Question

please help me please don't scam​

Answer 1

I think the ratio of the area of the triangles is 1:4 and hence the area of PQR is 160cm^2. You may prove it by similarity.

Answer 2

$\large\text{\underline{Let's begin.}}$

We start by constructing three perpendicular lines. (Refer to the first attachment.)

$\red{\bigstar E\text{ to }\overline{QR}}$

$\red{\bigstar F\text{ to }\overline{RP}}$

$\red{\bigstar G\text{ to }\overline{PQ}}$

The reason can be explained if we draw one of the lines. (Refer to the second attachment.)

$\large\text{\underline{Solution}}$

The length of each perpendicular line is $\dfrac{1}{3}$ times the height.

However, since points $E,F,G$ divide the sides in a ratio of $1:2$, the length of each base is $\dfrac{2}{3}$ of the base.

Thus, each of the areas of $\triangle EFQ,\triangle FGR,\triangle GEP$ are $\dfrac{2}{9}$ of $\triangle PQR$. So, these triangles take $\dfrac{2}{3}$ of the whole triangle.

$\implies \triangle EFG=\dfrac{1}{3}\triangle PQR$

$\implies\triangle PQR=3\times40$

$\implies\triangle PQR=120\text{ units}^{2}$

$\large\text{\underline{Conclusion}}$

Therefore, the area of $\triangle PQR$ is 3 times the area of $\triangle EFG$, so the area is $120\text{ units}^{2}$.