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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 ?

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 52255 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, \(\frac{CD}{AC}=\frac{AC}{BC}\) --> \(\frac{CD}{4}=\frac{4}{3}\) --> \(CD=\frac{16}{3}\).

splittingtriangle.jpg [ 4.22 KiB | Viewed 47446 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.

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, \(\frac{CD}{AC}=\frac{AC}{BC}\) --> \(\frac{CD}{4}=\frac{4}{3}\) --> \(CD=\frac{16}{3}\).

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 47152 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.

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

Show Tags

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.

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

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20 Jun 2013, 07:18

2

This post received KUDOS

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)

Considering Triangle BAD - AB^2 + AD^2 = BD^2

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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09 Mar 2015, 03:01

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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

1

This post received KUDOS

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 thus, AD=sqrt(x^2+16) also in traingle ABD, BD^2=AB^2+AD^2 (x+3)^2=25+(x^2+16) x=32/6=16/3

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, \(\frac{CD}{AC}=\frac{AC}{BC}\) --> \(\frac{CD}{4}=\frac{4}{3}\) --> \(CD=\frac{16}{3}\).

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?

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, \(\frac{CD}{AC}=\frac{AC}{BC}\) --> \(\frac{CD}{4}=\frac{4}{3}\) --> \(CD=\frac{16}{3}\).

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. _________________

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

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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, \(\frac{CD}{AC}=\frac{AC}{BC}\) --> \(\frac{CD}{4}=\frac{4}{3}\) --> \(CD=\frac{16}{3}\).

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?

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 _________________

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