# a great challenge to solve.

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. Note by Naitik Sanghavi
6 years, 9 months ago

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$\triangle ABC$, 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

$AC=10 , BC=8, AB=6$

Great Challenge (?).... I'm a moderator, can edit titles of only problems, can't edit the titles of Notes...

- 6 years, 9 months ago

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

- 6 years, 9 months ago

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 $6,8,10$.

- 6 years, 9 months ago

- 6 years, 9 months ago

- 6 years, 9 months ago

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

- 6 years, 9 months ago

- 6 years, 9 months ago

Thanks very much...:-)

- 6 years, 9 months ago