Question

A 144 piece of wire is bent to form a triangle with integral lengths. How many isosceles triangles can be formed?

Answer 1

Answer:

If the sides of the isosceles triangle are a , a , b , we must have 2a+b=144 with a,b∈N and b<2a . Thus, b=144–2a<2a , so that a>36 , and 2a<144 , so that a<72 . We must exclude the case a=b , for that gives an equilateral triangle. Since a=b implies a=48 , a∈{37,38,39,…,71}∖{48} .

There are 71–37=34 isosceles triangles with integer sides and perimeter 144 . ■

Answer 2

Given :- A 144m piece of wire is bent to form a triangle with integral lengths. How many isosceles triangles can be formed ?

Solution :-

→ Length of wire = 144m .

since length will remain same .

so,

→ Perimeter of isosceles ∆ = 144 m .

let us assume that, equal sides are x m and third side is y m .

then,

→ x + x + y = 144

→ 2x + y = 144

→ y = (144 - 2x) -------- Eqn.(1)

now,

→ 2x > y ---------- Eqn.(2) { sum of any two sides is greater than the third side. }

and,

→ 2x < 144

→ x < 72 --------- Eqn.(3)

putting Eqn.(2) in Eqn.(1)

→ 2x > 144 - 2x

→ 2x + 2x > 144

→ 4x > 144

→ x > 36 -------- Eqn.(4)

then , from Eqn.(3) and Eqn.(4)

→ 36 < x < 72

as given that, x is an integer .

therefore,

→ Number of Possible values of x = (72 - 36) - 1 = 35 (Ans.)

Learn more :-

let a and b positive integers such that 90 less than a+b less than 99 and 0.9 less than a/b less than 0.91. Find ab/46

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