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# In triangle ABC, if BC = 3 and AC = 4, then what is the

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In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink]

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Updated on: 19 Dec 2012, 02:41
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In triangle ABC, if BC = 3 and AC = 4, then what is the length of segment CD?

A. 3
B. 15/4
C. 5
D. 16/3
E. 20/3

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Originally posted by enigma123 on 01 Feb 2012, 16:55.
Last edited by Bunuel on 19 Dec 2012, 02:41, edited 2 times in total.
Edited the question and added the figure
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Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink]

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26 Feb 2012, 09:42
Hi

I need a quick clarification on the concept of perpendicular bisector.
With a perpendicular bisector, the bisector always crosses the line segment at right angles
If any line cuts another line at 90 then it should be a perpendicular bisector right - i.e it divided the line segment into equal halves at 90 ?

So here BC should be the perpendicular bisector and the AC=CD=3 right ?

Please let me know what am missing here.

I do understand the explanations in the other thread mentioned,but can someone clarify as to why AC is not the perpendicular bisector ?
Math Expert
Joined: 02 Sep 2009
Posts: 45423
Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink]

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26 Feb 2012, 12:07
5
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Expert's post
6
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rvinodhini wrote:
Hi

I need a quick clarification on the concept of perpendicular bisector.
With a perpendicular bisector, the bisector always crosses the line segment at right angles
If any line cuts another line at 90 then it should be a perpendicular bisector right - i.e it divided the line segment into equal halves at 90 ?

So here BC should be the perpendicular bisector and the AC=CD=3 right ?

Please let me know what am missing here.

I do understand the explanations in the other thread mentioned,but can someone clarify as to why AC is not the perpendicular bisector ?

A perpendicular bisector is a line which cuts a line segment into two equal parts at 90°. So AC to be a perpendicular bisector of BD it must not only cut it at 90° (which it does) but also cut it into two equal parts. Now, in order AC to cut BD into two equal parts right triangle ABD must be isosceles, which, as it turns out after some math, it is not.

Complete solution:
In triangle ABC, if BC = 3 and AC = 4, then what is the length of segment CD?
A. 3
B. 15/4
C. 5
D. 16/3
E. 20/3
Attachment:

splittingtriangle.jpg [ 4.22 KiB | Viewed 295338 times ]

Important property: perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.

Thus, the perpendicular AC divides right triangle ABD into two similar triangles ACB and DCA (which are also similar to big triangle ABD). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles marked with red and blue on the diagram).

So, $$\frac{CD}{AC}=\frac{AC}{BC}$$ --> $$\frac{CD}{4}=\frac{4}{3}$$ --> $$CD=\frac{16}{3}$$.

For more on this subject please check Triangles chapter of Math Book: math-triangles-87197.html

Hope it helps.
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Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink]

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26 Feb 2012, 21:09
Now I get it ..wonderful explanation !!
thanks
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Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink]

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27 Feb 2012, 09:11
thankx for the explanation..
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In triangle ABC to the right, if BC = 3 and AC = 4, then [#permalink]

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21 Apr 2012, 09:26
MBAhereIcome wrote:
all three triangles abc, acd & abd are similar.
so, $$\frac{4}{3} = \frac{cd}{4}$$ ... cd = $$\frac{16}{3}$$

it can't be $$\frac{4}{3} = \frac{4}{cd}$$ ... cd = 3, because in that case $$5^2+5^2 = 6^2$$ is not true

If the 3 triangles are proportional, why can't I solve using the ratio:

$$\frac{AD}{AB} = \frac{AB}{(X+3)}$$
$$\frac{4}{5} = \frac{5}{(X+3)}$$
4(X+3) = 25
4X + 12 = 25
X = $$\frac{13}{4}$$
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Posts: 45423
Re: In triangle ABC to the right, if BC = 3 and AC = 4, then [#permalink]

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21 Apr 2012, 10:35
pubchum wrote:
MBAhereIcome wrote:
all three triangles abc, acd & abd are similar.
so, $$\frac{4}{3} = \frac{cd}{4}$$ ... cd = $$\frac{16}{3}$$

it can't be $$\frac{4}{3} = \frac{4}{cd}$$ ... cd = 3, because in that case $$5^2+5^2 = 6^2$$ is not true

If the 3 triangles are proportional, why can't I solve using the ratio:

$$\frac{AD}{AB} = \frac{AB}{(X+3)}$$
$$\frac{4}{5} = \frac{5}{(X+3)}$$
4(X+3) = 25
4X + 12 = 25
X = $$\frac{13}{4}$$

Attachment:

splittingtriangle.jpg [ 4.22 KiB | Viewed 256240 times ]

In similar triangles the ratio of the corresponding sides are equal (corresponding sides are the sides opposite the same angles marked with red and blue on the diagram).

