What is the area of the circle (approximately) inscribed in a triangle with side lengths 12 cm, 16 cm and 20 cm ?
50 square cm
Finding the Area of a Circle Inscribed in a Triangle
Let's find the area of the circle inscribed in a triangle with side lengths 12 cm, 16 cm, and 20 cm. To do this, we first need to determine the radius of the inscribed circle. The radius of the inscribed circle, often called the inradius, depends on the area and the perimeter of the triangle.
Identify the Type of Triangle
The given side lengths are 12 cm, 16 cm, and 20 cm. Let's check if this is a right-angled triangle by using the Pythagorean theorem (\(a^2 + b^2 = c^2\)). Let the sides be \(a=12\), \(b=16\), and \(c=20\).
Calculate the square of the two shorter sides and add them:
\[ 12^2 + 16^2 = 144 + 256 = 400 \]Calculate the square of the longest side:
\[ 20^2 = 400 \]Since \(12^2 + 16^2 = 20^2\), the triangle satisfies the Pythagorean theorem. Therefore, it is a right-angled triangle. The sides of lengths 12 cm and 16 cm are the legs, and the side of length 20 cm is the hypotenuse.
Calculate the Area of the Triangle
For a right-angled triangle, the area is half the product of the lengths of the two legs.
\[ \text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} \]Using the legs as base and height:
\[ \text{Area} = \frac{1}{2} \times 12 \, \text{cm} \times 16 \, \text{cm} = 6 \, \text{cm} \times 16 \, \text{cm} = 96 \, \text{square cm} \]The area of the triangle is 96 square cm.
Calculate the Semi-Perimeter of the Triangle
The semi-perimeter (\(s\)) of a triangle is half of its perimeter. The perimeter is the sum of the lengths of all sides.
\[ \text{Perimeter} = 12 \, \text{cm} + 16 \, \text{cm} + 20 \, \text{cm} = 48 \, \text{cm} \] \[ \text{Semi-perimeter} \, (s) = \frac{\text{Perimeter}}{2} = \frac{48 \, \text{cm}}{2} = 24 \, \text{cm} \]The semi-perimeter is 24 cm.
Find the Inradius of the Inscribed Circle
The radius (\(r\)) of the inscribed circle (inradius) in any triangle can be found using the formula:
\[ r = \frac{\text{Area of triangle}}{\text{Semi-perimeter of triangle}} \]Using the values we calculated:
\[ r = \frac{96 \, \text{square cm}}{24 \, \text{cm}} = 4 \, \text{cm} \]Alternatively, for a right-angled triangle with legs \(a\) and \(b\) and hypotenuse \(c\), the inradius is given by a simpler formula:
\[ r = \frac{a + b - c}{2} \]Using this formula with \(a=12\), \(b=16\), and \(c=20\):
\[ r = \frac{12 \, \text{cm} + 16 \, \text{cm} - 20 \, \text{cm}}{2} = \frac{28 \, \text{cm} - 20 \, \text{cm}}{2} = \frac{8 \, \text{cm}}{2} = 4 \, \text{cm} \]Both methods give the same inradius, which is 4 cm.
Calculate the Area of the Inscribed Circle
Now that we have the radius of the inscribed circle (\(r = 4\) cm), we can calculate its area using the formula for the area of a circle:
\[ \text{Area of circle} = \pi r^2 \]Substitute the value of \(r\):
\[ \text{Area} = \pi (4 \, \text{cm})^2 = 16\pi \, \text{square cm} \]To find the approximate area, we use the value of \(\pi \approx 3.14159\).
\[ \text{Area} \approx 16 \times 3.14159 \, \text{square cm} \approx 50.26544 \, \text{square cm} \]Comparing with Options
The calculated approximate area of the inscribed circle is 50.26544 square cm. Let's compare this value with the given options:
- Option 1: 48 square cm
- Option 2: 50 square cm
- Option 3: 52 square cm
- Option 4: 54 square cm
The value 50.26544 square cm is closest to 50 square cm.
Revision Table: Key Concepts for Inscribed Circle Area
| Concept | Formula/Definition | Relevance to Problem |
|---|---|---|
| Right Triangle | Triangle satisfying \(a^2 + b^2 = c^2\) | Given sides 12, 16, 20 form a right triangle. |
| Area of Triangle | \(\frac{1}{2} \times \text{base} \times \text{height}\) (for right triangle) or \(\sqrt{s(s-a)(s-b)(s-c)}\) (Heron's formula) | Needed to calculate inradius (\(r\)). Calculated as 96 sq cm. |
| Semi-perimeter (\(s\)) | \(\frac{a+b+c}{2}\) | Needed to calculate inradius (\(r\)). Calculated as 24 cm. |
| Inradius (\(r\)) | \(\frac{\text{Area}}{s}\) or \(\frac{a+b-c}{2}\) (for right triangle) | Radius of the inscribed circle. Calculated as 4 cm. |
| Area of Circle | \(\pi r^2\) | Final step to find the area of the inscribed circle. Calculated as \(16\pi\) or approx. 50.26 sq cm. |
Additional Information on Inscribed Circles and Inradius
An inscribed circle is the largest possible circle that can be drawn inside a triangle such that it touches all three sides. The center of the inscribed circle is called the incenter. The incenter is the point where the angle bisectors of the triangle intersect. The radius of the inscribed circle is the perpendicular distance from the incenter to any of the triangle's sides.
The formula \(r = \frac{\text{Area}}{s}\) for the inradius is a general formula applicable to all types of triangles (acute, obtuse, or right). The area can be found using various methods, such as \(\frac{1}{2}ab\sin C\) or Heron's formula if side lengths are known. The semi-perimeter \(s\) is always half the sum of the side lengths.
For a right-angled triangle, the formula \(r = \frac{a+b-c}{2}\) provides a quicker way to find the inradius directly from the side lengths (\(a\) and \(b\) are legs, \(c\) is the hypotenuse).
The area of the inscribed circle is a direct application of the standard circle area formula, \(\pi r^2\), once the inradius \(r\) is determined.
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