Number of obtuse (or acute) triangles with sides $8$, $15$, and integer $x$

Question

Consider obtuse-angled triangles with sides $8$ cm, $15$ cm, and $x$ cm. If $x$ is an integer, then how many such triangles exist?

I approached this problem by assuming three cases :-

• 1st Case where $x<8<15$ then as per triangle's side rules $x+8 > 15 \Rightarrow x>7$.

• 2nd Case where $8<x<15$ then as per triangle's side rules $x+8 > 15 \Rightarrow x>7$.

• 3rd Case where $8<15<x$ then as per triangle's side rules $15+8 > x \Rightarrow x<23$.

So combining all the cases I can say that $7<x<23$ and there could be $15$ possible triangles but the answer given is $10$. What am I doing wrong?

And also how one can find the number of acute-angled triangles with sides $8$ cm, $15$ cm, and} $x$ cm given the same condition for $x$.

Please help!

Thanks in advance !

Answer 1

If in a triangle, $a \leq b \leq c$, $c^2 = a^2 + b^2$ gives you right angled triangle. $a^2 + b^2 \gt c^2$ gives you acute angled triangle and $a^2+b^2 \lt c^2$ gives you obtuse angled triangle (you can see it by law of cosine or using geometry).

When $7 \lt x \leq 15$, $8^2 + x^2 \lt 15^2$ gives you obtuse angled triangles. So you get $7 \lt x \lt 13$.

When $15 \lt x \lt 23$, $8^2 + 15^2 \lt x^2$ gives you obtuse angled triangles. So you get $17 \lt x \lt 23$.

Answer 2

Another approach:

Let AC=b=15 , BC=a=8 and AB=c=x we may write . Suppose $\angle ABC=90^o$ then :

$x^2=15^2-8^2\rightarrow x=12.68$

Therefore $ 15-8=7<x<12$, so numbers 8, 9, 10, 11 and 12 can be candidates, that is there is 5 possibility.Now suppose $\angle ACB=90^o$ then:

$AC^2=x^2=15^2+8^2=289\rightarrow x=17$

In this case $17<x<15+8=23$ and numbers 18, 19, 20,21 and 22 can be right for x.Hence total possibility is 10.

For acute triangle we must have $12<x<17$ which gives $x=13, 14, 15, 16$ as four possibility.