The ratios you are equating are not of corresponding sides. If you want to equate AD/AB then it should be AD/AB=AC/BC --> AD/5=4/3 --> AD=20/3. Also AD/AB=CD/AC --> (20/3)/5=CD/4 --> CD=16/3.

I merged this thread with an earlier discussion of the same question, so check this post: in-triangle-abc-if-bc-3-and-ac-4-then-what-is-the-126937.html#p1050155 it might help to clear your doubts.

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Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink]

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17 May 2012, 05:46
Hi Bunuel,

How did you infer that angle BAC = angle ADC. Could you please explain that part.. I know this can be proved thro similarity, however I wanted to understand from your concept, which you have mentioned below.
Pls explain.

Thanks
H
Bunuel wrote:
rvinodhini wrote:
Hi

I need a quick clarification on the concept of perpendicular bisector.
With a perpendicular bisector, the bisector always crosses the line segment at right angles
If any line cuts another line at 90 then it should be a perpendicular bisector right - i.e it divided the line segment into equal halves at 90 ?

So here BC should be the perpendicular bisector and the AC=CD=3 right ?

Please let me know what am missing here.

I do understand the explanations in the other thread mentioned,but can someone clarify as to why AC is not the perpendicular bisector ?

A perpendicular bisector is a line which cuts a line segment into two equal parts at 90°. So AC to be a perpendicular bisector of BD it must not only cut it at 90° (which it does) but also cut it into two equal parts. Now, in order AC to cut BD into two equal parts right triangle ABD must be isosceles, which, as it turns out after some math, it is not.

Complete solution:
In triangle ABC, if BC = 3 and AC = 4, then what is the length of segment CD?
A. 3
B. 15/4
C. 5
D. 16/3
E. 20/3
Attachment:
splittingtriangle.jpg

Important property: perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.

Thus, the perpendicular AC divides right triangle ABD into two similar triangles ACB and DCA (which are also similar to big triangle ABD). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles marked with red and blue on the diagram).

So, $$\frac{CD}{AC}=\frac{AC}{BC}$$ --> $$\frac{CD}{4}=\frac{4}{3}$$ --> $$CD=\frac{16}{3}$$.

For more on this subject please check Triangles chapter of Math Book: math-triangles-87197.html

Hope it helps.
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Joined: 02 Sep 2009
Posts: 45423
Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink]

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17 May 2012, 05:59
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Expert's post
imhimanshu wrote:
Hi Bunuel,

How did you infer that angle BAC = angle ADC. Could you please explain that part.. I know this can be proved thro similarity, however I wanted to understand from your concept, which you have mentioned below.
Pls explain.

Thanks
H

Attachment:

splittingtriangle.jpg [ 4.22 KiB | Viewed 255916 times ]

<B+<D+<A=180, since <A=90 then <B+<D=90;
Similarly in triangle ABC: <B+<BAC=90 since <B=90-<D then (90-<D)+<BAC=90 --> <BAC=<D.

Hope it's clear.
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Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink]

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17 May 2012, 07:36
Thanks for the solution. I just wanted to know whether it requires any calculation or is it just a corollary of the property described above. Its clear now.

Bunuel wrote:
imhimanshu wrote:
Hi Bunuel,

How did you infer that angle BAC = angle ADC. Could you please explain that part.. I know this can be proved thro similarity, however I wanted to understand from your concept, which you have mentioned below.
Pls explain.

Thanks
H

Attachment:
splittingtriangle.jpg

<B+<D+<A=180, since <A=90 then <B+<D=90;
Similarly in triangle ABC: <B+<BAC=90 since <B=90-<D then (90-<D)+<BAC=90 --> <BAC=<D.

Hope it's clear.
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Posts: 45423
Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink]

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20 Jun 2013, 06:39
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink]

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20 Jun 2013, 07:18
3
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Well i have a problem with similar triangles coz i sometimes make mistakes on the common sides. hence alternative approach for this problem..

Considering Triangle ACD - AC^2 + CD^2 = AD^2

Considering Triangle ABD - AB = sqrt(4^2 + 3^2) = 5 (Pythagorean triplet so you dont really have to do the math on the test)

25 + ac^2 + cd^2 = (3 + cd)^2

=> 25+16+cd^2= 9 + 6cd +cd^2

=>32 = 6cd

cd = 16/3
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Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink]

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22 Jun 2013, 08:23
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enigma123 wrote:
Attachment:
Triangle.jpg
In triangle ABC, if BC = 3 and AC = 4, then what is the length of segment CD?

