Question

As shown in the given figure, $\Delta$ABC is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The area of four of $6$ triangles are as indicated, then the area of $\Delta$ ABC is __________?

• A. $238$
• B. $464$
• C. $315$
• D. $412$

Answer 1

The correct option is C $315$

Let the interior point be p, let the points on BC, CA and AB be D, E and F, respectively. Let $x$ be the area of $△APE$ and $y$ be the area of $△CPD$. Note that $△APF$ and $△BPF$ share the same altitude from p, so the ratio of their areas is the same as the ratio of their bases. Similarly, $△ACF$ and $△BCF$ share the same altitude from c, so the ratio of their areas is the same as the ratio of their bases. Moreover, the two pairs of bases are actually the same, and thus in the same ratio.

As a result, we have:

$\frac{40}{30}=\frac{124+x}{65+y}$

or equivalently $372+3x=260+4y$

and so $4y=3x+112$.

Applying identical reasoning to the triangles with bases CD and BD, we get

$\frac{y}{35}=\frac{x+y+84}{105}$

so that $3y=x+y+84$

and $2y=x+84$.

Substituting from this equation into the previous one gives $x=56$, from which we get $y=70$ and so the area of $△ABC$ is $56+40+30+35+70+84=315$