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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] New post 13 Jun 2010, 04:04
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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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 26 Dec 2013, 02:47, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: Doubtful PS [#permalink] New post 13 Jun 2010, 05:17
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] New post 13 Jun 2010, 05:27
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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] New post 13 Jun 2010, 06:41
xy=56 & x+y=16,, what will be the values of x & y??

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Re: Doubtful PS [#permalink] New post 13 Jun 2010, 06:54
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Hussain15 wrote:
xy=56 & x+y=16,, what will be the values of x & y??

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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] New post 13 Jun 2010, 08:12
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] New post 21 May 2011, 17:58
Apologies

Can someone explain why we are adding 2ab?

I don't understand that part.

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Re: Doubtful PS [#permalink] New post 21 May 2011, 22:33
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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] New post 23 May 2011, 22:14
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] New post 27 Aug 2011, 07:06
thank you that was very helpful
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Re: Doubtful PS [#permalink] New post 27 Aug 2011, 08:29
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] New post 28 Jun 2015, 09:50
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Re: If a right triangle has area 28 and hypotenuse 12, what is   [#permalink] 28 Jun 2015, 09:50
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