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In the triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropp

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In the triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropp  [#permalink]

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New post 16 Apr 2010, 06:03
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In the triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropped from B, meets the side AC at D. A circle of radius BD (with center B) is drawn. If the circle cuts AB and BC at P and Q respectively, the AP:QC is equal to

A. 1:1
B. 3:2
C. 4:1
D. 3:8

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Re: In the triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropp  [#permalink]

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New post 16 Apr 2010, 06:40
1
is answer choice 4?

Well since you know the radis from 36-25 = 11 r= sqrt(11) you can find Ap QC with 6-sqrt(11) and 8 - sqrt(11). Also, another thing to notice is that PB and BQ will be the same value since they are radius for circle B. since AB is shorter than BC you know the ratio will be a smaller value to a larger value so 4 seems to fit the bill..
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Re: In the triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropp  [#permalink]

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New post 16 Apr 2010, 07:08
1
if we can know the length of BD we can find out AP and QC
since BP=BQ=BD=radius
now there are two ways to find out the length of BD
one is using pythagoras theorem another is using area(hero's formula)
pythagoras :
BD^2=AB^2-AD^2=BC^2-DC^2{DC=AC-AD}
using the above eqn we will get BD=24/5
AP=6-24/5=6/5
QC=8-24/5=16/5
ratio=6/16=3/8

hero's formula
s=a+b+c/2=(6+8+10)/2=12
area=Sqrt(s(s-a)(s-b)(s-c)=24
area is also=1/2 * BD*AC=5BD
so BD=24/5
rest is same
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Re: In the triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropp  [#permalink]

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New post 16 Apr 2010, 11:10
Hi,

I tried to first draw the diagram and when u take a close look u can easily tell what would be the answer 8-) .
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Re: In the triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropp  [#permalink]

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New post 16 Apr 2010, 11:33
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1
:idea: apparently, the answer can be found just by observing the choices.

we know that BC > AB => (BC - BD) > (AB - BD) => QC > AP

ie, the only logical choice is 4.
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Re: In the triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropp  [#permalink]

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New post 17 Apr 2010, 05:26
shaselai wrote:
is answer choice 4?

Well since you know the radis from 36-25 = 11 r= sqrt(11) you can find Ap QC with 6-sqrt(11) and 8 - sqrt(11). Also, another thing to notice is that PB and BQ will be the same value since they are radius for circle B. since AB is shorter than BC you know the ratio will be a smaller value to a larger value so 4 seems to fit the bill..



Where you get 25 from ........... (above in red)
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Re: In the triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropp  [#permalink]

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New post 17 Apr 2010, 05:29
xcusemeplz2009 wrote:
if we can know the length of BD we can find out AP and QC
since BP=BQ=BD=radius
now there are two ways to find out the length of BD
one is using pythagoras theorem another is using area(hero's formula)
pythagoras :
BD^2=AB^2-AD^2=BC^2-DC^2{DC=AC-AD}
using the above eqn we will get BD=24/5

AP=6-24/5=6/5
QC=8-24/5=16/5
ratio=6/16=3/8

hero's formula
s=a+b+c/2=(6+8+10)/2=12
area=Sqrt(s(s-a)(s-b)(s-c)=24
area is also=1/2 * BD*AC=5BD
so BD=24/5
rest is same



how do we know the value of AD and DC?
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Re: In the triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropp  [#permalink]

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New post 17 Apr 2010, 13:19
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ans should be 3/8

as per question abc is right angled at B and using similar triangle in triangle ABC and ABD

we get x = 24/5 , join line DP and DQ , both will be the radius, thus x = DP = DQ

so AP = 6 -24/5 = 6/5

QC = 8-25/5 = 16/5

ratio AP:QC = 6/16 = 3/8
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Re: In the triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropp  [#permalink]

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New post 17 Apr 2010, 13:54
i still dont get it
Can you please explain in more details how you get 24/5

This question is testing my patience!!!!
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Re: In the triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropp  [#permalink]

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New post 17 Apr 2010, 14:03
Draw a triangle ABC with b =90 and draw a perpendicular from b to D as given

now take BD = x

Since ABC and ABD are similar triangles.
We take ratio of the sides of equal angles
angle opp x is 90-c and opp BC is 90-c
similarly angle opp 90 in ABD is AB and in ABC is AC

x/BC = AB/AC => X = 24/5

I hope this helps.
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Re: In the triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropp  [#permalink]

