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ABC is a right triangle with right angle at point B. If the ratio of

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ABC is a right triangle with right angle at point B. If the ratio of AB to BC is 8 to 15, what is the area of the triangle?

(1) The height from point B to the hypotenuse is 120.
(2) The perimeter of the triangle is 680.
[Reveal] Spoiler: OA

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Re: ABC is a right triangle with right angle at point B. If the ratio of [#permalink]

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Bunuel wrote:
ABC is a right triangle with right angle at point B. If the ratio of AB to BC is 8 to 15, what is the area of the triangle?

(1) The height from point B to the hypotenuse is 120.
(2) The perimeter of the triangle is 680.


Answer is D

using hypotenuse = X(89)^1/2
we get 1/2B*H

St 1
that is 1/2*8X*5X = 1/2*X(89)^1/2*120
calculate X and get the area from there.

St 2
8X + 5X + X(89)^1/2 = 680
calculate X and obtain area.

hence both statements individually Suff.

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ABC is a right triangle with right angle at point B. If the ratio of [#permalink]

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shallow9323 wrote:
Bunuel wrote:
ABC is a right triangle with right angle at point B. If the ratio of AB to BC is 8 to 15, what is the area of the triangle?

(1) The height from point B to the hypotenuse is 120.
(2) The perimeter of the triangle is 680.


Answer is D

using hypotenuse = X(89)^1/2
we get 1/2B*H

St 1
that is 1/2*8X*5X = 1/2*X(89)^1/2*120
calculate X and get the area from there.

St 2
8X + 5X + X(89)^1/2 = 680
calculate X and obtain area.

hence both statements individually Suff.



I too believe the answer is D, but just wanted to correct a typo. The question stem says the ratio is 8:15 and your solution shows 8:5.

So for statement 2 we would have 8x+15x+x*(289)^1/2 =680.

Answer is D.
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Re: ABC is a right triangle with right angle at point B. If the ratio of [#permalink]

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New post 21 May 2015, 06:19
Oh yea,! thanks.
good luck

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Re: ABC is a right triangle with right angle at point B. If the ratio of [#permalink]

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New post 21 May 2015, 10:52
Could you guys explain it a little in detail as to why is statement 1 sufficient. I did not get it.

Warm Regards,

Rudraksh.

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Re: ABC is a right triangle with right angle at point B. If the ratio of [#permalink]

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New post 21 May 2015, 11:20
using any side as the base, the area should be the same, but the corresponding height from that base needs to be used.
half base into height.

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Re: ABC is a right triangle with right angle at point B. If the ratio of [#permalink]

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New post 21 May 2015, 11:31
Hi All,

When a Geometry question does not include a picture, you have to be careful about the assumptions you make about the shape(s) involved.

Here, do you KNOW which side of the triangle is the height: AB or BC.....? If you drew 2 different pictures, and did the necessary work, would the answer to the question change?

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Re: ABC is a right triangle with right angle at point B. If the ratio of [#permalink]

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New post 21 May 2015, 11:49
ABC is right angle at B
hence, any line touching B cant be the hypotenuse.

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Re: ABC is a right triangle with right angle at point B. If the ratio of [#permalink]

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RudeyboyZ wrote:
Could you guys explain it a little in detail as to why is statement 1 sufficient. I did not get it.

Warm Regards,

Rudraksh.


Dear RudeyboyZ

Please refer to the image below. It represents the information that we were given in the Question Statement and in St. 1.Also, if AB = 8x and BC = 15x, can you calculate AC? Sure, using Pythagoras Theorem, you know that it's 17x.

Image



Now, area of triangle ABC =\(\frac{1}{2}*AB*BC\) = \(\frac{1}{2}*8x*15x\)

But, suppose you take the base as AC. Then, what is the height? BD. And St. 1 tells us that BD = 120

So, the area of triangle ABC can also be written as = \(\frac{1}{2}*BD*AC\) = \(\frac{1}{2}*120*17x\)

By equating the two expressions for area of triangle ABC, you can find the value of x and therefore, you can find a unique value of the area of triangle ABC. So, St. 1 is sufficient.

Note that I could analyze Statement 1 till its logical conclusion because:

1. I drew a neat diagram that represented all the pieces of information that were given in Question statement + St. 1 and also the information that I had inferred (AC = 17x, using Pythagoras Theorem)

2. I was able to mentally rotate the triangle ABC such that AC became its base. By doing so, I could see that the area of triangle ABC can also be written as \(\frac{1}{2}*BD*AC\). This ability to mentally rotate a given image comes in handy in many questions.

