Bunuel
All possible obtuse angled triangle with sides of integers lengths are constructed , such that two of the sides have lengths 7 and 14.How many such triangles exist ?
A. 13
B. 11
C. 10
D. 8
E. 7
Two properties of triangles that can help us solve this question -
1) The third side of a triangle is always greater than the positive difference between the other two sides and is less than the sum of the other two sides.
In this case
14 - 7 < Third Side < 14+7
7 < Third Side < 21
2) In an obtuse-angled triangle, the square of the sum of the longest side is greater than the sum of the other two sides
(Longest Side) ^ 2 > (Side 1)^2 + (Side 2)^2
Let's apply these two properties -
Case 1: Assume the side with length 14 is the longest side
\(14^2 > 7^2 + x^2\)
x is the length of the third side
\(x^2 < 14^2 - 7^2\)
\(x^2 < (14+7)(14-7)\)
\(x^2 < 147\)
x < 12.XX (x is greater than 12 but less than 13)
Possible values of x = 8, 9, 10 , 11, 12
Case 2: Assume the unknown side is the longest side
\(x^2 > 14^2 + 7^2\)
\(x^2 > 196 + 49\)
\(x^2 > 245\)
\(x > 15.XX\)
Possible values of x = 16, 17, 18, 19, 20
Note that we cannot exceed 20, as the side should be less than 21
Number of possible values of the third side = 10
Option C