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# For triangle ABC, angle ABC = 90 degrees, and side AC has a length of

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Re: For triangle ABC, angle ABC = 90 degrees, and side AC has a length of [#permalink]
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ans C.. a right angle triangle given with length of hyp...
1) it tells us the other two sides are equal and gives the length of other two sides.. however D can be anywhere on line AC ..insufficient
2)it just tells us that BD is altitude.. no use unless sides are known..insufficient

combined sufficient as allsides known and alt bd can be found out
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Re: For triangle ABC, angle ABC = 90 degrees, and side AC has a length of [#permalink]
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Bunuel wrote:
For triangle ABC, angle ABC = 90 degrees, and side AC has a length of 15. If point D lies on side AC, and a line is drawn from point B to point D, what is the length of line segment BD?

(1) Triangle ABC is isosceles.

(2) Line segment BD is perpendicular to side AC.

Kudos for a correct solution.

Statement 1: We know that side AB and side BC are of the same length.
However, the segment BD can be of various lengths based on where the point D is located along side AC.
Insufficient.

Statement 2: segment BD forms a 90 deg angle with side AC. However, we still do not know how side AC is broken down or what the length is of any other side.
Insufficient.

Combined, we know right triangle ABC is isosceles with BD as its altitude and bisector. We can now apply pythagorean theorem to find the length of segment BD.

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Re: For triangle ABC, angle ABC = 90 degrees, and side AC has a length of [#permalink]
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Quote:
For triangle ABC, angle ABC = 90 degrees, and side AC has a length of 15. If point D lies on side AC, and a line is drawn from point B to point D, what is the length of line segment BD?

(1) Triangle ABC is isosceles.

(2) Line segment BD is perpendicular to side AC.

the length of AD and CD we can not use the similar triangle theorem, (sid/alt=alt/side)
not suff.
(1)triangle is a isosceles. but we do not know where is the point D on AC.

(1/2): since BD is perpendicular, we know that the isosceles ABC with °ABC=90 has BCA=45 and CAB=45. with AC = 15 we got AD=7,5 and CD=7,5. We could either use x:x:x*root(2) to determine AB / CB and afterwars BD (with pythagoras).
or we know that the perpendicular will build 2 triangles with 45-45-90 but as the ground we will get x, and as the legs we will get 0,5*15. so its suff
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Re: For triangle ABC, angle ABC = 90 degrees, and side AC has a length of [#permalink]
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Bunuel wrote:
For triangle ABC, angle ABC = 90 degrees, and side AC has a length of 15. If point D lies on side AC, and a line is drawn from point B to point D, what is the length of line segment BD?

(1) Triangle ABC is isosceles.

(2) Line segment BD is perpendicular to side AC.

Kudos for a correct solution.

1) If ABC is isosceles AB=BC but doesnt help us calculate BD.
2) If BD is perpendicular to AC we still dont know BD.

Combining, we know AB = BC from one so each of them can be calculated and the triangle is defined. So BD is calculable.
Or since area of triangle is constant AB*BC=AC*BD so BD can be calculated.

Press kudos if I am right.
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Re: For triangle ABC, angle ABC = 90 degrees, and side AC has a length of [#permalink]
Hi Bunuel,

I have understood your explanation. But I need one clarification.
What If I say, statement 2 alone is sufficient. Because we know hypotenuse is 15 and ABC is a right angled triangle and right angled at B. with Pythagoras tripplets, other two sides are 12 and 9. We know if we know 3 sides we can find out BD length.

with formula BD = (AB * BC) / AC.

With this I feel something I am missing. either I am assuming more than the required or carrying extra information. Can you tell me what is my mistake ?
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Re: For triangle ABC, angle ABC = 90 degrees, and side AC has a length of [#permalink]
dharan wrote:
Hi Bunuel,

I have understood your explanation. But I need one clarification.
What If I say, statement 2 alone is sufficient. Because we know hypotenuse is 15 and ABC is a right angled triangle and right angled at B. with Pythagoras tripplets, other two sides are 12 and 9. We know if we know 3 sides we can find out BD length.

with formula BD = (AB * BC) / AC.

With this I feel something I am missing. either I am assuming more than the required or carrying extra information. Can you tell me what is my mistake ?

