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by kvcpk » Thu Jun 17, 2010 2:41 am
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?
Last edited by kvcpk on Thu Jun 17, 2010 3:34 am, edited 1 time in total.
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by selango » Thu Jun 17, 2010 2:44 am
Sum of 2 sides of tri >third side.

10+x>12,Possible

10+12>x,Possible.

12+x>10,Possible.

So 3 ways.
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by kvcpk » Thu Jun 17, 2010 2:49 am
selango wrote:Sum of 2 sides of tri >third side.

10+x>12,Possible

10+12>x,Possible.

12+x>10,Possible.

So 3 ways.
Can you explain in detail?
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by selango » Thu Jun 17, 2010 2:55 am
oops my mistake.

Sum of 2 sides >third side

so 10+12>x

-->x<22,So 21 integers values(1,2......21)

12+x>10,any value of x(from 1 to 21)

10+x>12,For this x>2

So to satisfy the 3rd equation,2>x<22

So 19 integer values for x are possible which will satisfy all the 3 equations.
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by kvcpk » Thu Jun 17, 2010 3:05 am
selango wrote:oops my mistake.

Sum of 2 sides >third side

so 10+12>x

-->x<22,So 21 integers values(1,2......21)

12+x>10,any value of x(from 1 to 21)

10+x>12,For this x>2

So to satisfy the 3rd equation,2>x<22

So 19 integer values for x are possible which will satisfy all the 3 equations.
Where did you make use of 'acute angled triangle' ?
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by selango » Thu Jun 17, 2010 3:10 am
kvcpk,

I cannot understand ur question.


Acute angles triangle means all angles are less than 90.
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by kvcpk » Thu Jun 17, 2010 3:14 am
I meant to say that, You have given all possible values for X, so that 10,12,X can form a triangle..

But you didnot make use of the 'acute angled triangle' given in question.

The triangles that you are forming can be obtuse angled too...
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by selango » Thu Jun 17, 2010 3:19 am
I applied the property of triangle(Sum of 2 sides>3rd side) to find the value of x.

This property ll apply to all triangle whether its acute or obtuse.


francoisph,

what is OA?
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by kvcpk » Thu Jun 17, 2010 3:27 am
selango wrote:I applied the property of triangle(Sum of 2 sides>3rd side) to find the value of x.

This property ll apply to all triangle whether its acute or obtuse.


francoisph,

what is OA?
"If 10, 12 and 'x' are sides of an acute angled triangle, how many integer values of 'x' are possible?"

here triangle is mentioned to be only aute angled.. so we need to eliminate those values of x for which the triangle becomes right angle or obtuse..
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by kvcpk » Thu Jun 17, 2010 3:32 am
case1:
x<sqrt(12^2 + 10 ^2)
therefore x<16

case2:

x>sqrt(12^2 - 10^2)
so x>6

so possible values of x are 7,8,9,10,11,12,13,14,15

so 9 vales are possible..
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by jube » Thu Jun 17, 2010 4:03 am
Based on the discussion above I think it should be 2<x<16 which would be 13 values.

x has to be less than 16 by the argument kvcpk has given above

At the same the sum of any 2 sides of a triangle should be greater than the third side i.e. x>2.

What is the OA?
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by Rich@VeritasPrep » Thu Jun 17, 2010 4:18 am
kvcpk has the correct solution mapped out rather nicely, so go to his initial response post.

And by the way, kvcpk, you did just fine! No way to make the solution to this problem any simpler :)
Rich Zwelling
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by kvcpk » Thu Jun 17, 2010 4:21 am
raz1024 wrote:kvcpk has the correct solution mapped out rather nicely, so go to his initial response post.

And by the way, kvcpk, you did just fine! No way to make the solution to this problem any simpler :)
Oh.. thanks Raz!! Glad to know that.. I have been trying different ways..
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by francoisph » Thu Jun 17, 2010 5:08 am
Finding the answer to this question requires one to know two rules in geometry.

Rule 1: For an acute angled triangle, the square of the LONGEST side MUST BE LESS than the sum of squares of the other two sides.

Rule 2: For any triangle, sum of any two sides must be greater than the third side.

The sides are 10, 12 and 'x'.

From Rule 2, x can take the following values: 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21 - A total of 19 values.

When x = 3 or x = 4 or x = 5 or x = 6, the triangle is an OBTUSE angled triangle (Rule 1 is NOT satisfied).

The smallest value of x that satisfies BOTH conditions is 7. (102 + 72 > 122).

The highest value of x that satisfies BOTH conditions is 15. (102 + 122 > 152).

When x = 16 or x = 17 or x = 18 or x = 19 or x = 20 or x = 21, the triangle is an OBTUSE angled triangle (Rule 1 is NOT satisfied).

Hence, the values of x that satisfy both the rules are x = 7, 8, 9, 10, 11, 12, 13, 14, 15. A total of 9 values.
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