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If a right triangle has area 28 and hypotenuse 12, what is

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If a right triangle has area 28 and hypotenuse 12, what is  [#permalink]

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If a right triangle has area 28 and hypotenuse 12, what is its perimeter?

A. 20
B. 24
C. 28
D. 32
E. 36

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Originally posted by Hussain15 on 13 Jun 2010, 05:04.
Last edited by Bunuel on 26 Dec 2013, 03:47, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: Doubtful PS  [#permalink]

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New post 13 Jun 2010, 06:27
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2
Hussain15 wrote:
If a right triangle has area 28 and hypotenuse 12, what is its perimeter?

A. 20
B. 24
C. 28
D. 32
E. 36


Let the legs of this right triangle be \(x\) and \(y\).

Given: \(area=\frac{xy}{2}=28\) --> \(xy=56\) and \(hypotenuse=x^2+y^2=12^2\).
Question: \(P=x+y+12=?\), so we should calculate the value of \(x+y\).

Square \(x+y\) --> \((x+y)^2=x^2+2xy+y^2\). As \(xy=56\) and \(x^2+y^2=12^2\), then: \((x+y)^2=x^2+2xy+y^2=12^2+2*56=256\) --> \(x+y=\sqrt{256}=16\).

\(P=x+y+12=16+12=28\).

Answer: C.
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Re: Doubtful PS  [#permalink]

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New post 13 Jun 2010, 06:17
1
2
Hussain15 wrote:
If a right triangle has area 28 and hypotenuse 12, what is its perimeter?

A. 20
B. 24
C. 28
D. 32
E. 36


C - 28
let the other 2 sides be a and b then we have 1/2 * a*b = 28 or a*b =56
also we have a^2 + b^2 = 144

adding 2ab to a^2 + b^2 we have a^2 + b^2 + 2ab = 144 + 2*56 = 256 => (a+b)^2 = 256 or a+ b = 16. So perimeter is a+b+ hypotenuse = 16+ 12 = 28
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Re: Doubtful PS  [#permalink]

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New post 13 Jun 2010, 07:41
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xy=56 & x+y=16,, what will be the values of x & y??

Posted from my mobile device
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Re: Doubtful PS  [#permalink]

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New post 13 Jun 2010, 07:54
Hussain15 wrote:
xy=56 & x+y=16,, what will be the values of x & y??

Posted from my mobile device


We have the final answer without calculating the exact values of \(x\) and \(y\). So it doesn't matter. But if you are interested:

\(xy=56\) and \(x+y=16\), \(y=16-x\):

\(x(16-x)=56\) --> \(x^2-16x+56=0\) --> \(x=8-2\sqrt{2}\) and \(y=16-x=8+2\sqrt{2}\) OR \(x=8+2\sqrt{2}\) and \(y=16-x=8-2\sqrt{2}\).

Hope it helps.
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Re: Doubtful PS  [#permalink]

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New post 13 Jun 2010, 09:12
Thanks Bunuel!! Actually I started to solve this problem by using 3 4 5 formula of right triangle I.e 3^2+4^2= 5^2.This ended no where!! :(

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Re: Doubtful PS  [#permalink]

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New post 21 May 2011, 18:58
Apologies

Can someone explain why we are adding 2ab?

I don't understand that part.

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Re: Doubtful PS  [#permalink]

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New post 21 May 2011, 23:33
1
k4lnamja wrote:
Apologies

Can someone explain why we are adding 2ab?

I don't understand that part.

Cheers


No need to be apologetic k4lnamja. You can ask any question so far it's related to the topic.

We are dealing with a right angle triangle;

We are given its area and the hypotenuse and we are asked for perimeter.

Right angle triangle has three sides; one of which is hypotenuse. If we know the length of the other two, we will have the perimeter. However, there is no way to find out the length of the other two sides individually. Thus, our intention is to find the combined length of the other two sides and add it up with the hypotenuse to get the perimeter.

How can we use the information to know the combined length of the other two sides. Here's how.

Hypotenuse = c = 12
Let the other two sides of the right angle triangle be "a" and "b" and we know these two sides are perpendicular to each other.

