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
5 years ago

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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

naitik sanghavi - 5 years ago

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Done . Next time, give a reasonable heading to your note.

Aditya Raut - 5 years ago

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Thanks very much...:-)

naitik sanghavi - 5 years ago

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ABC\triangle ABC, let it be right angled at BB. (Hypotenuse ACAC).

Let ABBCAB\leq BC.

From 1st1^{st} statement, AB+BC+CA=24AB+BC+CA=24

From 2nd2^{nd} statement, 3×BC2×AB=AC+23\times BC -2\times AB = AC+2

By Pythagoras' theorem AC2=AB2+BC2AC^2=AB^2+BC^2.

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

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

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

Aditya Raut - 5 years ago

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

naitik sanghavi - 5 years ago

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As fast as possiblePlease help me

naitik sanghavi - 5 years ago

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

naitik sanghavi - 5 years ago

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Whenever you have nn equations and nn 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=24a+b+c=24; (ii) 3a2b=c+23a-2b=c+2; (iii) a2+b2=c2a^2+b^2=c^2.

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

(i) 3a2b=24(a+b)+23a-2b=24-(a+b)+2; (ii)a2+b2=(24ab)2a^2+b^2 = (24-a-b)^2

i.e. 4ab=264a-b=26 , 576+2ab48a48b=0576+2ab-48a-48b=0 i.e. 24a+24bab=28824a+24b-ab=288

Now do substitution a=26+b4a = \dfrac{26+b}{4}

24(26+b4)+24b(26+b4)b=288156+30bb2+26b4=28824\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 bb, and has positive root 66.

Put bb value in a=26+b4=26+64=8a=\dfrac{26+b}{4} = \dfrac{26+6}{4} = 8

And finally, c=10c=10

Sides 6,8,106,8,10.

Aditya Raut - 5 years ago

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