Track / Stage 3 / 39 of 260 #39 of 2000

Problem 39

AMC 10/12, early questions

Geometry Difficulty 3.1 Find the answer

Three vertices of a cube are $P=(7,12,10)$ , $Q=(8,8,1)$ , and $R=(11,3,9)$ . What is the surface area of the cube?

Your answer

Official solution

$PQ=\sqrt{(8-7)^2+(8-12)^2+(1-10)^2}=\sqrt{98}$
$PR=\sqrt{(11-7)^2+(3-12)^2+(9-10)^2}=\sqrt{98}$
$QR=\sqrt{(11-8)^2+(3-8)^2+(9-1)^2}=\sqrt{98}$
So, $PQR$ is an equilateral triangle. Let the side of the cube be $a$ .
$a\sqrt{2}=\sqrt{98}$
So, $a=7$ , and hence the surface area is $6a^2=\framebox{294}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.