A. 3
B. 15/4
C. 5
D. 16/3
E. 20/3

Learn this property:

AC^2=BC*DC

=>DC=16/3
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Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink]

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23 Jun 2013, 00:00
1
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In the triangles ABD, <CAB= <CDA
----> tan <CAB = tan <CD
----> 3/4=4/CD
----> CD=16/3
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Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink]

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12 Mar 2015, 00:43
2
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the given solution is neat and simple but it didnt strike me when solving
anyways here is another alternate way though i admit its lengthy calculation
let us assume CD=x
(x+3)^2=25+(x^2+16)
x=32/6=16/3

(D)
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Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink]

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24 Mar 2015, 17:13
Bunuel wrote:
rvinodhini wrote:
Hi

I need a quick clarification on the concept of perpendicular bisector.
With a perpendicular bisector, the bisector always crosses the line segment at right angles
If any line cuts another line at 90 then it should be a perpendicular bisector right - i.e it divided the line segment into equal halves at 90 ?

So here BC should be the perpendicular bisector and the AC=CD=3 right ?

Please let me know what am missing here.

I do understand the explanations in the other thread mentioned,but can someone clarify as to why AC is not the perpendicular bisector ?

A perpendicular bisector is a line which cuts a line segment into two equal parts at 90°. So AC to be a perpendicular bisector of BD it must not only cut it at 90° (which it does) but also cut it into two equal parts. Now, in order AC to cut BD into two equal parts right triangle ABD must be isosceles, which, as it turns out after some math, it is not.

Complete solution:
In triangle ABC, if BC = 3 and AC = 4, then what is the length of segment CD?
A. 3
B. 15/4
C. 5
D. 16/3
E. 20/3
Attachment:
splittingtriangle.jpg

Important property: perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.

Thus, the perpendicular AC divides right triangle ABD into two similar triangles ACB and DCA (which are also similar to big triangle ABD). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles marked with red and blue on the diagram).

So, $$\frac{CD}{AC}=\frac{AC}{BC}$$ --> $$\frac{CD}{4}=\frac{4}{3}$$ --> $$CD=\frac{16}{3}$$.

For more on this subject please check Triangles chapter of Math Book: math-triangles-87197.html

Hope it helps.

This part is clear: triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles)

However, how does one determine which angles are equal? Except 90 degree angles of both triangles, i could not seem to follow how exactly other angles became equal?
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Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink]

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24 Mar 2015, 21:22
earnit wrote:
Bunuel wrote:
rvinodhini wrote:
Hi

I need a quick clarification on the concept of perpendicular bisector.
With a perpendicular bisector, the bisector always crosses the line segment at right angles
If any line cuts another line at 90 then it should be a perpendicular bisector right - i.e it divided the line segment into equal halves at 90 ?

So here BC should be the perpendicular bisector and the AC=CD=3 right ?

Please let me know what am missing here.

I do understand the explanations in the other thread mentioned,but can someone clarify as to why AC is not the perpendicular bisector ?

A perpendicular bisector is a line which cuts a line segment into two equal parts at 90°. So AC to be a perpendicular bisector of BD it must not only cut it at 90° (which it does) but also cut it into two equal parts. Now, in order AC to cut BD into two equal parts right triangle ABD must be isosceles, which, as it turns out after some math, it is not.

Complete solution:
In triangle ABC, if BC = 3 and AC = 4, then what is the length of segment CD?
A. 3
B. 15/4
C. 5
D. 16/3
E. 20/3
Attachment:
splittingtriangle.jpg

Important property: perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.

Thus, the perpendicular AC divides right triangle ABD into two similar triangles ACB and DCA (which are also similar to big triangle ABD). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles marked with red and blue on the diagram).

So, $$\frac{CD}{AC}=\frac{AC}{BC}$$ --> $$\frac{CD}{4}=\frac{4}{3}$$ --> $$CD=\frac{16}{3}$$.

For more on this subject please check Triangles chapter of Math Book: math-triangles-87197.html

Hope it helps.

This part is clear: triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles)

However, how does one determine which angles are equal? Except 90 degree angles of both triangles, i could not seem to follow how exactly other angles became equal?

Both smaller triangles are similar to the large triangle. So they are similar to each other too.

