The side of a triangle are 5,12, and n. write an inequality that expresses the interval of values that n amy have
Answers
Answered by
55
Heya User,
--> By triangle inequality which states :->
" Sum of any two sides of a Δ > the third side "
--> We can write :->
5 + 12 > n => 17 > n --> ( i )
--> 5 + n > 12 => n > 7 --> ( ii )
--> 12 + n > 5 => 7 +n > 0
However, from ( i ) and ( ii ) -->
---> n > 7 and 17 > n
=> 7 < n < 17
--> +_+ There I found the Interval ...
--> By triangle inequality which states :->
" Sum of any two sides of a Δ > the third side "
--> We can write :->
5 + 12 > n => 17 > n --> ( i )
--> 5 + n > 12 => n > 7 --> ( ii )
--> 12 + n > 5 => 7 +n > 0
However, from ( i ) and ( ii ) -->
---> n > 7 and 17 > n
=> 7 < n < 17
--> +_+ There I found the Interval ...
Answered by
3
Answer:
The correct option is c) 7 < n < 17
Step-by-step explanation:
From the above question,
They have given :
Question :
The sides of a triangle are 5, 12, and n. Write an inequality that expresses the interval of values that n may have.
a) 5 < n < 12
b) 5 < n < 17
c) 7 < n < 17
d) 7 < n < 12
The correct option is c) 7 < n < 17
Since the sum of third side of a triangle is smaller than the sum of other two sides 7 < n < 17
The 3rd side must be less than the sum of the other two sides:
n < 12 + 5 = 17
The 3rd side must be greater than the difference:
12 - 5 = 7
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