Consider the following for the next three (03) items that follow: A triangle ABC with sides AB = 15 cm, BC = 9 cm, CA = 12 cm is inscribed in a circle.
What is the radius of the circle ?
Finding the Radius of the Circumscribed Circle for a Triangle
We are given a triangle ABC with side lengths AB = 15 cm, BC = 9 cm, and CA = 12 cm. This triangle is inscribed in a circle, and we need to find the radius of this circle, which is also known as the circumradius.
The first step is to determine the type of triangle ABC. We can check if it is a right-angled triangle by using the Pythagorean theorem. Let's compare the square of the longest side with the sum of the squares of the other two sides.
- Longest side: AB = 15 cm
- Other sides: BC = 9 cm, CA = 12 cm
Calculate the squares of the side lengths:
- \(\text{AB}^2 = 15^2 = 225\)
- \(\text{BC}^2 = 9^2 = 81\)
- \(\text{CA}^2 = 12^2 = 144\)
Now, let's check if the sum of the squares of the two shorter sides equals the square of the longest side:
\(\text{BC}^2 + \text{CA}^2 = 81 + 144 = 225\)
We see that \(\text{BC}^2 + \text{CA}^2 = \text{AB}^2\) (\(225 = 225\)). This confirms that triangle ABC is a right-angled triangle, with the right angle at vertex C (opposite the hypotenuse AB).
Circumradius of a Right-Angled Triangle
A key property of a right-angled triangle inscribed in a circle is that its hypotenuse is the diameter of the circle. The vertices of the triangle lie on the circle, and the hypotenuse is the longest chord, passing through the center of the circle.
In triangle ABC, the hypotenuse is AB, which has a length of 15 cm.
So, the diameter of the circumscribed circle is equal to the length of the hypotenuse:
Diameter (D) = AB = 15 cm
The radius (R) of a circle is half of its diameter.
\(R = \frac{D}{2}\)
Substituting the value of the diameter:
\(R = \frac{15 \text{ cm}}{2}\)
\(R = 7.5 \text{ cm}\)
Therefore, the radius of the circle in which the triangle ABC is inscribed is 7.5 cm.
Let's quickly verify this with the options provided:
- 4.5 cm
- 6 cm
- 7.5 cm
- 15 cm
Our calculated radius is 7.5 cm, which matches one of the options.
Revision Table: Triangle and Circle Properties
| Concept | Description | Relevance to Problem |
|---|---|---|
| Pythagorean Theorem | In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides (\(a^2 + b^2 = c^2\)). | Used to identify if triangle ABC is a right-angled triangle. |
| Right-Angled Triangle Inscribed in a Circle | If a right-angled triangle is inscribed in a circle, the hypotenuse of the triangle is the diameter of the circle. The circumcenter is the midpoint of the hypotenuse. | Crucial property used to find the diameter (and thus the radius) of the circle. |
| Circumradius (R) | The radius of the circumscribed circle (the circle passing through all vertices of a triangle). | The quantity we needed to calculate. For a right triangle, \(R = \frac{\text{Hypotenuse}}{2}\). For any triangle, \(R = \frac{abc}{4K}\), where a, b, c are side lengths and K is the area. |
Additional Information: Circumradius Calculation Methods
While for a right-angled triangle, finding the circumradius is straightforward using the hypotenuse, there are general formulas for calculating the circumradius (R) of any triangle ABC with sides a, b, c:
- Using the side lengths and area: \(R = \frac{abc}{4K}\) Where a, b, c are the lengths of the sides opposite vertices A, B, C respectively, and K is the area of the triangle. The area K can be calculated using Heron's formula if side lengths are known: \(K = \sqrt{s(s-a)(s-b)(s-c)}\), where \(s = \frac{a+b+c}{2}\) is the semi-perimeter.
- Using a side length and the sine of the opposite angle: By the Law of Sines, \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R\) So, \(R = \frac{a}{2 \sin A} = \frac{b}{2 \sin B} = \frac{c}{2 \sin C}\) This method requires knowing at least one angle and its opposite side.
In our specific problem, since we identified it as a right-angled triangle, the simplest method is to use the property that the hypotenuse is the diameter, avoiding the need for area calculation or trigonometry (though those methods would also yield the same result).
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