We are asked to solve three separate math problems: Problem 4: Given a circle with a diameter of 4 cm, find the radius. Problem 5: In a circle, if $AY=16$ cm, find $AC$ and $BX$. Problem 6: Given a circle with a radius of 9 m, find the exact circumference in terms of $\pi$ and approximate the circumference using a calculator, rounded to the nearest hundredth.

1. Problem Description

We are asked to solve three separate math problems:

Problem 4: Given a circle with a diameter of 4 cm, find the radius.

Problem 5: In a circle, if $AY=16$ cm, find $AC$ and $BX$.

Problem 6: Given a circle with a radius of 9 m, find the exact circumference in terms of $\pi$ and approximate the circumference using a calculator, rounded to the nearest hundredth.

2. Solution Steps

Problem 4:

The radius of a circle is half of its diameter. Given that the diameter is 4 cm, the radius is $\frac{4}{2}$ cm.

Problem 5:

$AY$ is a chord and $AC$ is a diameter. Also, $X$ is on the circle and thus $AXC$ is an inscribed angle subtending a diameter, hence it is a right angle.

Because $AY$ is a chord and $AC$ is a diameter, we can use properties of right triangles inscribed in circles. However, without more information about $Y$, $B$, or $X$ relationship to the circle, we cannot simply determine AC and BX. Given $AY=16$cm, consider if $AY=AX$. Since the angle subtended by $AC$ is 90 degrees, then $AYC$ forms a right triangle. If $AY=YC$, then angle $YAC=45$ degrees. Thus $AC = AY\sqrt{2}$.

$AC = 16\sqrt{2} \approx 22.63$.

$AC$ is also a diameter.

$BX$ is part of the diameter. If B is the center of the circle, $BX$ is the radius which is half of $AC$.

$BX=\frac{16\sqrt{2}}{2}=8\sqrt{2} \approx 11.31$.

Assuming $B$ is the center, and since we are given that $AY = 16$ cm, we need to find $AC$ and $BX$.

Since points $A, X, C$ lie on the circle, and the angle $\angle AXC = 90^{\circ}$, then $AC$ is a diameter. The triangle $AXC$ is a right triangle. Since we are only given $AY = 16$, we cannot directly find the diameter $AC$ or $BX$ which is the radius. This problem may be missing key information. It looks like we can infer $AY = AX$.

If we assume that $AX = 16$, then $AC = 16\sqrt{2}$. So $AC \approx 22.63$ cm.

Then $BX = \frac{AC}{2}$, so $BX = 8\sqrt{2} \approx 11.31$ cm.

Problem 6:

The radius of the circle is 9 m.

(a) The formula for the circumference $C$ of a circle is $C = 2\pi r$, where $r$ is the radius.

In this case, $r = 9$ m, so the circumference is $C = 2\pi(9) = 18\pi$ m.

(b) Using a calculator, we can approximate $18\pi$.

$18\pi \approx 56.5486677646...$

Rounding to the nearest hundredth, we get $56.55$ m.

3. Final Answer

Problem 4: 2 cm

Problem 5:

$AC = 16\sqrt{2}$ cm $\approx 22.63$ cm

$BX = 8\sqrt{2}$ cm $\approx 11.31$ cm

Problem 6:

(a) $18\pi$ m

(b) 56.55 m