• MHB
Monoxdifly
MHB
In an ABCD.EFGH cuboid with AB = 4 cm, BC = 3 cm, and CG = 5 cm there is a parallelogram OBFPH with O is located at the center of ABCD and P is located at the center of EFGH. The distance between the lines HO and PB is ...
A. $$\displaystyle 5\sqrt3$$ cm
B. $$\displaystyle 5\sqrt2$$ cm
C. $$\displaystyle \sqrt5$$ cm
D. $$\displaystyle \frac{5}{2}\sqrt2$$ cm
E. $$\displaystyle \frac{5}{3}\sqrt3$$ cm

By making use of the parallelogram formula, I got $$\displaystyle \frac{1}{5}\sqrt5$$. Do you guys get the same answer as me or any of the options?

Gold Member
MHB
In an ABCD.EFGH cuboid with AB = 4 cm, BC = 3 cm, and CG = 5 cm there is a parallelogram OBFPH with O is located at the center of ABCD and P is located at the center of EFGH. The distance between the lines HO and PB is ...
A. $$\displaystyle 5\sqrt3$$ cm
B. $$\displaystyle 5\sqrt2$$ cm
C. $$\displaystyle \sqrt5$$ cm
D. $$\displaystyle \frac{5}{2}\sqrt2$$ cm
E. $$\displaystyle \frac{5}{3}\sqrt3$$ cm

By making use of the parallelogram formula, I got $$\displaystyle \frac{1}{5}\sqrt5$$. Do you guys get the same answer as me or any of the options?
OBFPH is not a parallelogram. Did you mean OBPH? If so, then I agree with your answer $\frac15\sqrt5$. But maybe you misread the question?

Monoxdifly
MHB
Sorry, I meant OBPH.