kvcpk wrote:

francoisph wrote:If 10, 12 and 'x' are sides of an acute angled triangle, how many integer values of 'x' are possible?

Given two sides as 10, 12

If the triangle has to be acute angled, no angles hud be greater than 90.
therefore, let us see what if one angles is 90 degrees.

x^2 = 10^2 + 12^2
so x = sqrt(244) = less than 16

Also, 12^2 = 10^2 + x^2 can also be possible
In this case we will get x as greater than 6

sum of two sides shud be grater than third side.. so x cannot be less than 2.

so possible vales of X are 7,8,9.....15
Not sure if this is the rite procedure.. Will think in a different way and try again..

By the way what is OA?

ok so might be a stupid question but kvcpk why is x < 16 & >6 and not the other way around, etc.?

jube wrote: x^2 = 10^2 + 12^2
so x = sqrt(244) = less than 16

Also, 12^2 = 10^2 + x^2 can also be possible
In this case we will get x as greater than 6

ok so might be a stupid question but kvcpk why is x < 16 & >6 and not the other way around, etc.?

Hi Jube,

We know that the triangle should not be Obtuse or Right angle..
We know that the sides of the triangle are 12,10,x

The angle opposite to the longest side will be the largest angle.

So, assume 12 is the longest side.. it means that angle opposite to it should not be obtuse or right angle..
Assume it as right angle..
then, 12^2 = 10^2 + x^2
x^2 = 12^2 - 10^2
On solving we get x = 6.63
So for all values of X less than or equal to 6, angle opposite to side 12 keeps increasing and will be obtuse or right angle..

but, we need only acute..so possible values are 7,8,9,..............

similarly, now let us see the condition when X is the longest side..
for simial reason as above,
x^2 = 12^2 + 10^2
on solving x = 15.62
So for all values of X greater than or equal to 16, angle opposite to side X keeps increasing and will be obtuse or right angle..

but, we need only acute..so possible values are 15,14,13......

There cannot be a case where 10 is the greatest side.

so Intersection of these two sets gives us the possible values for X. which are 7,8,9....15

Hope this helps!!

Praveen

kvcpk wrote: Hi Jube,

We know that the triangle should not be Obtuse or Right angle..
We know that the sides of the triangle are 12,10,x

The angle opposite to the longest side will be the largest angle.

So, assume 12 is the longest side.. it means that angle opposite to it should not be obtuse or right angle..
Assume it as right angle..
then, 12^2 = 10^2 + x^2
x^2 = 12^2 - 10^2
On solving we get x = 6.63
So for all values of X less than or equal to 6, angle opposite to side 12 keeps increasing and will be obtuse or right angle..

but, we need only acute..so possible values are 7,8,9,..............

similarly, now let us see the condition when X is the longest side..
for simial reason as above,
x^2 = 12^2 + 10^2
on solving x = 15.62
So for all values of X greater than or equal to 16, angle opposite to side X keeps increasing and will be obtuse or right angle..

but, we need only acute..so possible values are 15,14,13......

There cannot be a case where 10 is the greatest side.

so Intersection of these two sets gives us the possible values for X. which are 7,8,9....15

Hope this helps!!

Praveen

Perfect! I see the point I was missing out on! Thanks.

explanation of kvcpk is fast and good!

thks!

francoisph wrote:explanation of kvcpk is fast and good!

thks!

Glad to hear that francoisph!!