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. Isn't it true that the hypotenuse of a right angled triangle, if 10, will have sides 6-8? Or is this not always the case? I know that if the triangle is not Right angled, then it is not the case;

but here: if the diagonal is 10, and since it is a rectangle, the 2 sides should be 6-8 and thus we can find the perimeter. Of course this is the same answer that you get by substituting values and using 2).

But OA says that we cannot know the sides from 1) or 2) and thus we need 1) and 2).

Concentration: International Business, General Management

GPA: 3.86

WE: Accounting (Commercial Banking)

Re: If the perimeter of rectangle Q - Confused about simple Q [#permalink]

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19 Nov 2011, 13:38

+1 C

from 1 we only know about the diagonal lenght and e do not get any information of sided so 1 not sufficient,

From two we know area 2*l*b=48 but l and b can have various values i.e for example 8*6 or12*4 so we cannot get exact values of l and B so 2 not sufficient

Now combining 1 and 2 we get area and diagonal i.e hypotenuse so now solving

we now know that diagonal which has lenght=10 and with two sides of the rectangle form a right angled triangle, and so when you consider above l or B values it only satifies for 8 and 6 or 6 and 8 as 6-8-10 form a right angle traingle, and now we know l and B values so you can find perimeter which is 2(l+B)

Re: If the perimeter of rectangle Q - Confused about simple Q [#permalink]

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19 Nov 2011, 13:57

kotela wrote:

+1 C

from 1 we only know about the diagonal lenght and e do not get any information of sided so 1 not sufficient,

From two we know area 2*l*b=48 but l and b can have various values i.e for example 8*6 or12*4 so we cannot get exact values of l and B so 2 not sufficient

Now combining 1 and 2 we get area and diagonal i.e hypotenuse so now solving

we now know that diagonal which has lenght=10 and with two sides of the rectangle form a right angled triangle, and so when you consider above l or B values it only satifies for 8 and 6 or 6 and 8 as 6-8-10 form a right angle traingle, and now we know l and B values so you can find perimeter which is 2(l+B)

Hope this clarifies

Thanks. I understand how to solve it.

My question was simply that, if a right angled triangle is given with a hypotenuse 5 or 10 for instance; can we not automatically assume that it is a 3-4-5 or a 6-8-10 triangle? I suppose you can make fractional values to satisfy this.

Nevermind, I just realized my mistake. I was doing another difficult question earlier where I came across a situation like this, but it was on a coordinate plane, and each value "HAD" to be an integer. Hence with a hypotenuse of 10, the only values for the 2 sides could have been 6 and 8. But I suppose if the integer condition is not specified, then the 2 sides of the right triangle can take any values.

. Isn't it true that the hypotenuse of a right angled triangle, if 10, will have sides 6-8? Or is this not always the case? I know that if the triangle is not Right angled, then it is not the case;

but here: if the diagonal is 10, and since it is a rectangle, the 2 sides should be 6-8 and thus we can find the perimeter. Of course this is the same answer that you get by substituting values and using 2).

But OA says that we cannot know the sides from 1) or 2) and thus we need 1) and 2).

Some explanation please.

Given: \(2*(l+b)=p\) S1: \(l^2+b^2=10^2\) (Pythagoras theorem) With this, we can't even solve l and b, so no chance of getting the p.

S2:\(l*b=48\). With this, we can't even solve l and b, so no chance of getting the p.

What we need to solve is p, which is

\(p=2*(l+b)\) Note that \((l+b)^2=l^2+b^2+2*l*b\)

Using S1 and S2: \((l+b)^2=10^2+2*48\), Thus \((l+b)\) can be found.

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