Why does [QAC] = QA/QR · QC/QP [PQR]?

Because both triangles share the angle at vertex Q: Area = 1/2 ab sin Q, so the ratio of areas is the product of side ratios sharing that angle.

JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Ratio of triangle areas formed by dividing sides internally

Let PQR be a triangle. The points A, B and C are on the sides QR, RP and PQ respectively such that QA/AR = RB/BP = PC/CQ = 1/2. Then (Area (ΔPQR))/(Area (ΔABC)) is equal to

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Step-by-step solutionView

Correct answer

The ratio of $Area (Δ PQR)$ to $Area (Δ ABC)$ is $3$ .

Option analysis

Why each option works or fails

A · 4

Assumes each corner triangle has area equal to $(1 / 3) \times (1 / 3) = 1 / 9$ of $Δ PQR$ , yielding $Area (Δ ABC) = (1 - 3 \times \frac{1}{9}) Δ = \frac{2}{3} Δ$ or miscalculating $Area (Δ ABC) / Δ = 1 / 4$ . Recall that if points divide sides in ratio $1 : 2$ , the adjacent side fractions forming each corner triangle are $1 / 3$ and $2 / 3$ , giving each corner an area of $\frac{1}{3} \times \frac{2}{3} = \frac{2}{9}$ of the whole.

B · 3

The area of each of the three corner triangles ( $Δ QAC$ , $Δ RBA$ , $Δ PCB$ ) is $\frac{1}{3} \times \frac{2}{3} = \frac{2}{9}$ of $Δ PQR$ , leaving $Area (Δ ABC) = (1 - 3 \times \frac{2}{9}) = \frac{1}{3}$ of the total area. Correctly determined that $\frac{Area (Δ PQR)}{Area (Δ ABC)} = \frac{1}{1 / 3} = 3$ .

C · 2

Confuses the side ratio $1 : 2$ with edge fractions $1 / 2$ and $1 / 2$ , resulting in midpoint triangle geometry where the remaining area fraction is incorrect, or subtracts $2 / 9$ only twice. A side divided in the ratio $1 : 2$ is partitioned into parts of length $1 / 3$ and $2 / 3$ of the total side, not $1 / 2$ .

Reviewed route

Solution

StepWorking

01Given

In $Δ PQR$ , points $A, B, C$ divide sides $Q R, R P, P Q$ such that $\frac{Q A}{A R} = \frac{R B}{B P} = \frac{P C}{C Q} = \frac{1}{2}$ .

02Approach

Choose vertex $Q$ as origin $\vec{0}$ . Let the position vector of $P$ be $\vec{p}$ and $R$ be $\vec{r}$ . Express position vectors of $A, B, C$ using the section formula, then compute the cross product area formula for $Δ ABC$ relative to $Δ PQR$ .

03Execute

Since $Q$ is origin, $\vec{a} = \frac{1 \vec{r} + 2 \vec{0}}{3} = \frac{\vec{r}}{3}$ . Point $B$ divides $R P$ in ratio $1 : 2$ , so $\vec{b} = \frac{1 \vec{p} + 2 \vec{r}}{3}$ . Point $C$ divides $P Q$ in ratio $1 : 2$ from $P$ , so $C$ divides $Q P$ in ratio $2 : 1$ , which gives $\vec{c} = \frac{2 \vec{p} + 1 \vec{0}}{3} = \frac{2 \vec{p}}{3}$ .

04Execute

The area of $Δ PQR$ is $\frac{1}{2} ∣ \vec{r} \times \vec{p} ∣$ . The area of $Δ ABC$ is $\frac{1}{2} ∣ (\vec{b} - \vec{a}) \times (\vec{c} - \vec{a}) ∣$ . We have $\vec{b} - \vec{a} = \frac{\vec{p} + \vec{r}}{3}$ and $\vec{c} - \vec{a} = \frac{2 \vec{p} - \vec{r}}{3}$ . Thus $(\vec{b} - \vec{a}) \times (\vec{c} - \vec{a}) = \frac{1}{9} [(\vec{p} + \vec{r}) \times (2 \vec{p} - \vec{r})] = \frac{1}{9} [- \vec{p} \times \vec{r} + 2 \vec{r} \times \vec{p}] = \frac{1}{9} [3 (\vec{r} \times \vec{p})] = \frac{1}{3} (\vec{r} \times \vec{p})$ . Hence $\frac{Area (Δ PQR)}{Area (Δ ABC)} = \frac{\frac{1}{2} ∣ \vec{r} \times \vec{p} ∣}{\frac{1}{6} ∣ \vec{r} \times \vec{p} ∣} = 3$ .

