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

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.

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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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
**********************

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.

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

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...

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.

I believe you meant 4 2/3 instead of 4 3/2

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.

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.

How do we know that one leg of the triangle must be the height?