Question

If \[k\] is an integer and \[2 < k < 7\], for how many different values of \[k\] is there a triangle with sides of lengths \[2\], \[k\] and \[7\]?
A.One
B.Two
C.Thre
D.Four
E. Five

Answer 1

Hint: We know that triangle has three sides and three angles, some of which may be the same. We know that the sum of the measures of the three angles of a triangle is always \[{180^0}\]. To solve this question we need the triangle inequality theorem. It states that the sum of the smaller two sides of a triangle not larger than the third.

Complete step-by-step answer:
Let us see the triangle inequality theorem.

We have
 \[A + B > C\]
 \[B + C > A\]
 \[A + C > B\].
Given, that \[k\] lies in between \[2\] and \[7\]. So \[k\] takes the values \[3\], \[4\], \[5\] and \[6\].
Seeing above diagram and inequality, applying we get that the only possible triangle is \[2\], \[7\] and \[k = 6\]. Because these sides will satisfy the triangle inequality theorem.
That is,\[2 + 6 > 7\].
While for \[k\] values \[3\], \[4\] and \[5\], the sum of the smaller two sides is not larger than the third side. Thus \[k = 6\] is the only possible value of \[k\] that satisfies the triangle inequality theorem.
So for only $one$ value of \[k\] is there a triangle with sides of lengths \[2\], \[k\] and\[7\].
So, the correct answer is “Option A”.

Note: You can check for different values of \[k\]. Let’s check for \[k = 3\] then following the above procedure we get, \[2 + 3 > 7\] which is a contradiction to the triangle inequality theorem. Similarly we can check for the remaining value of \[k\]. Note that in the triangle inequality theorem all the above three conditions must be satisfied.