Why do you assume that the sides should be Pythagorean Triples? Why do you assume that the sides should be integers at all?
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Re: For triangle ABC, angle ABC = 90 degrees, and side AC has a length of [#permalink]
Bunuel wrote:
For triangle ABC, angle ABC = 90 degrees, and side AC has a length of 15. If point D lies on side AC, and a line is drawn from point B to point D, what is the length of line segment BD?

(1) Triangle ABC is isosceles.

(2) Line segment BD is perpendicular to side AC.

Kudos for a correct solution.

I almost concluded D as the answer by confusing similarity for congruency and then realised that strict congruency can only be reached if apart from being isosceles and having BD as the common side BD is also a perpendicular bisector of the AC. But this condition was not found in statement 1.
Then i checked statement 2 and realised that statement 2 gives us exactly what was missing .. the condition for strict congruency and the both statement A and B are needed

Attached here is my wrong answer

Attached here is my WRONG answer

Attachments

IMG_20160630_213820.jpg [ 550.5 KiB | Viewed 302620 times ]

Originally posted by LogicGuru1 on 30 Jun 2016, 09:57.
Last edited by LogicGuru1 on 02 Sep 2016, 10:57, edited 1 time in total.
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Re: For triangle ABC, angle ABC = 90 degrees, and side AC has a length of [#permalink]
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LogicGuru1 wrote:
Bunuel wrote:
For triangle ABC, angle ABC = 90 degrees, and side AC has a length of 15. If point D lies on side AC, and a line is drawn from point B to point D, what is the length of line segment BD?

(1) Triangle ABC is isosceles.

(2) Line segment BD is perpendicular to side AC.

Kudos for a correct solution.

I almost concluded D as the answer by confusing similarity for congruency and then realised that strict congruency can only be reached if apart from being isosceles and having BD as the common side BD is also a perpendicular bisector of the AC. But this condition was not found in statement 1.
Then i checked statement 2 and realised that statement 2 gives us exactly what was missing .. the condition for strict congruency and the both statement A and B are needed

Attached here is my wrong answer

Attached here is my WRONG answer

Responding to a pm:

This is where you made a mistake: there is no SSA rule. The rule is SAS (two sides and the INCLUDED angle). If you say AB = BC and BD = BD, the angles which have to be same are DBA and DBC.
A special SSA rule works only in case of a right triangle - It is called the RHS rule. There has to be a right angle in both triangles, the hypotenuses have to be equal and any one set of sides should be equal.

Statement 1 gives you neither SAS nor RHS. So the two triangles may not be congruent.
The other congruency rules are SSS, ASA and AAS.
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Re: For triangle ABC, angle ABC = 90 degrees, and side AC has a length of [#permalink]
VeritasPrepKarishma wrote:
LogicGuru1 wrote:
Bunuel wrote:
For triangle ABC, angle ABC = 90 degrees, and side AC has a length of 15. If point D lies on side AC, and a line is drawn from point B to point D, what is the length of line segment BD?

(1) Triangle ABC is isosceles.

(2) Line segment BD is perpendicular to side AC.

Kudos for a correct solution.

I almost concluded D as the answer by confusing similarity for congruency and then realised that strict congruency can only be reached if apart from being isosceles and having BD as the common side BD is also a perpendicular bisector of the AC. But this condition was not found in statement 1.
Then i checked statement 2 and realised that statement 2 gives us exactly what was missing .. the condition for strict congruency and the both statement A and B are needed

Attached here is my wrong answer

Attached here is my WRONG answer

Responding to a pm:

This is where you made a mistake: there is no SSA rule. The rule is SAS (two sides and the INCLUDED angle). If you say AB = BC and BD = BD, the angles which have to be same are DBA and DBC.
A special SSA rule works only in case of a right triangle - It is called the RHS rule. There has to be a right angle in both triangles, the hypotenuses have to be equal and any one set of sides should be equal.

Statement 1 gives you neither SAS nor RHS. So the two triangles may not be congruent.
The other congruency rules are SSS, ASA and AAS.

Many thanks Karishma

I realised I am arriving at that wrong answer if I am choosing D.
You are a tremendous help..
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Re: For triangle ABC, angle ABC = 90 degrees, and side AC has a length of [#permalink]
1
Kudos
Bunuel wrote:
For triangle ABC, angle ABC = 90 degrees, and side AC has a length of 15. If point D lies on side AC, and a line is drawn from point B to point D, what is the length of line segment BD?

