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

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

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

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