Question

ABCD is a parallelogram. AB is divided at P and CD at Q such that AP : PB is 3 : 2 and CQ : QD is 4 : 1. if PQ meets AC at R, prove that AR= 3/7 AC ?

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Answer 1

Answer

$AR = \frac{3}{7} AC \quad Proved$

Step-by-step explanation:

Let, AB = CD = 5x according to the parallelogram definition that opposite sides are equal.

AP = 3x, PB = 2x  and  CQ = 4x, QD  = x

$\Rightarrow \quad \tri{ARP} \sim \tri{CRQ}$

Now, $\frac{AP}{CQ}\:=\:\frac{AR}{RC}$

$\frac{3x}{4x}\:=\:\frac{AR}{RC}$

$\frac{AR}{RC}\:=\:\frac{3}{4}$

According to this fraction, we have:

AR = 3 and RC = 4, so that AC = 3+4 = 7

Finally, $\frac{AR}{AC}\:=\:\frac{3}{7}$

$\Rightarrow \quad AR\:=\:\frac{3}{7}AC$

Answer 2

Answer:

Step-by-step explanation:

Let, AB = CD = 5x according to the parallelogram definition that opposite sides are equal.

AP = 3x, PB = 2x  and  CQ = 4x, QD  = x

Now,

According to this fraction, we have:

AR = 3 and RC = 4, so that AC =3+4=7

finally you have your answer as

AR=3\7AB

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