In a $△ A B C, P$ and $Q$ are respectively the mid-point of $A B$ and $B C$ and $R$ is the midpoint of $A P$ . Prove that:
$a r (△ P R Q) = \frac{1}{2} a r (△ A R C)$

Open in App

Solution

Consider $Δ A P Q$
$R$ is mid point of $A P$
So, $a r (Δ P R Q) = a r (Δ A R Q) \dots (1)$
Consider $Δ A R C$
$Q$ is mid point of $A C$
So, $a r (Δ A R Q) = \frac{1}{2} a r (Δ A R C) \dots (2)$
From $(1) (2)$
$⟹ a r (Δ P R Q) = \frac{1}{2} a r (Δ A R C)$

Suggest Corrections

Similar questions

P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that

(i) (ii)

(iii)

Q. P and Q are respectively the mid points of sides AB and BC of a triangle ABC and R is the midpoint of AP, show that
1)

a r (P R Q) = \frac{1}{2} a r (A R C)

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

a r (P B Q) = a r (A R C)

Q. Question 7 (i)
P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that:

(i) a r (P R Q) = \frac{1}{2} a r (A R C)

In a \triangle ABC, P, and Q are respectively, the mid points of AB and BC and R is the mid point of AP. Prove that
$(i) a r (△ P B Q) = a r (△ A R C) (i i) a r (P R Q) = \frac{1}{2} a r (△ A R C) (i i i) a r (△ R Q c) = \frac{3}{8} a r (△ A B C)$

Q. P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP,
Show that

a r (Δ P B Q) = a r (Δ A R C)

Join BYJU'S Learning Program

In a △ABC,P and Q are respectively the mid-point of AB and BC and R is the midpoint of AP. Prove that:ar(△PRQ)=12ar(△ARC)

Consider ΔAPQR is mid point of APSo, ar(ΔPRQ)=ar(ΔARQ)⋯(1)Consider ΔARCQ is mid point of ACSo,ar(ΔARQ)=12ar(ΔARC)⋯(2)From (1)(2)⟹ar(ΔPRQ)=12ar(ΔARC)

In a $△ A B C, P$ and $Q$ are respectively the mid-point of $A B$ and $B C$ and $R$ is the midpoint of $A P$ . Prove that:
$a r (△ P R Q) = \frac{1}{2} a r (△ A R C)$

Consider $Δ A P Q$
$R$ is mid point of $A P$
So, $a r (Δ P R Q) = a r (Δ A R Q) \dots (1)$
Consider $Δ A R C$
$Q$ is mid point of $A C$
So, $a r (Δ A R Q) = \frac{1}{2} a r (Δ A R C) \dots (2)$
From $(1) (2)$
$⟹ a r (Δ P R Q) = \frac{1}{2} a r (Δ A R C)$