Understanding the Problem: Calculating Circle Radii

The problem asks us to find the radii of two circles given the ratio of their diameters and the condition that the sum of their areas equals the area of a third circle with a known diameter. This involves using the formulas for the area of a circle and understanding ratios.

Setting Up the Ratios and Formulas

Let the two circles be Circle 1 and Circle 2, and the third circle be Circle 3.

• Let \(d_1\) and \(d_2\) be the diameters of Circle 1 and Circle 2, respectively.
• The ratio of their diameters is given as \(d_1 : d_2 = 12 : 5\).
• This implies that for some constant \(k\), \(d_1 = 12k\) and \(d_2 = 5k\).
• Let \(r_1\) and \(r_2\) be the radii of Circle 1 and Circle 2. The radius is half the diameter, so \(r_1 = d_1/2\) and \(r_2 = d_2/2\).
• The ratio of radii is also \(r_1 : r_2 = (d_1/2) : (d_2/2) = d_1 : d_2 = 12 : 5\).
• So, we can write \(r_1 = 12k'\) and \(r_2 = 5k'\) for some constant \(k'\). Note that \(k'\) here is \(k/2\) from the diameter relation, but we can just use a single constant for the ratio of radii. Let's stick to \(r_1 = 12m\) and \(r_2 = 5m\) for simplicity, where \(m\) is our constant.

The area of a circle is given by the formula \(A = \pi r^2\).

• Area of Circle 1: \(A_1 = \pi r_1^2 = \pi (12m)^2 = \pi (144m^2)\).
• Area of Circle 2: \(A_2 = \pi r_2^2 = \pi (5m)^2 = \pi (25m^2)\).

The third circle has a diameter of 65 cm. Let its diameter be \(d_3\) and radius be \(r_3\).

• \(d_3 = 65\) cm.
• \(r_3 = d_3/2 = 65/2 = 32.5\) cm.
• Area of Circle 3: \(A_3 = \pi r_3^2 = \pi (32.5)^2\).

Using the Sum of Areas Condition

The problem states that the sum of the areas of the first two circles is equal to the area of the third circle:

\(\quad A_1 + A_2 = A_3\)

Substitute the area formulas:

\(\quad \pi (144m^2) + \pi (25m^2) = \pi (32.5)^2\)

We can divide the entire equation by \(\pi\) (since \(\pi \ne 0\)):

\(\quad 144m^2 + 25m^2 = (32.5)^2\)

Combine the terms on the left side:

\(\quad (144 + 25)m^2 = (32.5)^2\)

\(\quad 169m^2 = (32.5)^2\)

To solve for \(m^2\), divide both sides by 169:

\(\quad m^2 = \frac{(32.5)^2}{169}\)

We know that \(32.5 = \frac{65}{2}\) and \(169 = 13^2\). Also, \(65 = 5 \times 13\).

\(\quad m^2 = \frac{(\frac{65}{2})^2}{13^2} = \frac{\frac{65^2}{2^2}}{13^2} = \frac{65^2}{4 \times 13^2}\)

\(\quad m^2 = \frac{(5 \times 13)^2}{4 \times 13^2} = \frac{5^2 \times 13^2}{4 \times 13^2}\)

Cancel out \(13^2\) from numerator and denominator:

\(\quad m^2 = \frac{5^2}{4} = \frac{25}{4}\)

Now, take the square root of both sides to find \(m\). Since radius must be positive, we take the positive root:

\(\quad m = \sqrt{\frac{25}{4}} = \frac{\sqrt{25}}{\sqrt{4}} = \frac{5}{2} = 2.5\)

Calculating the Radii

We defined the radii as \(r_1 = 12m\) and \(r_2 = 5m\). Substitute the value of \(m = 2.5\) cm:

• \(r_1 = 12 \times 2.5 = 12 \times \frac{5}{2} = 6 \times 5 = 30\) cm.
• \(r_2 = 5 \times 2.5 = 5 \times \frac{5}{2} = \frac{25}{2} = 12.5\) cm.

Thus, the radii of the two circles are 30 cm and 12.5 cm.

Verification

Let's check if the sum of their areas equals the area of the circle with 65 cm diameter.

• \(r_1 = 30\) cm, \(r_2 = 12.5\) cm.
• \(A_1 = \pi (30)^2 = 900\pi\) sq cm.
• \(A_2 = \pi (12.5)^2 = \pi (\frac{25}{2})^2 = \pi \frac{625}{4} = 156.25\pi\) sq cm.
• \(A_1 + A_2 = 900\pi + 156.25\pi = 1056.25\pi\) sq cm.

The third circle has diameter 65 cm, so radius \(r_3 = 32.5\) cm.

• \(A_3 = \pi (32.5)^2 = \pi (\frac{65}{2})^2 = \pi \frac{4225}{4} = 1056.25\pi\) sq cm.

Since \(A_1 + A_2 = A_3\), our calculated radii are correct.

Summary of Radii

Revision Table: Key Concepts

Additional Information: Scaling Areas

This problem demonstrates a key principle related to areas of similar figures (circles are always similar). If the ratio of corresponding linear dimensions (like radius or diameter) of two similar figures is \(a:b\), then the ratio of their areas is \(a^2:b^2\).

In our case, if \(r_1 : r_2 = 12 : 5\), then \(r_1 = 12m\) and \(r_2 = 5m\). The areas are proportional to the square of the radii: \(A_1 \propto (12m)^2 = 144m^2\) and \(A_2 \propto (5m)^2 = 25m^2\). So \(A_1 : A_2 = 144 : 25\).

The condition \(A_1 + A_2 = A_3\) implies \(\pi r_1^2 + \pi r_2^2 = \pi r_3^2\), which simplifies to \(r_1^2 + r_2^2 = r_3^2\). This equation is similar in form to the Pythagorean theorem (\(a^2 + b^2 = c^2\)), but it applies to the radii of circles whose areas sum up.

If the radii of two circles are \(r_1\) and \(r_2\), and the radius of a third circle is \(r_3\) such that \(A_1 + A_2 = A_3\), then \(r_1^2 + r_2^2 = r_3^2\). This means \((r_1, r_2, r_3)\) form a Pythagorean triple scaled by some factor. In this problem, we had \(r_1:r_2 = 12:5\) and \(r_3 = 32.5\). We found \(r_1=30\) and \(r_2=12.5\). Let's check the squares: \(30^2 = 900\), \(12.5^2 = 156.25\), \(32.5^2 = 1056.25\). Indeed, \(900 + 156.25 = 1056.25\). The underlying Pythagorean triple for the radii is \((12, 5, \sqrt{12^2+5^2}) = (12, 5, \sqrt{144+25}) = (12, 5, \sqrt{169}) = (12, 5, 13)\). Our radii are \((30, 12.5)\) and \(r_3=32.5\). Notice that \(30 = 12 \times 2.5\), \(12.5 = 5 \times 2.5\), and \(32.5 = 13 \times 2.5\). So the radii are indeed scaled by a factor of 2.5 from the \((12, 5, 13)\) triple, which corresponds to our value of \(m=2.5\). This confirms our approach and result.