Straight Lines
Ratio of areas of triangle and sub-triangle
nta_pyq_2023_jan
Grade 11
Question:
Let PQR be a triangle. The points A, B and C are on the sides QR, RP and PQ respectively such that $\dfrac{QA}{AR} = \dfrac{RB}{BP} = \dfrac{PC}{CQ} = \dfrac{1}{2}$. Then $\dfrac{\text{Area}(\Delta PQR)}{\text{Area}(\Delta ABC)}$ is equal to
Step-by-Step Solution
Key Concept: Use vectors. Place P at origin $\vec{0}$, Q at $\vec{q}$, R at $\vec{r}$. Then express A, B, C in terms of $\vec{q}$ and $\vec{r}$ using the given ratios. Compute areas using cross products.
A divides QR s.t. $QA:AR = 1:2$, so $A = \frac{2\vec{q}+\vec{r}}{3}$. Similarly $B = \frac{2\vec{r}}{3}$, $C = \frac{\vec{q}}{3}$. Area of $\Delta ABC = \frac{1}{6}|\vec{q}\times\vec{r}|$. Ratio $= \frac{\frac{1}{2}|\vec{q}\times\vec{r}|}{\frac{1}{6}|\vec{q}\times\vec{r}|} = 3$.
Correct Answer: 2