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

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

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New post Updated on: 18 Feb 2019, 05:07
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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


Attachment:
Triangle.jpg
Triangle.jpg [ 8.62 KiB | Viewed 258433 times ]

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?

Originally posted by kirankp on 16 Dec 2009, 05:22.
Last edited by Bunuel on 18 Feb 2019, 05:07, edited 5 times in total.
Edited the question.
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Re: In triangle ABC to the right, if BC = 3 and AC = 4, then what is the  [#permalink]

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New post 26 Feb 2012, 12:07
8
10
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
splittingtriangle.jpg [ 4.22 KiB | Viewed 462017 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, \(\frac{CD}{AC}=\frac{AC}{BC}\) --> \(\frac{CD}{4}=\frac{4}{3}\) --> \(CD=\frac{16}{3}\).

Answer: D.

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 to the right, if BC = 3 and AC = 4, then what is the  [#permalink]

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New post 20 Jun 2013, 07:18
7
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 to the right, if BC = 3 and AC = 4, then what is the  [#permalink]

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New post 16 Dec 2009, 06:24
6
1
kirankp wrote:
In triangle ABC to the right, if BC = 3 and AC = 4, then what is the length of segment CD? Image

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

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New post 24 Oct 2011, 22:09
5
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.

Thus, BC/CA = CA/CD.

Again, the correct answer is D.
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Re: In triangle ABC to the right, if BC = 3 and AC = 4, then what is the  [#permalink]

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New post Updated on: 03 Nov 2011, 14:20
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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
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Originally posted by MBAhereIcome on 03 Nov 2011, 02:06.
Last edited by MBAhereIcome on 03 Nov 2011, 14:20, edited 1 time in total.
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Re: In triangle ABC to the right, if BC = 3 and AC = 4, then what is the  [#permalink]

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New post 03 Nov 2011, 07:49
4
You can immediately tell that AB=5 because ABC is a 3-4-5 triangle. Label CD x, and AD y.

You get:

1) \(5^2+y^2=(3+x)^2\)
2) \(4^2+x^2=y^2\)

Plug in the definition of \(y^2\) from (2) into 1 and solve. You get 16/3.
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Re: In triangle ABC to the right, if BC = 3 and AC = 4, then what is the  [#permalink]

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

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

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New post 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
splittingtriangle.jpg [ 4.22 KiB | Viewed 414651 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.

Please ask if anything remains unclear.
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Re: In triangle ABC to the right, if BC = 3 and AC = 4, then what is the  [#permalink]

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

Answer: D.

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 to the right, if BC = 3 and AC = 4, then what is the  [#permalink]

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New post 17 May 2012, 05:59
4
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
splittingtriangle.jpg [ 4.22 KiB | Viewed 414261 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 to the right, if BC = 3 and AC = 4, then what is the  [#permalink]

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New post 22 Jun 2013, 08:23
3
1
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 to the right, if BC = 3 and AC = 4, then what is the  [#permalink]

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New post 12 Mar 2015, 00:43
2
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

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

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

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

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New post 24 Mar 2015, 21:22
1
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, \(\frac{CD}{AC}=\frac{AC}{BC}\) --> \(\frac{CD}{4}=\frac{4}{3}\) --> \(CD=\frac{16}{3}\).

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

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

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New post 30 Mar 2015, 01:40
2
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
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Re: In triangle ABC to the right, if BC = 3 and AC = 4, then what is the  [#permalink]

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New post 30 Mar 2015, 06:57
enigma123 wrote:
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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



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= \(\frac{16}{3}\)
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Re: In triangle ABC to the right, if BC = 3 and AC = 4, then what is the  [#permalink]

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New post 02 Jun 2015, 06:25
1
Hi Bunuel,

please explain me the correlation between 30-60-90(1:sq(3):2), 45-45-90(1:1:sq(2)) with pythagorean triplets(3-4-5, 7-24-25). I am confused because if we have a 90deg triangle, and 2 sides 3,4 we can put other side as 5? If so then it becomes 30-60-90 triangle right? then why we are not able to correlate 1:sq(3):2 with 3:4:5?
because we need to commonly multiply the ratio, so if 1*3 then sq(3) must also multiply by 3 which is not equal to 4..

Because I had tried an another approach which gives a wrong ans... Need clarification.. If ABC is a right triangle with 3,4 then other side must be 5. So Angle BAC will be 30 deg. which makes Ang CAD as 60 and CDA as 30.. So we have 30-60-90 triangle on ACD. If thats the case, then CD must be 4*sqrt(3) right? what's wrong in my approach?

Please help..
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Re: In triangle ABC to the right, if BC = 3 and AC = 4, then what is the   [#permalink] 02 Jun 2015, 06:25

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