An isosceles triangle (2 sides equal) whose equal sides are of length b units has a perimeter of 32 units. What is the length of all the sides if the third side is 11 units more than the other 2 sides?

8,8,19

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6,6,17

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7,7,18

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9,9,20

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Solution

The correct option is C 7,7,18
It is given that the equal sides are of length b units.
Third side is 11 units more than the other 2 sides, that is, Third side $= b + 11$
Perimeter of the triangle is given as 32, $b + b + b + 11 = 32$
$⟹ 3 b + 11 = 32$

Value of b L.H.S R.H.S Balanced

b=8 35 32 no

b=6 29 32 no

b=7 32 32 yes

Therefore third side $= b + 11 = 7 + 11 = 18 u n i t s$

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