Question

George is given two circles, circle O and circle X, as shown. If he wants toprove that the two circles are similar, what would be the correct second stepin his proof?Given: The radius of circle O is r, and the radius of circle X is .Prove: Circle Ois similar to circle X.1.C = 2rr and C'= 2rr' by the definition of circumference.2._3. d= 2rand d = 2r by the definition of diameter.d= 2r4. d'= 2r andby the division property of equality.5d=2rby the substitution property.and6. Circle Ois similar to circle Xbecause all the linear dimensions are in thesame proportion.

Answer 1

Answer: B

The correct second step for the given proof is;

$\begin{gathered} \frac{C}{C^{\prime}}=\frac{2\pi r}{2\pi r^{\prime}} \\ and\text{;} \\ \frac{C}{C^{\prime}}=\frac{r}{r^{\prime}} \end{gathered}$

by the division property of equality.

Explanation:

We want to find the correct second step for the given proof.

Recall that from division property of equality, when we have;

a=b and c=d, dividing the equations by each other, the equation will still be equal;

$\frac{a}{c}=\frac{b}{d}$

For the given Prove;

statement 1 states that the circumference circle 1 and 2 can be expressed as;

$\begin{gathered} C=2\pi r \\ \text{and} \\ C^{\prime}=2\pi r^{\prime} \end{gathered}$

So, for step 2;

Applying the division property of equality, we have;

$\begin{gathered} \frac{C}{C^{\prime}}=\frac{2\pi r}{2\pi r^{\prime}} \\ \text{which then equals;} \\ \frac{C}{C^{\prime}}=\frac{r}{r^{\prime}} \end{gathered}$

Therefore, the correct second step for the given proof is;

$\begin{gathered} \frac{C}{C^{\prime}}=\frac{2\pi r}{2\pi r^{\prime}} \\ and\text{;} \\ \frac{C}{C^{\prime}}=\frac{r}{r^{\prime}} \end{gathered}$

by the division property of equality.