Angle BAD = BCA (90 degrees)
and angle B is common in both
So by AA, triangles BAD and BCA are similar

angle BAD = ACD (90 degrees)
and angle D is common in both
So by AA, triangles BAD and ACD are similar

So triangle BAD is similar to triangle BCA and ACD so triangle BCA is similar to triangle ACD too.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 06 Mar 2015 Posts: 13 Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink] ### Show Tags 27 Mar 2015, 10:54 VeritasPrepKarishma wrote: Bunuel wrote: rvinodhini wrote: Hi I need a quick clarification on the concept of perpendicular bisector. With a perpendicular bisector, the bisector always crosses the line segment at right angles If any line cuts another line at 90 then it should be a perpendicular bisector right - i.e it divided the line segment into equal halves at 90 ? So here BC should be the perpendicular bisector and the AC=CD=3 right ? Please let me know what am missing here. I do understand the explanations in the other thread mentioned,but can someone clarify as to why AC is not the perpendicular bisector ? A perpendicular bisector is a line which cuts a line segment into two equal parts at 90°. So AC to be a perpendicular bisector of BD it must not only cut it at 90° (which it does) but also cut it into two equal parts. Now, in order AC to cut BD into two equal parts right triangle ABD must be isosceles, which, as it turns out after some math, it is not. Complete solution: In triangle ABC, if BC = 3 and AC = 4, then what is the length of segment CD? A. 3 B. 15/4 C. 5 D. 16/3 E. 20/3 Attachment: splittingtriangle.jpg Important property: perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. Thus, the perpendicular AC divides right triangle ABD into two similar triangles ACB and DCA (which are also similar to big triangle ABD). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles marked with red and blue on the diagram). So, $$\frac{CD}{AC}=\frac{AC}{BC}$$ --> $$\frac{CD}{4}=\frac{4}{3}$$ --> $$CD=\frac{16}{3}$$. Answer: D. For more on this subject please check Triangles chapter of Math Book: math-triangles-87197.html Hope it helps. This part is clear: triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles) However, how does one determine which angles are equal? Except 90 degree angles of both triangles, i could not seem to follow how exactly other angles became equal? Both smaller triangles are similar to the large triangle. So they are similar to each other too. In triangles BAD and BCA, Angle BAD = BCA (90 degrees) and angle B is common in both So by AA, triangles BAD and BCA are similar Similarly, in triangles BAD and ADC, angle BAD = ACD (90 degrees) and angle D is common in both So by AA, triangles BAD and ACD are similar So triangle BAD is similar to triangle BCA and ACD so triangle BCA is similar to triangle ACD too. Hi Karishma, Thanks for explaining this. I understand how 3 triangles are similar to each other. However how do we determine which side is similar to which in order to set up the ratio. for example, if we take two smaller triangles - Triangle ABC and ADC....I think side AB is corresponding to AD, BC is corresponding to CD and AC is common. Is that correct? if yes, how do I find which side is corresponding in the smaller triangle with respect to the bigger triangle? Many thanks in advance. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8079 Location: Pune, India Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink] ### Show Tags 30 Mar 2015, 01:40 1 This post received KUDOS Expert's post gmatkiller88 wrote: Thanks for explaining this. I understand how 3 triangles are similar to each other. However how do we determine which side is similar to which in order to set up the ratio. for example, if we take two smaller triangles - Triangle ABC and ADC....I think side AB is corresponding to AD, BC is corresponding to CD and AC is common. Is that correct? if yes, how do I find which side is corresponding in the smaller triangle with respect to the bigger triangle? Many thanks in advance. There is a very simple method of finding out the corresponding sides in similar triangles. Say, you have two triangles ABC and DEF. You find by AA that the triangles are similar. All you have to do is name the triangles the way the angles are equal. Say angle A = angle E, angle B = angle D and and hence angle C = angle F. Then we write: triangle ABC is similar to triangle EDF. Now you have the corresponding sides. That is, AB/ED = BC/DF = AC/EF In the question above, triangles BAD and BCA are similar So BA/BC = AD/CA = BD/BA triangles BAD and ACD are similar So BA/AC = AD/CD = BD/AD triangles BCA is similar to triangle ACD So BC/AC = CA/CD = BA/AD _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink]

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30 Mar 2015, 06:57
enigma123 wrote:
Attachment:
The attachment Triangle.jpg is no longer available
In triangle ABC, if BC = 3 and AC = 4, then what is the length of segment CD?

A. 3
B. 15/4
C. 5
D. 16/3
E. 20/3

such questions where there is a 90 degree angle in a triangle can always be solved easily by drawing a circle .
as we draw circle and extend AC to point E we will get a triangle BED , exact copy of triangle BAD --Eq1.

as shown in the attached image , Triangle ABC and Triangle ECD are similar as all angles are equal.
so $$\frac{EC}{BC} = \frac{CD}{AC} = \frac{ED}{AB}$$

3x=4y --> $$x=\frac{4}{3}*y$$
y=4 as shown in eq1
so x= $$\frac{16}{3}$$
Attachments

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In triangle ABC, if BC = 3 and AC = 4, then what is the   [#permalink] 30 Mar 2015, 06:57

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