In the given figure- ABCD is a parallelogram- Points P and Q on BC trisect BC-Prove that - i ar - APR - ar - DRQii ar - DPQ - 16 ar ABCDiii arARPB - arDRQC

You visited us 1 times! Enjoying our articles? Unlock Full Access!

A Parallelogram and a Triangle between the Same Parallels

In the given ...

Question

In the given figure, ABCD is a parallelogram. Points P and Q on BC trisect BC.
Prove that :
i) $a r (Δ A P R) = a r (Δ D R Q)$
ii) $a r (Δ D P Q) = \frac{1}{6} a r (A B C D)$
iii) $a r (A R P B) = a r (D R Q C)$

Open in App

Solution

$B P = P Q + Q C = \frac{1}{3} B C$
$A B C D$ is a parallelogram.
(i) $a r (△ A P R) = a r (△ D R Q)$ .
ina parallellogram, triangle inscribed with the
same bak have equal access.
$\Rightarrow a r (△ A P D) = a r (△ A Q D)$
$\Rightarrow a r (△ A R D) + a r (△ A P R) = a r (△ A R D) + a r (△ Q R D)$
$a r (△ A P R) = a r (D R Q)$ .....(1)
$\Rightarrow (△ D P Q) \frac{1}{2} \times h \times (P Q) = \frac{1}{2} \times h r (\frac{B C}{3})$
h=height of parallelogram.
$a r (△ D P Q) \frac{(h \times B C)}{6}$
area of parallellogram $A B C D = (h \times B C)$
$∴ \Rightarrow a r (△ D P Q) \frac{1}{6} \times a r (A B C D)$
(iii) $a r (△ A B P) = \frac{1}{2} \times h \times P B = \frac{1}{2} \times h \times \frac{B C}{3} = \frac{1}{6} (h \times B C)$
$a r (△ D Q C) \frac{1}{2} \times h \times (Q C) = \frac{1}{2} \times h \times \frac{B C}{3} = \frac{1}{6} (h \times B C)$
$\Rightarrow a r (△ A B P) = \frac{1}{6} (h \times B C) = a r (△ D Q C)$
$a r (△ A B P) = a r (△ D Q C)$ ....(3)
add (1) and (3)
$a r (A B P) a r (△ A P R) = a r (△ D Q C) + a r (D Q R)$
$a r (A R P B) = a r (D R Q C)$

Suggest Corrections

Similar questions

A B C D

is a parallelogram, points

P

and

Q

B C

trisects

B C

. Prove that :-

a r (△ A P Q) = a r (△ D P Q) = \frac{1}{6} a r (p a r a l l e l o g r a m A B C D)

Q. In the adjoining figure

, A B C D

is a parallelogram. Any line through A cuts

D C

at a point

P

and

B C

produced at

Q .

Prove that :

a r (Δ B P C) = a r (Δ D P Q) .

1293878_f8df6de59fdb466294cd8e3124285b76.1293878-Q

The given figure shows parallelogram ABCD. Points P and Q trisects the side AB.
To prove: Area of $△$ DPQ = $\frac{1}{6}$ × Area of (parallelogram ABCD)

In the figure, given below, ABCD is a parallelogram. P is a point on BC such that BP : PC = 1 : 2. DP produced meets AB produced at Q. Given the area of triangle CPQ = 20 $c m^{2}$ .
Calculate : (i) area of triangle CDP, (ii) area of parallelogram ABCD.

Q. The given figure shows a parallelogram ABCD. Point P and Q trisect the side AB. Show that : area (

Δ D P Q

) = area (BQC) =

\frac{1}{6} \times a r e a (/ /_{g m} A B C D)

Join BYJU'S Learning Program

In the given figure, ABCD is a parallelogram. Points P and Q on BC trisect BC.Prove that : i) ar(ΔAPR)=ar(ΔDRQ)ii) ar(ΔDPQ)=16ar(ABCD)iii) ar(ARPB)=ar(DRQC)

In the given figure, ABCD is a parallelogram. Points P and Q on BC trisect BC.
Prove that :
i) $a r (Δ A P R) = a r (Δ D R Q)$
ii) $a r (Δ D P Q) = \frac{1}{6} a r (A B C D)$
iii) $a r (A R P B) = a r (D R Q C)$