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) $a r (P R Q) = \frac{1}{2} a r (A R C)$
2) $a r (R Q C) = \frac{3}{8} a r (, A B C)$
3) $a r (P B Q) = a r (A R C)$

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Solution

Let ABC is a triangle. P and Q are the mid points of AB and BC respectively and R is the mid-point of AP.
Join PQ, QR, AQ, PC, RC as shown in the figure.
Now we know that median of a triangle divides it into two triangles of equal areas.
In triangle CAP, Cr is the mid point.
So $A r e a (Δ C R A) = (\frac{1}{2}) \times A r e a (Δ C A P)$ ....1
Again in triangle CAB, CP is the mid point.
So $A r e a (Δ C A P) = (\frac{1}{2}) \times A r e a (Δ C P B)$ ....2
from eqaution 1 and 2, we get
$A r e a (Δ C A P) = (\frac{1}{2}) \times A r e a (Δ C P B)$ .....3
Again in triangle PBC, PQ is the mid point.
So $(\frac{1}{2}) \times A r e a (Δ C P B) = A r e a (Δ P B Q)$ ....4
From equation 3 and 4, we get
$A r e a (Δ A R C) = A r e a (Δ P B Q)$ ....5
Now QP, And QR are the medians of triangle QAB and QAP respectively
So $A r e a (Δ Q A P) = A r e a (Δ Q B P)$ ............6
and $A r e a (Δ Q A P) = 2 \times A r e a (Δ Q R P)$ ............7
from equation 6 and 7, we get
$A r e a (Δ P R Q) = (\frac{1}{2}) \times A r e a (Δ P B Q)$ ............8
from equation 5 and 8, we get
$A r e a (Δ P R Q) = (\frac{1}{2}) \times A r e a (Δ A R C)$
Now CR is the median of triangle CAP
So $A r e a (Δ A R C) = (\frac{1}{2}) \times A r e a (Δ C A P)$
$= (\frac{1}{2}) \times (\frac{1}{2}) \times A r e a (Δ A B C)$
$= (\frac{1}{4}) \times A r e a (Δ A B C)$
Again RQ is the median of triangle RBC
So $A r e a (Δ R Q C) = (\frac{1}{2}) \times A r e a (Δ R B C)$
$= (\frac{1}{2}) \times A r e a (Δ A B C) - (\frac{1}{2}) \times A r e a (Δ A R C)$
$= (\frac{1}{2}) \times A r e a (Δ A B C) - (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times A r e a (Δ A B C)$
$= (\frac{1}{2}) \times A r e a (Δ A B C) - (\frac{1}{8}) \times A r e a (Δ A B C)$
$= (\frac{3}{8}) \times A r e a (Δ A B C)$
So $A r e a (Δ R Q C) = (\frac{3}{8}) \times A r e a (Δ A B C)$

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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)

△ A B C, P

Q

A B

B C

R

A P

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

a r (Δ P B Q) = a r (Δ A R C)

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

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

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