Question

A chord of a circle has length 3n,n is a positive integer . the segment cutoff by the chard has height n,what is the smallest value of n for which the radius of circle is also a positive integer

Answer 1

Let radius of the circle is a positive integer e.g. r
Given, chord length ,PQ = 3n
so, PB = BQ = PQ/2 = 3n/2 = 1.5n
OB = OA - AB = r - n
OQ = radius of circle = r

now from Pythagoras theorem,
in ∆OBQ , OB² + BQ² = OQ²
now, (r - n)² + (1.5n)² = r²
r² + n² - 2rn + 2.25n² = r²
3.25n² - 2rn = 0
3.25n - 2r = 0
r = 3.25n/2 = 1.625n

because r is positive integer so, we have to choose a smallest positive integer of n.
let's choose n = 4
then, r = 1.625 × 4 = 6.5 ≠ integer
take n = 8 , r = 1.625 × 8 = 13 = a positive integer .

hence, smallest intger value of $\bf{n = 8}$ by which we get positive integer of r e.g., 13 unit.

Answer 2

Thank you for asking this question, here is your answer:

We will consider the radius to be r

Chord length ,PQ = 3n

PB = BQ = PQ/2 = 3n/2 = 1.5n

OB = OA - AB = r - n  

OQ = radius of circle = r

Here we will use the Pythagoras theorem

in ∆OBQ , OB² + BQ² = OQ²  

now, (r - n)² + (1.5n)² = r²  

r² + n² - 2rn + 2.25n² = r²  

3.25n² - 2rn = 0

3.25n - 2r = 0  

r = 3.25n/2 = 1.625n

Now we will chose the smallest positive integer for n and that would be 4

r = 1.625 × 4 = 6.5 ≠ integer

n = 8 , r = 1.625 × 8 = 13 = a positive integer

So we know that the smallest integer value is 8 now.

If there is any confusion please leave a comment below.