Triangle PQR...

**foreverbrokenpromises** · May 25th 2009, 12:58 PM

In triangle PQR, <p = 30 degrees
<Q = 80 degrees
<R = 70 degreees
QH is an altitude and RM is a median and H is on the line PR,
the measure of <MHP is:

a) 15 degrees or 20 degrees or 30 degrees or 40 degrees or 45 degreess ?

diagram is attatched

**I-Think** · May 25th 2009, 04:49 PM

Multiple hellos. It is 30 degrees.

Solution: (All angular measurements in degrees)
$\angle P = 30$ degrees
$\angle Q = 80$ degrees
$\angle R = 70$ degrees
$QH$ is an altitude and $RM$ is a median and $H$ is on the line $PR$

If $QH$ is an altitude, then $\angle QHR$ and $\angle QHP$ are $90$ .
Filling in relevant angles angles: $\angle PHQ=60$ ,

Area of $\triangle PQR=\frac{PRQH}{2}$

A median splits a triangle into two triangles equal in area.
Construct altitude $MO$ . (dotted line in diagram, forgot to label O)

Two similar triangles are discovered, $\triangle MOP$ and $\triangle PHQ<br />$ (as $MO$ and $QH$ are parallel)
Calculate area of $\triangle PMR=\frac{MOPR}{2}=\frac{PRQH}{4}$
(Because a median splits a triangle into two triangles equal in area.)

Hence, $MO=\frac{QH}{2}$
Which means that $PO=\frac{PH}{2}$ and $PM=\frac{PQ}{2}$

Thus $OH=PH-\frac{PH}{2}=\frac{PH}{2}$
Hence $\triangle MPO$ and $\triangle MOH$ are identical
Hence $\angle MHP=30$

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