Area = 28
Area of a triangle = 1/2*base*height = 1/2*a*b

1/2*a*b=28
a*b=56
c=12

As per pythagoras:
a^2+b^2=c^2
(a+b)^2-2ab=c^2
(a+b)^2-2*56=12^2
(a+b)^2-112=144
(a+b)^2=256
a+b=16

Thus, we know the sum of other two sides.
a+b=16
c=12
a+b+c=16+12=28
******************

Ans: "C"

**********************
Just to expand the formula used:
(a+b)^2=a^2+b^2+2ab
(a+b)^2-2ab=a^2+b^2
(a+b)^2-2ab=c^2
**********************
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Re: Doubtful PS  [#permalink]

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New post 23 May 2011, 23:14
1
1/2 * 12 * altitude = 28
altitude = 7

using similar triangle

7/x = x/12 gives x^2 = 84

12^2 - 84 = 60

thus 60 ^ (1/2) + 84 ^(1/2) + 12 = 28.7 approx.

Hence C.
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Re: Doubtful PS  [#permalink]

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New post 27 Aug 2011, 09:29
area = (1/2)bh = 28 => bh=56

hypotenuse = sqrt(b^2+h^2) = 12 => b^2+h^2 = 144

perimeter = b+h+sqrt(b^2+h^2)

we know that (b+h)^2 = b^2+h^2+2bh

= 144+2(56)

=> b+h = 16

=> perimeter = 16+12 = 28
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Re: If a right triangle has area 28 and hypotenuse 12, what is  [#permalink]

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New post 16 Aug 2015, 18:08
amit2k9 wrote:
1/2 * 12 * altitude = 28
altitude = 7

using similar triangle

7/x = x/12 gives x^2 = 84

12^2 - 84 = 60

thus 60 ^ (1/2) + 84 ^(1/2) + 12 = 28.7 approx.

Hence C.


I know it's been a long time since this post was made, but is anyone able to explain where 7/x = x/12 gives x^2 = 84 comes from??? I would like to try and understand this alternate approach that uses the hypotenuse as the base.

Thanks! :)
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Re: If a right triangle has area 28 and hypotenuse 12, what is  [#permalink]

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New post 16 Aug 2015, 18:12
DropBear wrote:
amit2k9 wrote:
1/2 * 12 * altitude = 28
altitude = 7

using similar triangle

7/x = x/12 gives x^2 = 84

12^2 - 84 = 60

thus 60 ^ (1/2) + 84 ^(1/2) + 12 = 28.7 approx.

Hence C.


I know it's been a long time since this post was made, but is anyone able to explain where 7/x = x/12 gives x^2 = 84 comes from??? I would like to try and understand this alternate approach that uses the hypotenuse as the base.

Thanks! :)


Cross-multiply \(\frac{7}{x} = \frac{x}{12}\) to get \(7*12=x*x\) --> \(84=x^2\).
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Re: If a right triangle has area 28 and hypotenuse 12, what is  [#permalink]

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New post 16 Aug 2015, 18:20
Bunuel wrote:
DropBear wrote:
amit2k9 wrote:
1/2 * 12 * altitude = 28
altitude = 7

using similar triangle

7/x = x/12 gives x^2 = 84

12^2 - 84 = 60

thus 60 ^ (1/2) + 84 ^(1/2) + 12 = 28.7 approx.

Hence C.


I know it's been a long time since this post was made, but is anyone able to explain where 7/x = x/12 gives x^2 = 84 comes from??? I would like to try and understand this alternate approach that uses the hypotenuse as the base.

Thanks! :)


Cross-multiply \(\frac{7}{x} = \frac{x}{12}\) to get \(7*12=x*x\) --> \(84=x^2\).


Hi Bunuel,

Thanks for the very quick reply. I understand the calculation, just not sure about how we use similar triangles to arrive at that line in the first place? Don't understand why we are doing \(\frac{7}{x} = \frac{x}{12}\) in the first place... Sorry if this seems rudimentary... :(
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Re: If a right triangle has area 28 and hypotenuse 12, what is  [#permalink]

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New post 16 Aug 2015, 18:24
DropBear wrote:
amit2k9 wrote:
1/2 * 12 * altitude = 28
altitude = 7


using similar triangle

7/x = x/12 gives x^2 = 84

12^2 - 84 = 60

thus 60 ^ (1/2) + 84 ^(1/2) + 12 = 28.7 approx.

Hence C.


