Math, asked by jafarsadik3, 1 year ago

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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Answered by CarlynBronk
6

Solution:

A Parallelogram A B CD in which , two points P and Q lies on side AB and CD Such that AP: PB=3:2 and C Q:Q D=4:1.Diagonal AC is intersected by Segment P Q at Point R.

Consider two triangles in the parallelogram ,ΔA RP and Δ C R Q

∠ARP=∠CRQ→→[Vertically Opposite Angles]

∠PAR=∠RCQ→→[ Alternate Interior Angles]

ΔA RP ~ Δ C R Q→→[AA Similarity]

Keep in Mind , when two triangles are similar , their corresponding sides are proportional.

\frac{AP}{CQ}=\frac{AR}{CR}

AB= CD→→ [Opposite sides of a Parallelogram are equal]

Let AP and PB be 3 k and 2 k respectively.and CQ and QD be 4 x and x respectively.

3 k + 2 k = 3 x + 2 x

5 k= 5 x

k = x

\frac{3}{4}=\frac{AR}{RC}\\\\  \frac{4}{3}+1=\frac{CR}{RA}+1 \\\\ \frac{7}{3}=\frac{AC}{RA}\\\\  AR= \frac{3}{7} \times AC




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