✓Verify

Alternatively, by complementary area: $Area (Δ Q A C) = \frac{1}{3} \times \frac{1}{3} Δ = \frac{1}{9} Δ$ . But $Q C = \frac{2}{3} Q P$ and $Q A = \frac{1}{3} Q R$ , so $Area (Δ Q A C) = \frac{1}{3} \cdot \frac{2}{3} Δ = \frac{2}{9} Δ$ . By symmetry, each corner triangle has area $\frac{2}{9} Δ$ . The inner area is $Δ - 3 \times \frac{2}{9} Δ = \frac{1}{3} Δ$ . Thus the ratio is $\frac{Δ}{\frac{1}{3} Δ} = 3$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Does setting vertex Q as the origin work for any general triangle?

Yes, area ratios are affine invariants, so fixing an origin at any convenient vertex does not change the ratio.

Why does

[Q A C] = \frac{Q A}{Q R} \cdot \frac{Q C}{Q P} [P Q R]

Because both triangles share the angle at vertex $Q$ : $Area = \frac{1}{2} a b \sin Q$ , so the ratio of areas is the product of side ratios sharing that angle.

Answer

The ratio of $Area (Δ PQR)$ to $Area (Δ ABC)$ is $3$ .

Why each option works or fails

A: 4 - Assumes each corner triangle has area equal to $(1 / 3) \times (1 / 3) = 1 / 9$ of $Δ PQR$ , yielding $Area (Δ ABC) = (1 - 3 \times \frac{1}{9}) Δ = \frac{2}{3} Δ$ or miscalculating $Area (Δ ABC) / Δ = 1 / 4$ . Recall that if points divide sides in ratio $1 : 2$ , the adjacent side fractions forming each corner triangle are $1 / 3$ and $2 / 3$ , giving each corner an area of $\frac{1}{3} \times \frac{2}{3} = \frac{2}{9}$ of the whole.
B · correct: 3 - The area of each of the three corner triangles ( $Δ QAC$ , $Δ RBA$ , $Δ PCB$ ) is $\frac{1}{3} \times \frac{2}{3} = \frac{2}{9}$ of $Δ PQR$ , leaving $Area (Δ ABC) = (1 - 3 \times \frac{2}{9}) = \frac{1}{3}$ of the total area. Correctly determined that $\frac{Area (Δ PQR)}{Area (Δ ABC)} = \frac{1}{1 / 3} = 3$ .
C: 2 - Confuses the side ratio $1 : 2$ with edge fractions $1 / 2$ and $1 / 2$ , resulting in midpoint triangle geometry where the remaining area fraction is incorrect, or subtracts $2 / 9$ only twice. A side divided in the ratio $1 : 2$ is partitioned into parts of length $1 / 3$ and $2 / 3$ of the total side, not $1 / 2$ .

