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An isosceles triangle has two sides of equal length, [tex]a[/tex], and a base, [tex]b[/tex]. The perimeter of the triangle is 15.7 inches, so the equation to solve is [tex]2a + b = 15.7[/tex].

If we recall that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side, which lengths make sense for possible values of [tex]b[/tex]? Select two options.

- 2 in.
- 0 in.
- 0.5 in.
- 2 in.
- 7.9 in.

Sagot :

To solve this problem, we need to check which possible base lengths [tex]\( b \)[/tex] satisfy the constraints given by the triangle inequality theorem, given the perimeter of 15.7 inches.

The perimeter of the isosceles triangle with two equal sides [tex]\( a \)[/tex] and base [tex]\( b \)[/tex] is given by:
[tex]\[ 2a + b = 15.7 \][/tex]

From this, we can express [tex]\( a \)[/tex] in terms of [tex]\( b \)[/tex]:
[tex]\[ a = \frac{15.7 - b}{2} \][/tex]

Next, we should check each possible value of [tex]\( b \)[/tex] to see if it meets the triangle inequality conditions, specifically:
1. [tex]\( a + a > b \)[/tex]
2. [tex]\( a + b > a \)[/tex]
3. [tex]\( b + a > a \)[/tex]

However, for an isosceles triangle, if the first condition [tex]\( 2a > b \)[/tex] is ensured and the values are positive, the other two conditions will naturally be satisfied. Let's evaluate each possible value of [tex]\( b \)[/tex]:

1. [tex]\( b = -2 \)[/tex] inches:
[tex]\[ a = \frac{15.7 - (-2)}{2} = \frac{15.7 + 2}{2} = \frac{17.7}{2} = 8.85 \][/tex]
- Check [tex]\( 2a > b \)[/tex]:
[tex]\[ 2(8.85) > -2 \][/tex]
[tex]\[ 17.7 > -2 \][/tex]
- This condition is valid, however, a base length cannot be negative.

Therefore, [tex]\( b = -2 \)[/tex] is not a valid value for a base.

2. [tex]\( b = 0 \)[/tex] inches:
[tex]\[ a = \frac{15.7 - 0}{2} = \frac{15.7}{2} = 7.85 \][/tex]
- Check [tex]\( 2a > b \)[/tex]:
[tex]\[ 2(7.85) > 0 \][/tex]
[tex]\[ 15.7 > 0 \][/tex]
- This condition is valid, but a base length cannot be zero for a valid triangle.

Therefore, [tex]\( b = 0 \)[/tex] is not a valid value for a base.

3. [tex]\( b = 0.5 \)[/tex] inches:
[tex]\[ a = \frac{15.7 - 0.5}{2} = \frac{15.2}{2} = 7.6 \][/tex]
- Check [tex]\( 2a > b \)[/tex]:
[tex]\[ 2(7.6) > 0.5 \][/tex]
[tex]\[ 15.2 > 0.5 \][/tex]
- This condition is valid.

Therefore, [tex]\( b = 0.5 \)[/tex] is a valid value for a base.

4. [tex]\( b = 2 \)[/tex] inches:
[tex]\[ a = \frac{15.7 - 2}{2} = \frac{13.7}{2} = 6.85 \][/tex]
- Check [tex]\( 2a > b \)[/tex]:
[tex]\[ 2(6.85) > 2 \][/tex]
[tex]\[ 13.7 > 2 \][/tex]
- This condition is valid.

Therefore, [tex]\( b = 2 \)[/tex] is a valid value for a base.

5. [tex]\( b = 7.9 \)[/tex] inches:
[tex]\[ a = \frac{15.7 - 7.9}{2} = \frac{7.8}{2} = 3.9 \][/tex]
- Check [tex]\( 2a > b \)[/tex]:
[tex]\[ 2(3.9) > 7.9 \][/tex]
[tex]\[ 7.8 > 7.9 \][/tex]
- This condition is not valid.

Therefore, [tex]\( b = 7.9 \)[/tex] is not a valid value for a base.

The possible valid values for the base [tex]\( b \)[/tex] are:

[tex]\[ \boxed{0.5 \text{ inches and } 2 \text{ inches}} \][/tex]
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