In an isosceles triangle the base is a whole number and is 4 feet less than the sum of the two equal sides. the perimeter is a whole number between 0 and 75 feet. find the possible lengths of the equal sides

1 < x < 19

Step-by-step explanation:

The three sides of the isosceles triangle would be

x, x, and 2x-4

Since the perimeter is between 0 and 75 we can make an inequality

0 < x+x+(2x-4) < 75

0 < 4x-4 < 75

4< 4x < 79

1 < x < 79/4

1 < x < 19.75

(whole number)

1 < x <19

We have duplicated side length a and base b.  Since we have a triangle, we have the triangle inequalities:

a > 0

b > 0

2a > b

And a+b > a which doesn't add any information.

Since the perimeter p=2a+b is a whole number and b is a whole number we conclude p-b=2a is a whole number.  That doesn't mean a is a whole number, it could have a fractional part of 1/2.

The perimeter constraint is

0 < 2a + b ≤ 75

The relation between a and b is

b = 2a - 4

That automatically means 2a > b so we don't have to worry about that one.

0 < 2a + 2a - 4  ≤ 75

4 < 4a ≤ 79

1 < a  ≤ 79/4

1 < a  ≤ 19.75

So far that gives possible a values of 1.5, 2, 2.5, ..., 19, 19.5

But we still have our other constraints, in particular

b > 0

2a  - 4 > 0

2a > 4

a > 2

So our possible lengths of a are:

2.5, 3, 3.5, 4, 4.5, 5, 5.5, 6, 6.5, 7, 7.5, 8, 8.5, 9, 9.5, 10, 10.5, 11, 11.5, 12, 12.5, 13, 13.5, 14, 14.5, 15, 15.5, 16, 16.5, 17, 17.5, 18, 18.5, 19 and 19.5

Let's check a=2.5 and a=19.5

a=2.5, b= 2a-4 = 5-4 = 1, perimeter p=2.5+2.5+1=6,  seems good

a=19.5, b=2a-4=35, p=19.5+19.5+35=74, seems good

let's go outside the bounds,

a=2, b=2(2)-4=0, not a triangle, good

a=20, b=2(20)-4=36, p=20+20+36=76, over the limit, good

Seems right.

answer

false

explanation

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