girisshhh84 wrote:
HI Karishma ,
Thank you , i could understand the explanation as you told me .
But Still have a doubt with 'bangalorian2000's statement "As this is acute angled triangle x must be largest hence x > 12 and x < 22 so total of 9 possible values hence C."
To help me understand , pls tell me when we have
2 < x < 22 i.e. Total of 19 possible values.
How many possible values of x when obtuse triangle ?
How many values of x when right triangle.
Ok. A side of a triangle is less than the sum of other two sides.
Also a side of a triangle is greater than the difference of other two sides. Since the other two sides are 10 and 12, x should be less than 22 and more than 2.
Now we only have to consider acute angled triangles. Look at the figures below.
Attachment:
Ques.jpg [ 17.39 KiB | Viewed 39881 times ]
In the first figure when x is very small, till the angle goes to 90, the triangle is obtuse, which is not allowed.
When x becomes \(\sqrt{(12^2 - 10^2)}\) = root 44 = 6.something, the angle is right angled. Now x greater than this value is allowed since we will get acute triangles.
Values of 3, 4, 5 and 6 give obtuse angled triangles.
Values of 7, 8, 9 ... give acute triangles so allowed.
Now look at the second figure. Finally x is \(\sqrt{(12^2 + 10^2)}\) = root 244 = 15.something
So values of x till 15.something make acute triangles. Therefore 7, 8, 9, 10, 11, 12, 13, 14 and 15 (9 values) are allowed.
When x is even bigger (16, 17, 18, 19, 20, 21), it will again make an obtuse angled triangle.
And, as we saw, there are two values of x (6.something and 15.something) when we get a right angled triangle.
As for bangalorian2000's statement, he was probably mistaken.
_________________
Karishma
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