(1) Triangle ABC is isosceles.

(2) Line segment BD is perpendicular to side AC.

Kudos for a correct solution.

Responding to a pm:
Quote:
How come if angles ABC and ADB are equal at 90, and angle A is shared between triangles ABC and ADB, and these triangles also share AB as one of their sides...why aren't they similar making B sufficient?

Triangles ABC and ADB are similar using AA rule (two equal angles) - no problem there.
The point is - how do you get the length of the line segment BD?
All you know is that AC is 15. You don't know the ratio of the corresponding sides of the two triangles. You don't know AB/BC/AD. How will you find BD?
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Re: For triangle ABC, angle ABC = 90 degrees, and side AC has a length of [#permalink]
dharan wrote:
Hi Bunuel,

I have understood your explanation. But I need one clarification.
What If I say, statement 2 alone is sufficient. Because we know hypotenuse is 15 and ABC is a right angled triangle and right angled at B. with Pythagoras tripplets, other two sides are 12 and 9. We know if we know 3 sides we can find out BD length.

with formula BD = (AB * BC) / AC.

With this I feel something I am missing. either I am assuming more than the required or carrying extra information. Can you tell me what is my mistake ?

Hi Bunuel
Just curious which is the above highlighted formulae. Can u plz share its applicability.Thanks
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Re: For triangle ABC, angle ABC = 90 degrees, and side AC has a length of [#permalink]
cruiseav wrote:
dharan wrote:
Hi Bunuel,

I have understood your explanation. But I need one clarification.
What If I say, statement 2 alone is sufficient. Because we know hypotenuse is 15 and ABC is a right angled triangle and right angled at B. with Pythagoras tripplets, other two sides are 12 and 9. We know if we know 3 sides we can find out BD length.

with formula BD = (AB * BC) / AC.

With this I feel something I am missing. either I am assuming more than the required or carrying extra information. Can you tell me what is my mistake ?

Hi Bunuel
Just curious which is the above highlighted formulae. Can u plz share its applicability.Thanks

Hello

Consider a triangle ABC, which is 90 degrees at B. Now area of this triangle = 1/2 * base * height = 1/2 * AB * BC (AB, BC as base/height)
Now lets draw a perpendicular BD to side AC. Taking AC as base, BD would be the height. So Area can also be = 1/2 * BD * AC

Equating the two, AB*BC = BD*AC or BD = (AB * BC) / AC
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Re: For triangle ABC, angle ABC = 90 degrees, and side AC has a length of [#permalink]
cruiseav wrote:
dharan wrote:
Hi Bunuel,

I have understood your explanation. But I need one clarification.
What If I say, statement 2 alone is sufficient. Because we know hypotenuse is 15 and ABC is a right angled triangle and right angled at B. with Pythagoras tripplets, other two sides are 12 and 9. We know if we know 3 sides we can find out BD length.

with formula BD = (AB * BC) / AC.

With this I feel something I am missing. either I am assuming more than the required or carrying extra information. Can you tell me what is my mistake ?

Hi Bunuel
Just curious which is the above highlighted formulae. Can u plz share its applicability.Thanks

You can find the area of a righ triangle ABC (right angled at B) in several ways.

1. Area = leg1*leg2/2 = AB*BC/2
2. Area = height*base/2 = BD*AC/2 (where BD is altitude from B to the hypotenuse and hypotenuse AC is the base).

Equate: AB*BC/2 = BD*AC/2;

AB*BC = BD*AC;

BD = AB*BC/AC

Hope it's clear.
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Re: For triangle ABC, angle ABC = 90 degrees, and side AC has a length of [#permalink]
For triangle ABC, angle ABC = 90 degrees, and side AC has a length of 15. If point D lies on side AC, and a line is drawn from point B to point D, what is the length of line segment BD?

(1) Triangle ABC is isosceles.
Insufficient b/c we don't know where specifically D is located on AC.

(2) Line segment BD is perpendicular to side AC.
Insufficient b/c we don't know what type of triangle we are looking at.

Combined: Sufficient.

We have a 45-45-90 triangle and D is the midpoint of AC.
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Re: For triangle ABC, angle ABC = 90 degrees, and side AC has a length of [#permalink]
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Re: For triangle ABC, angle ABC = 90 degrees, and side AC has a length of [#permalink]
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