The dimension of a triangle are given by (x+2) cm, (2x+7) cm, and (4x+1). Find the possible values of x that are integers. Use the formula a + b >c
Answers
Topic :-
Inequalities
Given :-
The dimension of a triangle are given by (x + 2) cm, (2x + 7) cm, and (4x + 1).
To Find :-
The possible values of 'x' that are integers.
Solution :-
According to triangle inequality theorem, "The sum of the lengths of any two sides of a triangle is always greater than the length of the third side."
Using this inequality theorem, three cases are obtained for value of 'x'.
Case 1 :
First two sides of triangle are (x + 2) cm and (2x + 7) cm.
Third side of triangle is (4x + 1) cm.
Applying Theorem,
→ (x + 2) + (2x + 7) > (4x + 1)
→ (x + 2x) + (2 + 7) > 4x + 1
→ 3x + 9 > 4x + 1
→ 4x - 3x < 9 - 1
→ x < 8
∴ x ∈ (-∞, 8)
Case 2 :
First two sides of triangle are (2x + 7) cm and (4x + 1) cm.
Third side of triangle is (x + 2) cm.
Applying Theorem,
→ (2x + 7) + (4x + 1) > (x + 2)
→ (2x + 4x) + (7 + 1) > x + 2
→ 6x + 8 > x + 2
→ 6x - x > 2 - 8
→ 5x > -6
→ x > -6/5
∴ x ∈ (-6/5, ∞)
Case 3 :
First two sides of triangle are (4x + 1) cm and (x + 2) cm.
Third side of triangle is (2x + 7) cm.
Applying Theorem,
→ (4x + 1) + (x + 2) > (2x + 7)
→ (4x + x) + (1 + 2) > 2x + 7
→ 5x + 3 > 2x + 7
→ 5x - 2x > 7 - 3
→ 3x > 4
→ x > 4/3
∴ x ∈ (4/3, ∞)
Taking intersection of obtained intervals,
→ (-∞, 8) ∩ (-6/5, ∞) ∩ (4/3, ∞)
→ (4/3, 8)
So, integers present in the obtained range would be possible values of 'x'.
Writing integers present in (4/3, 8),
Integers which are greater than 4/3 and lesser than 8 would be required answer. Thus, 2, 3, 4, 5, 6 and 7 are possible values of 'x'.
Answer :-
The possible values of x that are integers are 2, 3, 4, 5, 6 and 7 or
x ∈ {2, 3, 4, 5, 6, 7}.