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New post 17 Apr 2010, 19:27
ykaiim wrote:
In the triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropped from B, meets the side AC at D. A circle of radius BD (with center B) is drawn. If the circle cuts AB and BC at P and Q respectively, the AP:QC is equal to
1. 1:1
2. 3:2
3. 4:1
4. 3:8

Is there any smarter/shorter way?

try this :)
since it is right angle triangle, altitude BD is given by formula BD = BC*AB/AC => BD = 6*8/10 = 24/5
also from question stem, BP = BQ = BD = 24/5 ( radius of circle )
hence, AP = AB-BP = 6-24/5 = 6/5 and QC = BC-BQ = 8-24/5 = 16/5
so, AP:QC = 6/5:16/5 = 3:8
for the formula and picture please refer following link:
http://gmatclub.com/forum/what-to-remember-in-gmat-quant-all-time-92840.html
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Re: In the triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropp  [#permalink]

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New post 17 Apr 2010, 23:47
hardnstrong wrote:
xcusemeplz2009 wrote:
if we can know the length of BD we can find out AP and QC
since BP=BQ=BD=radius
now there are two ways to find out the length of BD
one is using pythagoras theorem another is using area(hero's formula)
pythagoras :
BD^2=AB^2-AD^2=BC^2-DC^2{DC=AC-AD}
using the above eqn we will get BD=24/5

AP=6-24/5=6/5
QC=8-24/5=16/5
ratio=6/16=3/8
how do we know the value of AD and DC?

no need to find AD and DC
we know that AD+DC=10 -----> DC=10-AD and have eqn AB^2-AD^2=BC^2-DC^2 put DC=10-AD
or AB^2-AD^2=BC^2-(10-AD)^2----->solving we get BD=24/5
HTH
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Re: In the triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropp  [#permalink]

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New post 18 Apr 2010, 04:59
ykaiim wrote:
In the triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropped from B, meets the side AC at D. A circle of radius BD (with center B) is drawn. If the circle cuts AB and BC at P and Q respectively, the AP:QC is equal to
1. 1:1
2. 3:2
3. 4:1
4. 3:8

Is there any smarter/shorter way?


First of all we should spot that ABC is a right triangle with right angle at B, as 6-8-10 is a Pythagorean triplet and AC, the longest side, is the hypotenuse.

Next in right triangle: \(\frac{height}{leg_1}=\frac{leg_2}{hypotenuse}\) --> \(\frac{BD}{AB}=\frac{BC}{AC}\) --> \(\frac{BD}{6}=\frac{8}{10}\) --> \(BD=\frac{24}{5}\).

Now, \(BD=BP=BQ=radius=\frac{24}{5}\).

\(\frac{AP}{QC}=\frac{BA-BP}{BC-BQ}=\frac{6-\frac{24}{5}}{8-\frac{24}{5}}=\frac{3}{8}\).

Answer: D (\(\frac{3}{8}\))
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Re: In the triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropp  [#permalink]

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New post 19 Apr 2010, 08:29
One more approach to solve such problems:

Triangle ABC is a right angled triangle. So, we can use the area calculation method:
1/2×BC×AB=1/2×BD×AC >>> 6×8=BD×10

This results BD = 4.8 and BP= BQ= 4.8
So, AP=AB–BP=6–4.8=1.2 and similarly, CQ=BC–BQ=8–4.8=3.2.

Finally, we can derive the result by taking ratio:
AP:CQ=1.2:3.2=3:8
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Re: In the triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropp  [#permalink]

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New post 20 Apr 2010, 04:05
Got it now - Thanks Bunuel (Didnt know about this property of right triangle)
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Re: In the triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropp  [#permalink]

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New post 13 Oct 2011, 14:15
Ans 4, as QC is greater than AP
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Re: In the triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropp  [#permalink]

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New post 18 Oct 2011, 04:04
let r be the radius.

AP=6-x; CQ=8-x
CQ is greater than AP so the only possible ans is 3/8
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Re: In the triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropp  [#permalink]

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Re: In the triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropp   [#permalink] 30 Dec 2018, 13:26
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