# Finding the area of a triangle using fractions? To find the area of the triangle do you use Pythagorean theorem from what you have? Could this use similar triangles.

• The image doesn't exist at the location you refer to. – gnometorule Mar 3 '13 at 20:33
• Could you clarify? – Jon Mar 3 '13 at 20:37
• @:LittleJon: As you already have an answer by AN who must have seen your image, I assume it was my antivirus that blocked the image location, so only a problem for me (I get 404 if I go to the location). So just ignore what I said. – gnometorule Mar 3 '13 at 20:39

The area of $\triangle PST$ is the sum of the areas of $\triangle PSV$ and $\triangle VST$.

\begin{align} \text{The area of }\triangle PSV &= \frac12 PV\cdot SV\\ &=\frac12 PV\cdot\frac23 QV\\ &=\frac13 PV\cdot QV, \end{align} \begin{align} \text{and the area of }\triangle VST &= \frac12VT\cdot SV\\ &=\frac12 VT\cdot \frac23QV\\ &=\frac13VT\cdot QV. \end{align} \begin{align} \text{So the area of }\triangle PST &= \frac13QV(PV+VT)\\ &=\frac13QV\cdot PT\\ &=\frac13QV\cdot\frac34PR\\ &=\frac14QV\cdot PR. \end{align}
The area of $\triangle PQR = \frac12QV\cdot PR$.
\begin{align} \text{So the ratio }\frac{\triangle PST}{\triangle PQR} &= \frac{\frac14 QV\cdot PR}{\frac12QV\cdot PR}\\ &=\frac{\frac14}{\frac12}\\ &=\frac12. \end{align}

Draw the line $QT$. Note that the area of $\triangle PQT$ is $\frac{3}{4}$ times the area of the big triangle $PQR$: same height, base $\frac{3}{4}$ as big.

Note also that the area of $\triangle PST$ is $\frac{2}{3}$ times the area of $\triangle PQT$. The argument is essentially the same. The two triangles have the same base, and the height of $\triangle PST$ is $\frac{2}{3}$ times the height of $\triangle PQT$.

Thus the desired ratio of areas is $\frac{3}{4}\cdot \frac{2}{3}=\frac{1}{2}$.

Remark: A more detailed analysis shows that we do not need to know that the line $QV$ is perpendicular to $PR$. The only advantage of perpendicularity is that it makes the fact that the height of $\triangle PST$ is $\frac{2}{3}$ times the height of $\triangle PQT$ obvious. But the relationship holds even without perpendicularity.