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**Monox D. I-Fly****Member**- From: Indonesia
- Registered: 2015-12-02
- Posts: 1,269

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.

B. cm

C. cm

D. cm

E. cm

By making use of the parallelogram formula, I got

. Do you guys get the same answer as me or any of the options?Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away. May his adventurous soul rest in peace at heaven.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,400

hi

??? FPH is a straight line and parallelograms only have four not five vertices. I think OBFPH is a trapezium.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Monox D. I-Fly****Member**- From: Indonesia
- Registered: 2015-12-02
- Posts: 1,269

Sorry, I meant OBPH.

Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away. May his adventurous soul rest in peace at heaven.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,400

Arhh. That's better.

FP = PH = 2.5 by Pythag. PO = 5 and angle OPH is 90 degrees so OH is given by

Angle PHO = PBO = alpha

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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