IIT JEE , AIEEE Forum - Mathematics , Trignometry

meghana2007

a triangle has sides whose lengths are consecutive integers. Its area is a multiple of 20. Find the smallest triangle in which the above conditions satisfies.

Srijan_theDON

the sides are of lengths 2^1/2 -1, 2^1/2,2^1/2 +1 respectively

meghana2007

can u answer me in detail plz...

Srijan_theDON

Is the answer correct ?

meghana2007

actually...no!

master_purav

But the sides are integers. How can the side be 2^1/2???

master_purav

meghana2007

can u tell how to solve?

amar.gupta

dear,

please check and confirm whether it is a objective type question or there is some other condition which you might have forgotten to write in the question.

if it is objective type , mention the options too.

rsharma.ranchi_000

your question is wrong since triangle cannot be formed if the sides are of consecutive integer.

avinash.sharma

rsharma.ranchi_000 a triangle is possible with consecutive integer as 3,4,5 are three consecutive integers and make a right angled tringle as

3²+ 4² = 5²

But in the case of above question the reply is :

No such triangle exists. As the three consecutive numbers that satisfy the first and second conditions are as follows:

a	b	c	s=(a+b+c)/2	Area	Area/20 as (Multiple of 20)
1	2	3	3	0	0
2701	2702	2703	4053	3161340	158067

But both the points are not make a triangle as for both the cases a+b=c which is contrary to triangle definition.

In a traingle the sum of any two side must be greater than third side.

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