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we have 3 similar triangles the main triangle : ABD two other triangles BC and ADC . Now to find out CD we can use the later two triangles , so by similarity we have ,

Re: split triangles [#permalink]
16 Dec 2009, 04:52

1

This post received KUDOS

kirankp wrote:

In triangle ABC to the right, 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

For this problem the solution is :

we have 3 similar triangles the main triangle : ABD two other triangles BC and ADC . Now to find out CD we can use the later two triangles , so by similarity we have ,

BC/CA = CD/AC

which yields CD as 3.

but the answer is wrong. where have i gone wrong?

the problem in your appraoch is that u have assumed wrong angles to be similar to check if cd=3 then bd=6 and since ac=4 and bc=3 ab and ad will be 5 hence 5^2+5^2=6^2 which is not possible now the triangle similar are DCA,ACB, DCA from tri DAB and ACB ad/4=5/3 ad=20/3 now in tri DAB and DAC cd/ad=4/5 -->cd=4/5*ad =4/5*20/3=16/3

hence D OA?? _________________

GMAT is not a game for losers , and the moment u decide to appear for it u are no more a loser........ITS A BRAIN GAME

Re: split triangles [#permalink]
16 Dec 2009, 05:24

4

This post received KUDOS

kirankp wrote:

In triangle ABC to the right, 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

For this problem the solution is :

we have 3 similar triangles the main triangle : ABD two other triangles BC and ADC . Now to find out CD we can use the later two triangles , so by similarity we have ,

BC/CA = CD/AC

which yields CD as 3.

but the answer is wrong. where have i gone wrong?

D- 16/3

we can use pythagoras theorem to solve this. AB we will be 5. Let CD = x then AD = sqrt ( 16 + x^2) in Triangle BAD we have AB^2 + AD^2 = BD^2 => 25 + 16 + x^2 = (3+x)^2 solving the above we get x= 16/3

Re: split triangles [#permalink]
24 Oct 2011, 21:09

This is a MGMAT Question, OA is D and OE is as below. Hope it helps.

Because angles BAD and ACD are right angles, the figure above is composed of three similar right triangles: BAD, ACD and BCA. [Any time a height is dropped from the right angle vertex of a right triangle to the opposite side of that right triangle, the three triangles that result have the same 3 angle measures. This means that they are similar triangles. See below for further explanation.]

To solve for the length of side CD, we can set up a proportion, based on the relationship between the similar triangles ACD and BCA:

BC/CA = CA/CD

3/4 = 4/CD

CD = 16/3 The correct answer is D.

Addendum: Let's look at how we know that triangles ACD and BCA are similar.

1) Let's say that <CDA is x degrees, and <DAC is y degrees. Since <ACD is 90 degrees, and the sum of all the interior angles in a triangle is 180, we know that x + y = 90.

2) Now let's look at <BAC. We know that <BAC + <DAC = 90, since <BAD is labeled as a right angle. We also know that <DAC is y degrees (from step 1), and that x + y = 90. Putting these facts together, we know that <BAC is x degrees.

3) We know <ACB is a right angle, since <ACD is a right angle. Since <ACB is a right angle, <BAC + <CBA = 90. Given that <BAC is x degrees, <CBA must be y degrees.

4) To summarize, <CAB has the same measure as <CDA (x degrees) , and <CBA has the same measure as <DAC (y degrees). This means that in similar triangles CAB and CAD, side BC of CAB corresponds to side CA of CAD, and side CA of CAB corresponds to side CD of CAD.

Re: split triangles [#permalink]
18 Dec 2012, 15:30

arjunbt wrote:

This is a MGMAT Question, OA is D and OE is as below. Hope it helps.

Because angles BAD and ACD are right angles, the figure above is composed of three similar right triangles: BAD, ACD and BCA. [Any time a height is dropped from the right angle vertex of a right triangle to the opposite side of that right triangle, the three triangles that result have the same 3 angle measures. This means that they are similar triangles. See below for further explanation.]

To solve for the length of side CD, we can set up a proportion, based on the relationship between the similar triangles ACD and BCA:

BC/CA = CA/CD

3/4 = 4/CD

CD = 16/3 The correct answer is D.

Addendum: Let's look at how we know that triangles ACD and BCA are similar.

1) Let's say that <CDA is x degrees, and <DAC is y degrees. Since <ACD is 90 degrees, and the sum of all the interior angles in a triangle is 180, we know that x + y = 90.

2) Now let's look at <BAC. We know that <BAC + <DAC = 90, since <BAD is labeled as a right angle. We also know that <DAC is y degrees (from step 1), and that x + y = 90. Putting these facts together, we know that <BAC is x degrees.

3) We know <ACB is a right angle, since <ACD is a right angle. Since <ACB is a right angle, <BAC + <CBA = 90. Given that <BAC is x degrees, <CBA must be y degrees.

4) To summarize, <CAB has the same measure as <CDA (x degrees) , and <CBA has the same measure as <DAC (y degrees). This means that in similar triangles CAB and CAD, side BC of CAB corresponds to side CA of CAD, and side CA of CAB corresponds to side CD of CAD.

Re: split triangles [#permalink]
19 Dec 2012, 01:35

1

This post received KUDOS

Expert's post

AlyoshaKaramazov wrote:

where does the bolded element come from?

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

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}.

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...