Kubus ABCD.EFGH, P pada EG sehibgga EP =2PG.jika jarak titik E ke garis AP adalah a,panjang rusuk kubus tersebut adalah...

[tex]\displaystyle \text{EP}=\frac23\text{EG}\\\text{EP}=\frac23s\sqrt2\\\\\text{AP}^2=\text{EP}^2+\text{AE}^2\\\text{AP}^2=\frac89s^2+s^2\\\text{AP}^2=\frac{17}9s^2\\\text{AP}=\frac13s\sqrt{17}\\\\\frac{\text{AP}}{\text{EP}}=\frac{\text{AE}}{\text{EE}'}\\\frac{\frac13s\sqrt{17}}{\frac23s\sqrt2}=\frac{s}{a}\\\frac{\sqrt{17}}{2\sqrt2}=\frac{s}{a}\\\boxed{\boxed{\frac14a\sqrt{34}=s}}[/tex]

EP = 2 PG => EP/PG = 2/1
Misal rusuk kubus = r
EP : PG = 2 : 1
EP = 2/3 EG = 2/3 r√2

AP = √(AE^2 + AP^2)
= √(r^2 + (2/3 r√2)^2)
= √(r^2 + 4/9 . 2r^2)
= √(9r^2/9 + 8r^2/9)
= √(17r^2/9)
= 1/3 r √17

Jarak E ke AP = a
(EP . EA)/AP = a
(2/3 r√2 . r) / (1/3 r√17) = a
(2√2 r) / √17 = a
2√2 r = a√17
r = (a√17) / (2√2)
r = (a√17) / (2√2) . √2/√2
r = a√34 / 4
r = 1/4 a√34