

Achmad Damanhuri - 2 years, 3 months ago

A Point in Rectangle

2 solutions

0 pending reports