Question

In a right angle triangle, the length of the sides are in AP. If the lengths of the sides of the triangle are in integers, which of the following could be the length of the shortest side​

Answer 1

Answer 3,4,5

Let the terms be a,a-1,a+1

By Pythagoreans theorem

a^2+(a-1)^2=(a+1)^2

a^2+a^2+1-2a=a^2+1+2a

a^2-4a=0

a(a-4)=0

a=0

Or

a=4

So,

a+1=5

a=4

a-1=3

Answer 2

The length of the smallest side of such a triangle will be 3 units.

Explanation:

Given that the lengths of the side of a right angled triangle are integers in Arithmetic progression. So, we assume the length of the sides to be:

$a,\ (a+n),(a-n)$

where:

$a=$length of smallest side

$n=$common difference of AP

Now using Pythagoras theorem:

$(a+n)^2=a^2+(a-n)^2$

$a^2+n^2+2an=a^2+n^2-2an+a^2$

$4an=a^2$

$4n=a$

now according to the question we need integers, so we put the smallest integer as n=1 to get the side as an integer.

$a=4$

and other sides will be: 3 & 5

So the length of the smallest side of such a triangle will be 3 units.

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TOPIC: arithmetic progression.

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