Question

An isosceles triangle has two sides of equal length, a, and a base, b. The perimeter of the triangle is 15.7 inches, so the equation to solve is 2a + b = 15.7. 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 b? Select two options.
–2 in.
0 in.
0.5 in.
2 in.
7.9 in

Answer 1

Answer:

b=0.5 in

b=2 in

Step-by-step explanation:

we know that

The perimeter of triangle is equal to

$2a+b=15$

Solve for a

$a=\frac{15-b}{2}$ -----> equation A

Applying the Triangle Inequality Theorem

a+a > b

2a > b -----> inequality B

Verify each case

case 1) b=-2 in

This value not make sense, the length side cannot be a negative number

case 2) b=0 in

This value not make sense

case 3) b=0.5 in

substitute the value of b in the equation A and solve for a

$a=\frac{15-0.5}{2}=7.25\ in$

substitute the values of b and  a in the inequality B

2a > b

2(7.25) > 0.5

14.50 > 0.5 -----> is  true

therefore

b=0.5 in  make sense for possible values of b

case 4) b=2 in

substitute the value of b in the equation A and solve for a

$a=\frac{15-2}{2}=6.5\ in$

substitute the values of b and  a in the inequality B

2a > b

2(6.5) > 2

13 > 2 -----> is true

therefore

b=2 in  make sense for possible values of b

case 5) b=7.9 in

substitute the value of b in the equation A and solve for a

$a=\frac{15-7.9}{2}=3.55\ in$

substitute the values of b and  a in the inequality B

2a > b

2(3.55) >7.9

7.1 > 7.9 -----> is not true

therefore

b=7.9 in  not make sense for possible values of b

Answer 2

Answer:

0.5 in and 2 in.

Step-by-step explanation:

We have:

$2a+b=15.7$                subtract b from both sides

$2a+b-b=15.7-b$

$2a=15.7-b$             divide both sides by 2

$\dfrac{2a}{2}=\dfrac{15.7-b}{2}$

$a=\dfrac{15.7-b}{2}\qquad\bold{(*)}$

We know:

$2a>b\qquad\bold{(**)}$

Subtitute (*) to (**):

$2\!\!\!\!\diagup\cdot\dfrac{15.7-b}{2\!\!\!\!\diagup}>b$

$15.7-b>b$                 add  to both sides

$15.7-b+b>b+b$

$15.7>2b$                 divide both sides by 2

$\dfrac{15.7}{2}>\dfrac{2}{b}\\\\7.85>b\to \boxed{b0}$

Only lengths 0.5in and 2in satisfy this inequality.