I know it's been a long time since this post was made, but is anyone able to explain where 7/x = x/12 gives x^2 = 84 comes from??? I would like to try and understand this alternate approach that uses the hypotenuse as the base.

Thanks! :)


I believe Bunuel has answered your question.

As a matter of fact, the text in red above is incorrect. The altitude should be 14/3 and NOT 7 as it has been calculated. The final answer as well is an integer, dont know how is the poster getting a decimal value.

IMO, the method is a 'forced' one as I am having difficulty in coming to the same equation for 'x'. Not a good method.
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If a right triangle has area 28 and hypotenuse 12, what is  [#permalink]

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New post Updated on: 16 Aug 2015, 19:48
Engr2012 wrote:
DropBear wrote:
amit2k9 wrote:
1/2 * 12 * altitude = 28
altitude = 7


using similar triangle

7/x = x/12 gives x^2 = 84

12^2 - 84 = 60

thus 60 ^ (1/2) + 84 ^(1/2) + 12 = 28.7 approx.

Hence C.


I know it's been a long time since this post was made, but is anyone able to explain where 7/x = x/12 gives x^2 = 84 comes from??? I would like to try and understand this alternate approach that uses the hypotenuse as the base.

Thanks! :)


I believe Bunuel has answered your question.

As a matter of fact, the text in red above is incorrect. The altitude should be 14/3 and NOT 7 as it has been calculated. The final answer as well is an integer, dont know how is the poster getting a decimal value.

IMO, the method is a 'forced' one as I am having difficulty in coming to the same equation for 'x'. Not a good method.


I completely overlooked that part in red, in that case 1/2 * 12 * altitude = 28 altitude = 4 2/3. However I will take your advice and leave this one alone as I don't want to confuse myself.

Edit: Edited 4 3/2 to 4 2/3

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Originally posted by DropBear on 16 Aug 2015, 18:50.
Last edited by DropBear on 16 Aug 2015, 19:48, edited 1 time in total.
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Re: If a right triangle has area 28 and hypotenuse 12, what is  [#permalink]

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New post 16 Aug 2015, 19:46
DropBear wrote:


I completely overlooked that part in red, in that case 1/2 * 12 * altitude = 28 altitude = 4 3/2. However I will take your advice and leave this one alone as I don't want to confuse myself.


I believe you meant 4 2/3 instead of 4 3/2
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If a right triangle has area 28 and hypotenuse 12, what is  [#permalink]

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New post 26 Jul 2016, 14:16
1

Another approach:


GMAT loves special triangles and guess what this is a special triangle.

With an hypotenuse of 12, the first thing that comes to my mind is a 30:60:90 triangle (l:l√3:2l),

If this was the case, we would have:
2l= 12
l=6
l√3= 6√3

To test it we simply apply the pythagorean theorem, \(6^{2}+\left( 6\sqrt{3} \right)^{2}\; has\; to\; equal\; 12^{2}\; ->\; \sqrt{36\; +\; 108}\; =\; \sqrt{144}\; ->\; 12\; =\; 12\;\)

\(->\; The\; right\; triangle\; is\; a\; 30:60:90\; triangle.\)

Now that we know the measures of the sides we simply add them.
2p= 12+6*1,7 +6= 12 + 5*2 + 6 = 28.

Answer C.
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Re: If a right triangle has area 28 and hypotenuse 12, what is  [#permalink]

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New post 02 Oct 2016, 03:16
1
what is wrong with this approach?

(1/2)*b*h=28
b*h = 56
now 56 can be broken down into following pairs
1, 56
2, 28
4, 14
8, 7
Since two sides of a triangle must be bigger than third side we can use 8,7 so the perimeter is 8+7+12=27
BUT 8^2 + 7^2 does not equal 144.

Where did I get it wrong??
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Re: If a right triangle has area 28 and hypotenuse 12, what is  [#permalink]

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New post 02 Oct 2016, 05:23
1
digitalmohsin wrote:
what is wrong with this approach?

(1/2)*b*h=28
b*h = 56
now 56 can be broken down into following pairs
1, 56
2, 28
4, 14
8, 7
Since two sides of a triangle must be bigger than third side we can use 8,7 so the perimeter is 8+7+12=27
BUT 8^2 + 7^2 does not equal 144.

Where did I get it wrong??


We are not told that the legs have integer lengths, so bh = 56 does not mean that b and h must be the values you consider.
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