Geometry Problem

Geometry Level pending

$ABCD$ is a rectangle and $P$ is a point in the rectangle such that $PA=3$ , $PB=4$ , and $PC=5$ , find $PD$ .

2 \sqrt 5

3 \sqrt 3

3 \sqrt 2

\sqrt 5

3 solutions

A Former Brilliant Member
Jun 15, 2017

By the British Flag Theorem , we get

$(PA)^2+(PC)^2=(PB)^2+(PD)^2$

$3^2+5^2=4^2+(PD)^2$

$9+25=16+(PD)^2$

$(PD)^2=18$

$PD=\sqrt{18}=\sqrt{9 \times2}=$ $\boxed{\color{#D61F06}3\sqrt{2}}$

Brian Charlesworth
Apr 12, 2017

By the British Flag Theorem ,

$PA^{2} + PC^{2} = PB^{2} + PD^{2} \Longrightarrow 3^{2} + 5^{2} = 4^{2} + PD^{2} \Longrightarrow PD^{2} = 18 \Longrightarrow PD = \sqrt{18} = \boxed{3\sqrt{2}}$ .

A Former Brilliant Member
Apr 13, 2017

Using the British Flag Theorem , we have

$\overline{PD}~^2+\overline{PB}~^ 2=\overline{PA}~^2+\overline{PC}~^2$

$PD=\sqrt{9(2)}=$ $\boxed{\color{#D61F06}3\sqrt{2}}$ $\color{#69047E}\text{answer}$

Geometry Problem

3 solutions

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