SOLUTION:
Let's assume first that the given numbers on the diagram are the areas of the four smaller triangles. (because the numbers are on the middle)
We can solve the area of ΔABC by adding the areas of six smaller triangles together.
Now, let x and y be the unknown areas of two smaller triangles respectively.
Since ΔBDP and ΔCDP have the same heights, the ratio of their altitudes and areas are equal.
[tex]\sf \frac{a_{\triangle BDP}}{a_{\triangle CDP}} = \frac{Area_{\triangle BDP}}{Area_{\triangle CDP}} = \frac{40}{30} = \frac{4}{3}[/tex]
Since line segments AD, BE, CF intersect at point P, we have:
[tex]\sf \frac{BD}{DC} = \frac{\triangle AFP + x}{\triangle APE + y}[/tex]
[tex]\implies \sf \frac{84+x}{70+y} = \frac{4}{3} \rightarrow First \: eq.[/tex]
Similarly,
[tex]\sf \frac{AE}{EC} = \frac{Area_{\triangle APE}}{y} = \frac{Area_{\triangle AFP} + x}{Area_{\triangle BPC}}[/tex]
[tex]\implies \sf \frac{70}{y} = \frac{84+x}{70}[/tex]
[tex]\implies \sf \frac{84+x}{70}=\frac{70}{y} \rightarrow Second \: eq.[/tex]
Dividing first eq. by second eq.,
[tex]\sf \frac{\frac{84+x}{70+y} }{\frac{84+x}{70} } = \frac{\frac{4}{3} }{\frac{70}{y} }[/tex]
[tex]\sf \implies \frac{\cancel{84+x}}{70+y} \times \frac{70}{\cancel{84+x}} = \frac{4}{3} \times \frac{y}{70}[/tex]
[tex]\sf \implies \frac{70}{70+y} = \frac{4y}{210}[/tex]
[tex]\implies \sf y = 35[/tex]
So that from second eq. 2,
[tex]\sf \frac{84+x}{70}=\frac{70}{y}[/tex]
[tex]\implies \sf \frac{84+x}{70} = \frac{70}{35}[/tex]
[tex]\sf \implies \frac{84+x}{70} = 2[/tex]
[tex]\implies \sf x = 56[/tex]
Therefore, the area of ΔABC is 84 + 70 + 56 + 35 + 40 + 30 = 315.
ANSWER:
315 square units
