Basic Geometry and Area of the right triangle.

Question

Forgive me for the terrible formatting.Also i am having trouble uploading images.

A right angled Triangle

So in Triangle PRQ which expression is correct?

$$\dfrac 12 PR \cdot RQ \quad \text{or} \quad \dfrac 12 RM \cdot PQ ?$$

Answer 1

Michael's answer is perfect, but I'm just writting this to help you visualize this a little better to convince yourself.

Look at this:

Note that both rectangles (U=blue+black and V=red+black) have twice the area of your $PRQ$ triangle, so they are the same. The area of $U$ is $PQ\cdot RM$ and the area of $V$ is $PR\cdot RQ$, since they are the same, then $PQ\cdot RM=PR\cdot RQ$ and half of this is the area of your triangle. Sorry for my terrible drawing.

You could also see this as two separate triangles, $A=\Delta PRM$ and $B=\Delta RMQ$, then $area(A)=\frac{1}{2}PM\cdot RM$ and $area(B)=\frac{1}{2} MQ\cdot RM$, so $$area(\Delta PRQ)=area(A)+area(B)=\frac{1}{2}(PM\cdot RM+MQ\cdot RM)=\frac{1}{2}((PM+MQ)\cdot RM),$$ but $PM+MQ=PQ$.

Answer 2

Yes, you are right.

For right-angled triangle $PRQ$, where $\measuredangle PRQ=90^{\circ}$ and $RM$ is an altitude we have: $$S_{\Delta PQR}=\frac{1}{2}PR\cdot RQ=\frac{1}{2}RM\cdot PQ$$ because we can think also that $RQ$ is an altitude and $PR$ is the side for which we drew this altitude.