Question

$P$ and $Q$ are respectively the mid-points of sides $A B$ and $B C$ of a triangle $A B C$ and $R$ is the mid-point of $A P,$ show that

$a r (R Q C) = \frac{3}{8} a r (A B C)$

Accepted Answer

Consider a point $S$ on $A C$ such that $S$ is the mid-point of $A C$

Construction: Extend $P Q$ to $T$ such that $P Q = Q T$ , join $T C, Q S, P S,$ and $A Q$

In $△ A B C, P$ and $Q$ are the mid-points of $A B$ and $B C$ respectively. Hence, by using mid-point theorem, we obtain

$P Q ∥ A C$ and $P Q = \frac{1}{2} A C$

$\Rightarrow P Q ∥ A S$ and $P Q = A S$ (As $S$ is the mid-point of $A C$ )

$∴ P Q S A$ is a parallelogram. We know that diagonals of a parallelogram bisect it into equal areas of triangles.

$∴ a r (△ P A S) = a r (△ S Q P) = a r (△ P A Q) = a r (△ S Q A)$

Similarly, it is possible to prove that quadrilaterals $P S C Q, Q S C T$ and $P S Q B$ are also parallelograms.

Therefore,

$a r (△ P S Q) = a r (△ C Q S)$ (For parallelogram $P S C Q$ )

$a r (△ Q S C) = a r (△ C T Q)$ (For parallelogram $Q S C T$ )

$a r (△ P S Q) = a r (△ Q B P)$ (For parallelogram $P S Q B$ )

Thus,

$ar (△ PAS) = ar (△ SQP) = ar (△ PAQ) = ar (△ SQA) = ar (△ QSC) = ar (△ CTQ) = ar (△ QBP) . . . . . . (1)$

Also, $a r (△ A B C) = a r (△ P B Q) + a r (△ P A S) + a r (△ P Q S) + a r (△ Q S C)$

$\Rightarrow a r (△ A B C) = a r (△ P B Q) + a r (△ P B Q) + a r (△ P B Q) + a r (△ P B Q)$

$\Rightarrow ar (△ ABC) = 4 a r (△ P B Q)$

$\Rightarrow a r (△ P B Q) = \frac{1}{4} a r (△ A B C)$ $. . . (2)$

Join point $P$ to $C$

In $△ P A Q, Q R$ is the median.

$∴ a r (△ P R Q) = \frac{1}{2} a r (△ P A Q) = \frac{1}{2} \times \frac{1}{4} a r (△ A B C) = \frac{1}{8} a r (△ A B C)$ ...... $(3)$

In $△ A B C, P$ and $Q$ are the mid-points of $A B$ and $B C$ respectively. Hence, by using mid-point theorem, we obtain

$P Q = \frac{1}{2} A C$

$\Rightarrow A C = 2 P Q = P T$

Also, $P Q ∥ A C \Rightarrow P T ∥ A C$

Hence, $P A C T$ is a parallelogram.

Now, $a r (P A C T) = a r (P A C Q) + a r (△ Q T C)$

$\Rightarrow ar (PACT) = a r (P A C Q) + a r (△ P B Q)$ [Using equation $(1)$ ]

$∴ a r (P A C T) = a r (△ A B C)$ $. . . . (4)$

Now, $a r (△ A R C) = \frac{1}{2} a r (△ P A C)$ ( $C R$ is the median of $△ P A C$ )

$= \frac{1}{2} \times \frac{1}{2} a r (P A C T)$ ( $P C$ is the diagonal of parallelogram $P A C T$ )

$= \frac{1}{4} a r (△ P A C T) = \frac{1}{4} a r (△ A B C)$

$\Rightarrow ar (△ ARC) = \frac{1}{4} ar (△ ABC)$

$\Rightarrow \frac{1}{2} a r (△ A R C) = \frac{1}{8} a r (△ A B C)$

$\Rightarrow \frac{1}{2} a r (△ A R C) = a r (△ P R Q)$ [Using equation $(3)$ ] $. . . . (5)$

$a r (P A C T) = a r (△ P R Q) + a r (△ A R C) + a r (△ Q T C) + a r (△ R Q C)$

Putting the values from equations $(1), (2), (3), (4)$ and $(5)$ , we obtain

$a r (△ A B C) = \frac{1}{8} a r (△ A B C) + \frac{1}{4} a r (△ A B C) + \frac{1}{4} a r (△ A B C) + a r (△ R Q C)$

$\Rightarrow a r (△ A B C) = \frac{5}{8} a r (△ A B C) + a r (△ R Q C)$

$\Rightarrow a r (△ R Q C) = (1 - \frac{5}{8}) a r (△ A B C)$

$\Rightarrow a r (△ R Q C) = \frac{3}{8} a r (△ A B C)$

Hence proved.

Assam Board Solutions for Chapter: Areas of Parallelograms and Triangles, Exercise 4: EXERCISE (Optional)*

Assam Board Mathematics Solutions for Exercise - Assam Board Solutions for Chapter: Areas of Parallelograms and Triangles, Exercise 4: EXERCISE (Optional)*

Questions from Assam Board Solutions for Chapter: Areas of Parallelograms and Triangles, Exercise 4: EXERCISE (Optional)* with Hints & Solutions

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