Step-by-step solution

Given: In $Δ PQR$ , points $A, B, C$ divide sides $Q R, R P, P Q$ such that $\frac{Q A}{A R} = \frac{R B}{B P} = \frac{P C}{C Q} = \frac{1}{2}$ .
Approach: Choose vertex $Q$ as origin $\vec{0}$ . Let the position vector of $P$ be $\vec{p}$ and $R$ be $\vec{r}$ . Express position vectors of $A, B, C$ using the section formula, then compute the cross product area formula for $Δ ABC$ relative to $Δ PQR$ .
Execute: Since $Q$ is origin, $\vec{a} = \frac{1 \vec{r} + 2 \vec{0}}{3} = \frac{\vec{r}}{3}$ . Point $B$ divides $R P$ in ratio $1 : 2$ , so $\vec{b} = \frac{1 \vec{p} + 2 \vec{r}}{3}$ . Point $C$ divides $P Q$ in ratio $1 : 2$ from $P$ , so $C$ divides $Q P$ in ratio $2 : 1$ , which gives $\vec{c} = \frac{2 \vec{p} + 1 \vec{0}}{3} = \frac{2 \vec{p}}{3}$ .
Execute: The area of $Δ PQR$ is $\frac{1}{2} ∣ \vec{r} \times \vec{p} ∣$ . The area of $Δ ABC$ is $\frac{1}{2} ∣ (\vec{b} - \vec{a}) \times (\vec{c} - \vec{a}) ∣$ . We have $\vec{b} - \vec{a} = \frac{\vec{p} + \vec{r}}{3}$ and $\vec{c} - \vec{a} = \frac{2 \vec{p} - \vec{r}}{3}$ . Thus $(\vec{b} - \vec{a}) \times (\vec{c} - \vec{a}) = \frac{1}{9} [(\vec{p} + \vec{r}) \times (2 \vec{p} - \vec{r})] = \frac{1}{9} [- \vec{p} \times \vec{r} + 2 \vec{r} \times \vec{p}] = \frac{1}{9} [3 (\vec{r} \times \vec{p})] = \frac{1}{3} (\vec{r} \times \vec{p})$ . Hence $\frac{Area (Δ PQR)}{Area (Δ ABC)} = \frac{\frac{1}{2} ∣ \vec{r} \times \vec{p} ∣}{\frac{1}{6} ∣ \vec{r} \times \vec{p} ∣} = 3$ .
Verify: Alternatively, by complementary area: $Area (Δ Q A C) = \frac{1}{3} \times \frac{1}{3} Δ = \frac{1}{9} Δ$ . But $Q C = \frac{2}{3} Q P$ and $Q A = \frac{1}{3} Q R$ , so $Area (Δ Q A C) = \frac{1}{3} \cdot \frac{2}{3} Δ = \frac{2}{9} Δ$ . By symmetry, each corner triangle has area $\frac{2}{9} Δ$ . The inner area is $Δ - 3 \times \frac{2}{9} Δ = \frac{1}{3} Δ$ . Thus the ratio is $\frac{Δ}{\frac{1}{3} Δ} = 3$ .

Shortcut:

When to use it: Fastest conceptual method for triangles with dividing points on the edges.

Given: Points $A \in Q R$ , $B \in R P$ , $C \in P Q$ divide sides such that $Q A = \frac{1}{3} Q R$ , $A R = \frac{2}{3} Q R$ ; $R B = \frac{1}{3} R P$ , $B P = \frac{2}{3} R P$ ; $P C = \frac{1}{3} P Q$ , $C Q = \frac{2}{3} P Q$ .

Goal: Find the ratio $\frac{Area (Δ PQR)}{Area (Δ ABC)}$ .

Approach: Express the area of $Δ ABC$ by subtracting the three corner triangles $Δ Q A C$ , $Δ R B A$ , and $Δ P C B$ from $Δ PQR$ .

Execute: Let $[e x t i n c t i o n]$ denote area. $[Q A C] = \frac{Q A}{Q R} \cdot \frac{Q C}{Q P} [P Q R] = \frac{1}{3} \cdot \frac{2}{3} [P Q R] = \frac{2}{9} [P Q R]$ . Similarly, $[R B A] = \frac{R B}{R P} \cdot \frac{R A}{R Q} [P Q R] = \frac{1}{3} \cdot \frac{2}{3} [P Q R] = \frac{2}{9} [P Q R]$ . Also $[P C B] = \frac{P C}{P Q} \cdot \frac{P B}{P R} [P Q R] = \frac{1}{3} \cdot \frac{2}{3} [P Q R] = \frac{2}{9} [P Q R]$ . Thus, $[A B C] = [P Q R] - 3 \times \frac{2}{9} [P Q R] = (1 - \frac{2}{3}) [P Q R] = \frac{1}{3} [P Q R]$ . Hence $\frac{[P Q R]}{[A B C]} = 3$ .

Verify: Notice that if ratio was $1 : 1$ , midpoints would give $[A B C] = [P Q R] - 3 \times \frac{1}{4} [P Q R] = \frac{1}{4} [P Q R]$ , ratio = 4. Here ratio is $1 : 2$ , giving $\frac{1}{3} [P Q R]$ , which smoothly transitions between 1 (no cut) and $\frac{1}{4}$ (midpoints).

Ratio of triangle areas formed by dividing sides internally

What feels right?

The ratio of $Area (Δ PQR)$ to $Area (Δ ABC)$ is $3$ .

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

✓Verify

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