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# 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
3 years, 3 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...

- 3 years, 3 months ago

- 3 years, 3 months ago

- 3 years, 3 months ago

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

- 3 years, 3 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 $$6,8,10$$.

- 3 years, 3 months ago

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

- 3 years, 3 months ago

- 3 years, 3 months ago

Thanks very much...:-)

- 3 years, 3 months ago