p,b,c are the side lengths of a triangle ABC with b=4 units and c=8 units, then
A
4<p<12
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B
3<p<15
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C
4<p<15
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Solution
The correct option is A4<p<12 For the given triangle ABC,c=8&b=4 units.
Now, according to triangle inequality, sum of the lengths of any two sides is always greater than the length of the third side.
Now, there are three cases:
Case 1: b+p>c ⇒4+p>8 ⇒p>4
Case 2: p+c>b ⇒p+8>4
Which is true always, so we will ignore this case.
Case 3: b+c>p ⇒12>p
Now, combining all three cases, that are: p>4 and 12>p ∴ Combining both the cases, we get: 4<p<12