# problem in solid geometry

• August 31st 2012, 12:51 PM
mido22
problem in solid geometry
if ABC is a triangle where AB = 5cm , AC = 3cm , D is the midpoint of BC and draw AH = (6 *root 2)cm perpendicular to both AC and AB . find the length of DH

Attachment 24659
• August 31st 2012, 02:15 PM
MaxJasper
Re: problem in solid geometry
Image is wrong if AH=6*squrt(2)=8.485 cm. Is AH perpendicular to BC? You say it is prep to both AC & AB!!!!!!?!??!
• August 31st 2012, 07:39 PM
Soroban
Re: problem in solid geometry
Hello, mido22!

Quote:

ABC is a triangle where AB = 5cm, AC = 3cm, D is the midpoint of BC.
Draw AH = 6√2 cm, perpendicular to both AC and AB.
Find the length of DH.

The diagram looks like this.
Code:

            H             o             | *           *|  _*             |6√2 *           * |      *             |        *         *  o - - - - - o B           *A    5  *         * *      *         *3    o       **  *  D       * *       o       C
I don't see a unique length for $D\!H.$

It depends on $\theta = \angle B\!AC.$

If $\theta = 0^o,\:D\!H = 7\sqrt{2}$

If $\theta = 180^o,\:D\!H = 6\sqrt{2}$
• September 1st 2012, 02:51 PM
mido22
Re: problem in solid geometry
sorry i forgot to write that BC = 7 cm

<BAC not zero and not 180

i have a final answer that HD = 2.5 but i can't find a proof
• September 1st 2012, 02:53 PM
mido22
Re: problem in solid geometry
i can prove that AH is perpendicular to BC since it is perpendicular to both AB , AC but then i don't know >>:(