The perimeter of a right triangle is 24 cm. Three times the length of the longer side minus two times the length of the shorter side exceeds hypotenuse by 2 cm.what are the lengths of all three sides??state method too.

Whenever you have \(n\) equations and \(n\) variables, just go on substituting from one to other and finally you reach 1 variable which leads to all others.

Our current Example is,

(i) \(a+b+c=24\); (ii) \(3a-2b=c+2\); (iii) \(a^2+b^2=c^2\).

Use \(c=24-(a+b)\). So now you only have 2 equations and 2 variables.

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TopNewest\(\triangle ABC\), let it be right angled at \(B\). (Hypotenuse \(AC\)).

Let \(AB\leq BC\).

From \(1^{st}\) statement, \(AB+BC+CA=24\)

From \(2^{nd}\) statement, \(3\times BC -2\times AB = AC+2\)

By Pythagoras' theorem \(AC^2=AB^2+BC^2\).

Now it's simply 3 equations and 3 variables, solving simultaneously gives

\(AC=10 , BC=8, AB=6\)

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Please answer writing each step to the end..plse help me!!!

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As fast as possiblePlease help me

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Can u please write the whole calculation so I can understand because up to here I also did easily

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Whenever you have \(n\) equations and \(n\) variables, just go on substituting from one to other and finally you reach 1 variable which leads to all others.

Our current Example is,

(i) \(a+b+c=24\); (ii) \(3a-2b=c+2\); (iii) \(a^2+b^2=c^2\).

Use \(c=24-(a+b)\). So now you only have 2 equations and 2 variables.

(i) \(3a-2b=24-(a+b)+2\); (ii)\(a^2+b^2 = (24-a-b)^2\)

i.e. \(4a-b=26\) , \(576+2ab-48a-48b=0\) i.e. \(24a+24b-ab=288\)

Now do substitution \(a = \dfrac{26+b}{4}\)

\(24\biggl(\dfrac{26+b}{4}\biggr) +24b -\biggl(\dfrac{26+b}{4}\biggr) b = 288\\ \therefore 156+30b-\dfrac{b^2+26b}{4} =288\)

This is a quadratic in \(b\), and has positive root \(6\).

Put \(b\) value in \(a=\dfrac{26+b}{4} = \dfrac{26+6}{4} = 8\)

And finally, \(c=10\)

Sides \(6,8,10\).

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If someone understood the answer then please answer me writing it step by step upto the end in 12hrs plssssss.its urgent

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Done . Next time, give a

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Thanks very much...:-)

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