Which of these can never be the ratio of the sides of the triangle? a. 3:5:7 b. 3:5:3 C. 2:2:3 d. 2:5:8
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Answer:
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To Find :- Which of these can never be the ratio of the sides of the triangle ?
A) 3:5:7
B) 3:5:3
C) 2:2:3
D) 2:5:8
Concept used :-
- Sum of any two sides of a ∆ is greater than the third side .
- Difference between any two sides of a ∆ is smaller than the third side .
Solution :-
Checking all given options we get,
A) 3 : 5 : 7
Let us assume that, sides of given ∆ are 3x , 5x and 7x unit .
So,
Sum of any two sides is greater than third side :-
→ 3x + 5x > 7x => 8x > 7x
→ 5x + 7x > 3x => 12x > 3x
→ 3x + 7x > 5x => 10x > 5x
Difference between any two sides is smaller than the third side :-
→ 5x - 3x < 7x => 2x < 7x
→ 7x - 5x < 3x => 2x < 3x
→ 7x - 3x < 5x => 4x < 5x
Therefore, given ratio is possible .
B) 3 : 5 : 3
Let us assume that, sides of given ∆ are 3x , 5x and 3x unit .
So,
Sum of any two sides is greater than third side :-
→ 3x + 5x > 3x => 8x > 3x
→ 5x + 3x > 3x => 8x > 3x
→ 3x + 3x > 5x => 6x > 5x
Difference between any two sides is smaller than the third side :-
→ 5x - 3x < 3x => 2x < 3x
→ 5x - 3x < 3x => 2x < 3x
→ 3x - 3x < 5x => 0 < 5x
Therefore, given ratio is possible .
C) 2 : 2 : 3
Let us assume that, sides of given ∆ are 2x , 2x and 3x unit .
So,
Sum of any two sides is greater than third side :-
→ 2x + 2x > 3x => 4x > 3x
→ 2x + 3x > 2x => 5x > 2x
→ 2x + 3x > 2x => 5x > 2x
Difference between any two sides is smaller than the third side :-
→ 2x - 2x < 3x => 0 < 3x
→ 3x - 2x < 2x => x < 2x
→ 3x - 2x < 2x => x < 2x
Therefore, given ratio is possible .
D) 2 : 5 : 8
Let us assume that, sides of given ∆ are 2x , 5x and 8x unit .
So,
Sum of any two sides is greater than third side :-
→ 2x + 5x > 8x
→ 7x > 8x
But, 8x > 7x .
Also,
Difference between any two sides is smaller than the third side :-
→ 8x - 2x < 5x
→ 6x < 5x
But, 6x > 5x .
Therefore, Sides of a ∆ can't be 2x , 5x and 8x units.
Hence, we can conclude that, Option (D) 2 : 5 : 8 can never be the ratio of the sides of the triangle .
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