How to approach this question

Let midpoint of $OA$ be $P$ and $Q$ is the point on $BC$ such that $ BQ : QC = 3 : 1 $

Find the length of $PQ.$

Meanwhile $h = 4 \sqrt7$ cm

[ You are expected to use Pythagorean theorem only ]

#Geometry

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Comments

Let the center of the base square $ABCD$ be the origin $(0,0,0)$ and $A(4,-4,0)$ , $B(4,4,0)$ , $C(-4,4,0)$ , $D(-4,-4,0)$ , and $O(0,0,h) = O(0,0,4\sqrt 7)$ . Note that $h = \sqrt{12^2-(4\sqrt 2)^2} = 4\sqrt 7$ .

Then $P \left(\frac {4+0}2, \frac {-4+0}2, \frac {0+4\sqrt 7}2\right) = P (2,-2,2\sqrt 7)$ and $Q(-2,4,0)$ . By Pythagorean theorem:

$\begin{aligned} PQ^2 & = (2-(-2))^2 + (-2-4)^2 + (2\sqrt 7-0)^2 = 80 \\ \implies PQ & = \boxed{4\sqrt 5} \end{aligned}$

Chew-Seong Cheong - 1 year, 4 months ago

Nop. PQ = $4 \sqrt{5}$ but I don't know how

Syed Hamza Khalid - 1 year, 4 months ago

Note that $h$ cannot be $\text{12 cm}$ because the slanting sides $AO=BO=CO=DO=\text{12 cm}$ .

Chew-Seong Cheong - 1 year, 4 months ago

I have got it. $h = 4\sqrt 7$ .

Chew-Seong Cheong - 1 year, 4 months ago

@Chew-Seong Cheong – Yea I am sorry it is just as you said. I have edited the problem

Syed Hamza Khalid - 1 year, 4 months ago

@Chew-Seong Cheong – But still the ans for PQ is $4 \sqrt{5}$ how?

Syed Hamza Khalid - 1 year, 4 months ago

on point!