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In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1

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In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1  [#permalink]

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New post 19 Sep 2019, 21:25
1
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A
B
C
D
E

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  75% (hard)

Question Stats:

54% (02:11) correct 46% (02:49) wrong based on 65 sessions

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Re: In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1  [#permalink]

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New post 19 Sep 2019, 21:54
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Re: In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1  [#permalink]

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New post 19 Sep 2019, 22:10
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triangles, ABD, PQD are similar , AB=3CD , BD= BQ + QD

so AB/PQ = BD/QD



CDB,PQB are simialr

CD/PQ = BD/QB

SOLVING above we will get PQ/CD = 3/4
I.E CD/PQ = 1/0.75

OA:B
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Re: In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1  [#permalink]

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New post 19 Sep 2019, 23:20
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triangle BQP is similar to triangle BDC (AAA similarity)

\(\frac{BD}{BQ}\) = \(\frac{CD}{PQ}\)

BD = \(\frac{CD*BQ}{PQ}\)---(1)

triangle DBA is similar to triangle DQP (AAA similarity)

\(\frac{BD}{QD}\) = \(\frac{AB}{PQ}\)

BD = \(\frac{AB*QD}{PQ}\)---(2)

(1) = (2) we get

\(\frac{CD*BQ}{PQ}\) = \(\frac{AB*QD}{PQ}\)

\(\frac{BD}{BQ}\) = \(\frac{CD}{AB}\)

\(\frac{BD}{BQ}\) = \(\frac{1}{3}\) (given \(\frac{AB}{CD}\)=\(\frac{3}{1}\))

from triangle BQP and triangle BDC

\(\frac{BD}{BQ}\) = \(\frac{CD}{PQ}\)

so \(\frac{CD}{PQ}\)=\(\frac{1}{3}\)

E is the correct answer
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Re: In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1  [#permalink]

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New post 20 Sep 2019, 00:32
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B is correct

According to Thales theorem
CD/PQ=BD/BQ=BQ+QD/BQ=4/3
CD=1then PQ=0.75

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Re: In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1  [#permalink]

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New post 20 Sep 2019, 01:05
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In the given figure, since \(∠ABD = ∠CDB = ∠PQD = 90°\), \(AB || PQ || CD\).
BC and AD are two lines intersecting these lines. Thus, \(Δ ABP ˜ Δ DCP\)(AAA congruency) where \(∠A = ∠D, ∠B = ∠C\) and ∠P is common.

Hence
\(\frac{AB}{CD} = \frac{BP}{CP} = \frac{AP}{PD} = \frac{3}{1}\) → ①

Also \(Δ BQP ˜ Δ BDC\)(AAA congruency) where \(∠Q = ∠D, ∠P = ∠C\) and ∠B is common.
Thus, \(\frac{BQ}{BD} = \frac{BP}{BC} = \frac{PQ}{CD}\)

 \(\frac{PQ}{CD} = \frac{BP}{(BP + PC)}\)
 \(= \frac{3PC}{(3PC + PC)}\) [From ① BP = 3PC]
 \(= \frac{3PC}{4PC}\)
 \(= \frac{3}{4}\)
 \(= \frac{0.75}{1}\)

Answer (B).
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Re: In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1  [#permalink]

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New post 20 Sep 2019, 05:52
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In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1, the ratio of CD:PQ

A. 1:0.69
B. 1:0.75
C. 1:0.72
D. 1:0.50
E. None of these


ABD and CDB have the same base and their sides are in proportion; since, AB:CD=3:1, then let their base BD=4;
now, imagine this diagram in a plane, with AB at the origin;
point A(0,3) B(0,0) C(4,1) D(4,0), and point P is the place where both lines AD and BC intersect;
line AD: m=(0-3)/(4-0)=-3/4…y=-3x/4+b…1=-3(4)/4+b…b=3…y=-3x/4+3
line BC: m=(1-0)/(4-0)=1/4…y=x/4+b…1=(4)/4+b…b=0…y=x/4
equate lines: y=x/4 and y=-3x/4+3… x/4=-3x/4+3…x=3…y=x/4=(3/4)/4=3/4
point P(3,3/4) so CD:PQ=1:3/4

Answer (B)
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Re: In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1  [#permalink]

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New post 20 Sep 2019, 05:53
In the diagram given above,
∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1 --> the ratio of CD:PQ=???

∆ABD and ∆PQD are similar triangles
--> \(\frac{AB}{PQ}\)=\(\frac{BD}{QD}\) --> AB=\(\frac{PQ*BD}{QD}\)

∆CDB and ∆PQB are similar triangles
--> \(\frac{CD}{PQ}\)=\(\frac{BD}{(BD-QD)}\) --> CD=\(\frac{PQ*BD}{(BD-QD)}\)

\(\frac{AB}{CD}=\frac{3}{1}=(\frac{PQ*BD}{QD})/\frac{PQ*BD}{(BD-QD)}= \frac{(BD-QD)}{QD}\)
3=\(\frac{(BD-QD)}{QD}\)
--> BD=4*QD

CD=\(\frac{PQ*4*QD}{3*QD}\)=\(\frac{PQ*4}{3}\)
\(\frac{CD}{PQ}=\frac{4}{3}=\frac{1}{(3/4)}\)=1:0.75

The answer is B.
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Re: In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1  [#permalink]

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New post 21 Sep 2019, 19:22
The triangles ABP and PCD are similar becasue they have the same sized angles. Their heights will be in the same ratio as AB and CD are in. They heights are BQ and QD.The ratio of BQ :QD =3:1 since the ratio of their bases AB:CD is 3:1. Also one can see that CBD is similar to PBQ,meaning their bases which are the heights of ABP and PCD will have to be in the ratio of 4 :3 This allows us to say that the ratio of CD to PQ is also 4:3.

Answer is B
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Re: In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1  [#permalink]

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New post 28 Oct 2019, 23:43
Bunuel wrote:

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In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1, the ratio of CD:PQ

A. 1:0.69
B. 1:0.75
C. 1:0.72
D. 1:0.50
E. None of these

Attachment:
diagram.PNG


Since triangles BCD and BPQ are similar
CD:PQ = BD:BQ ----------1

Since triangles ABD and PQD are similar
AB:PQ = BD:QD -----------2

(2)/(1)
AB:CD = BQ:QD = 3:1
BD:BQ = 4:3 = 1: .75

IMO B
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Re: In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1   [#permalink] 28 Oct 2019, 23:43
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In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1

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