Here's an official question that involves rotation of a triangle:

http://gmatclub.com/forum/in-the-figure-above-triangle-abc-is-equilateral-and-point-143496.html

Hope this helped! :)

Japinder
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Re: ABC is a right triangle with right angle at point B. If the ratio of [#permalink]

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New post 22 May 2015, 02:42
EgmatQuantExpert wrote:
RudeyboyZ wrote:
Could you guys explain it a little in detail as to why is statement 1 sufficient. I did not get it.

Warm Regards,

Rudraksh.


Dear RudeyboyZ

Please refer to the image below. It represents the information that we were given in the Question Statement and in St. 1.Also, if AB = 8x and BC = 15x, can you calculate AC? Sure, using Pythagoras Theorem, you know that it's 17x.

Image



Now, area of triangle ABC =\(\frac{1}{2}*AB*BC\) = \(\frac{1}{2}*8x*15x\)

But, suppose you take the base as AC. Then, what is the height? BD. And St. 1 tells us that BD = 120

So, the area of triangle ABC can also be written as = \(\frac{1}{2}*BD*AC\) = \(\frac{1}{2}*120*17x\)

By equating the two expressions for area of triangle ABC, you can find the value of x and therefore, you can find a unique value of the area of triangle ABC. So, St. 1 is sufficient.

Note that I could analyze Statement 1 till its logical conclusion because:

1. I drew a neat diagram that represented all the pieces of information that were given in Question statement + St. 1 and also the information that I had inferred (AC = 17x, using Pythagoras Theorem)

2. I was able to mentally rotate the triangle ABC such that AC became its base. By doing so, I could see that the area of triangle ABC can also be written as \(\frac{1}{2}*BD*AC\). This ability to mentally rotate a given image comes in handy in many questions.

Here's an official question that involves rotation of a triangle:

http://gmatclub.com/forum/in-the-figure-above-triangle-abc-is-equilateral-and-point-143496.html

Hope this helped! :)

Japinder


Dear Japinder,

The detailed explanation is really helpful. Thank you!

Warm Regards,

Rudey.

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Re: ABC is a right triangle with right angle at point B. If the ratio of [#permalink]

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New post 22 May 2015, 02:43
EgmatQuantExpert wrote:
RudeyboyZ wrote:
Could you guys explain it a little in detail as to why is statement 1 sufficient. I did not get it.

Warm Regards,

Rudraksh.


Dear RudeyboyZ

Please refer to the image below. It represents the information that we were given in the Question Statement and in St. 1.Also, if AB = 8x and BC = 15x, can you calculate AC? Sure, using Pythagoras Theorem, you know that it's 17x.

Image



Now, area of triangle ABC =\(\frac{1}{2}*AB*BC\) = \(\frac{1}{2}*8x*15x\)

But, suppose you take the base as AC. Then, what is the height? BD. And St. 1 tells us that BD = 120

So, the area of triangle ABC can also be written as = \(\frac{1}{2}*BD*AC\) = \(\frac{1}{2}*120*17x\)

By equating the two expressions for area of triangle ABC, you can find the value of x and therefore, you can find a unique value of the area of triangle ABC. So, St. 1 is sufficient.

Note that I could analyze Statement 1 till its logical conclusion because:

1. I drew a neat diagram that represented all the pieces of information that were given in Question statement + St. 1 and also the information that I had inferred (AC = 17x, using Pythagoras Theorem)

2. I was able to mentally rotate the triangle ABC such that AC became its base. By doing so, I could see that the area of triangle ABC can also be written as \(\frac{1}{2}*BD*AC\). This ability to mentally rotate a given image comes in handy in many questions.

Here's an official question that involves rotation of a triangle:

http://gmatclub.com/forum/in-the-figure-above-triangle-abc-is-equilateral-and-point-143496.html

Hope this helped! :)

Japinder


Dear Japinder,

The detailed explanation is really helpful. Thank you!

Warm Regards,

Rudraksh.

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ABC is a right triangle with right angle at point B. If the ratio of [#permalink]

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Bunuel wrote:
ABC is a right triangle with right angle at point B. If the ratio of AB to BC is 8 to 15, what is the area of the triangle?

(1) The height from point B to the hypotenuse is 120.
(2) The perimeter of the triangle is 680.


OFFICIAL SOLUTION:

ABC is a right triangle with right angle at point B. If the ratio of AB to BC is 8 to 15, what is the area of the triangle?

First of all since the ration of AB to BC is 8x:15x, then the hypotenuse is \(\sqrt{(8x)^2+(15x)^2}=17x\).

(1) The height from point B to the hypotenuse is 120.

Consider the image below:
Image
The area of a right triangle equals to \(\frac{1}{2}*leg_1*leg_2=\frac{1}{2}*AB*BC\). But the area of a right triangle can also be found by \(\frac{1}{2}*(altitude \ from \ right \ angle)*(hypotenuse)=\frac{1}{2}*BD*AC\).

Equate these expressions:
\(\frac{1}{2}*AB*BC=\frac{1}{2}*BD*AC\);
\(\frac{1}{2}*8x*15x=\frac{1}{2}*120*17x\);
\(120x=120*17\);
\(x=17\).

The area = \(\frac{1}{2}*AB*BC=\frac{1}{2}*8x*15x=60x=60*17\). Sufficient.

(2) The perimeter of the triangle is 680.

The perimeter = \(8x+15x+17x=680\). We can find x, thus we can find the area. Sufficient.

Answer: D.

P.S. Thank you Japinder for the drawing.


[Reveal] Spoiler:
Attachment:
image.png
image.png [ 26.82 KiB | Viewed 2786 times ]

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Re: ABC is a right triangle with right angle at point B. If the ratio of [#permalink]

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New post 14 Aug 2015, 12:27
Bunuel wrote:
Bunuel wrote:
ABC is a right triangle with right angle at point B. If the ratio of AB to BC is 8 to 15, what is the area of the triangle?

(1) The height from point B to the hypotenuse is 120.
(2) The perimeter of the triangle is 680.


OFFICIAL SOLUTION:

ABC is a right triangle with right angle at point B. If the ratio of AB to BC is 8 to 15, what is the area of the triangle?

First of all since the ration of AB to BC is 8x:15x, then the hypotenuse is \(\sqrt{(8x)^2+(15x)^2}=17x\).

(1) The height from point B to the hypotenuse is 120.

Consider the image below:
Image
The area of a right triangle equals to \(\frac{1}{2}*leg_1*leg_2=\frac{1}{2}*AB*BC\). But the area of a right triangle can also be found by \(\frac{1}{2}*(altitude \ from \ right \ angle)*(hypotenuse)=\frac{1}{2}*BD*AC\).

Equate these expressions:
\(\frac{1}{2}*AB*BC=\frac{1}{2}*BD*AC\);
\(\frac{1}{2}*8x*15x=\frac{1}{2}*120*17x\);
\(120x=120*17\);
\(x=17\).

The area = \(\frac{1}{2}*AB*BC=\frac{1}{2}*8x*15x=60x=60*17\). Sufficient.

(2) The perimeter of the triangle is 680.

The perimeter = \(8x+15x+17x=680\). We can find x, thus we can find the area. Sufficient.

Answer: D.

P.S. Thank you Japinder for the drawing.


[Reveal] Spoiler:
Attachment:
image.png



Hi Bunuel,
Can you please help me with this question, i have a specific doubt.
St 1 says, The height from point B to the hypotenuse is 120; We know that Perpendicular to the hypotenuse(height from point B to the hypotenuse) will always divide the triangle into two triangles with the same properties as the original triangle and it'll be half the length of the hypotenuse? Which only goes on to prove hypotenuse = 2x120=240
St 2 says, perimeter of the triangle is 680; 8x+15x+17x=680 x=17 hypotenuse = 17x17 = 289
How can we have 2 different hypotenuse measures for the same triangle?

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Re: ABC is a right triangle with right angle at point B. If the ratio of [#permalink]

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New post 14 Aug 2015, 12:41
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Ted21 wrote:
Hi Bunuel,
Can you please help me with this question, i have a specific doubt.
St 1 says, The height from point B to the hypotenuse is 120; We know that Perpendicular to the hypotenuse(height from point B to the hypotenuse) will always divide the triangle into two triangles with the same properties as the original triangle and it'll be half the length of the hypotenuse? Which only goes on to prove hypotenuse = 2x120=240 St 2 says, perimeter of the triangle is 680; 8x+15x+17x=680 x=17 hypotenuse = 17x17 = 289
How can we have 2 different hypotenuse measures for the same triangle?



you are quoting statement 2 correctly but your interpretation of statement 1 is not correct.

The property that you have mentioned in the red text above is ONLY possible when the triangle is a 45-45-90 triangle and in this triangle AB = BC which is not the case provided to us.

Triangle ABC is NOT 45-45-90 triangle and hence you can not say that the hypotenuse will be divided into 2 by the height from B.

The way you will be able to calculate the impact of height is as follows:

Let the 2 sides of the triangle be 8k and 15k (as the ratio is 8:15)

Thus the hypotenuse is = ((8k)^2+(15k)^2)^0.5 = 17k.

Thus the area of the triangle = 0.5*AB*BC = 0.5*Height*Hypotenuse ---> 0.5*8k*15k = 0.5*120*17k ---> k = 17 and thus hypotenuse = 17*17= 289 same as Statement 2.

Hope this helps.
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