Consider the following for the next two (02) items that follow: A chord of length l of a circle makes an angle 90° at the centre of the circle.
What is the area of the major segment ?
Calculating the Area of the Major Segment
The question asks for the area of the major segment of a circle, given that a chord of length \(l\) makes an angle of 90° at the center of the circle.
Let the circle have its center at O and radius \(r\). Let the chord be AB, with length \(AB = l\). The angle subtended by the chord at the center is \(\angle AOB = 90^\circ\).
The chord AB and the two radii OA and OB form a triangle OAB. Since OA and OB are radii, \(OA = OB = r\). Given \(\angle AOB = 90^\circ\), triangle OAB is a right-angled triangle.
Relating Chord Length and Radius
In the right-angled triangle OAB, by the Pythagorean theorem:
\(OA^2 + OB^2 = AB^2\)
\(r^2 + r^2 = l^2\)
\(2r^2 = l^2\)
\(r^2 = \frac{l^2}{2}\)
This gives us the relationship between the radius squared and the chord length squared.
Area Calculations
To find the area of the major segment, we first need to find the area of the circle and the area of the minor segment. The area of the major segment is the area of the circle minus the area of the minor segment.
1. Area of the Circle
The area of the circle is given by \(\pi r^2\). Substituting the value of \(r^2\):
Area of circle = \(\pi \left(\frac{l^2}{2}\right) = \frac{\pi l^2}{2}\)
2. Area of the Sector
The sector OAB is formed by the radii OA, OB and the arc AB. The angle of this sector is 90°. The area of a sector with angle \(\theta\) is given by \(\frac{\theta}{360^\circ} \times \pi r^2\).
Area of sector OAB = \(\frac{90^\circ}{360^\circ} \times \pi r^2 = \frac{1}{4} \pi r^2\)
Substitute \(r^2 = \frac{l^2}{2}\):
Area of sector OAB = \(\frac{1}{4} \pi \left(\frac{l^2}{2}\right) = \frac{\pi l^2}{8}\)
3. Area of the Triangle
Triangle OAB is a right-angled triangle with base and height equal to \(r\). The area of triangle OAB is \(\frac{1}{2} \times base \times height\).
Area of triangle OAB = \(\frac{1}{2} \times r \times r = \frac{1}{2} r^2\)
Substitute \(r^2 = \frac{l^2}{2}\):
Area of triangle OAB = \(\frac{1}{2} \left(\frac{l^2}{2}\right) = \frac{l^2}{4}\)
4. Area of the Minor Segment
The minor segment is the region between the chord AB and the arc AB. Its area is the area of the sector OAB minus the area of the triangle OAB.
Area of minor segment = Area of sector OAB - Area of triangle OAB
Area of minor segment = \(\frac{\pi l^2}{8} - \frac{l^2}{4}\)
Area of minor segment = \(l^2 \left(\frac{\pi}{8} - \frac{1}{4}\right) = l^2 \left(\frac{\pi - 2}{8}\right)\)
5. Area of the Major Segment
The major segment is the larger region of the circle bounded by the chord AB and the major arc AB. Its area is the area of the entire circle minus the area of the minor segment.
Area of major segment = Area of circle - Area of minor segment
Area of major segment = \(\frac{\pi l^2}{2} - l^2 \left(\frac{\pi - 2}{8}\right)\)
Combine the terms:
Area of major segment = \(l^2 \left(\frac{\pi}{2} - \frac{\pi - 2}{8}\right)\)
Find a common denominator, which is 8:
Area of major segment = \(l^2 \left(\frac{4\pi}{8} - \frac{\pi - 2}{8}\right)\)
Area of major segment = \(l^2 \left(\frac{4\pi - (\pi - 2)}{8}\right)\)
Area of major segment = \(l^2 \left(\frac{4\pi - \pi + 2}{8}\right)\)
Area of major segment = \(l^2 \left(\frac{3\pi + 2}{8}\right)\)
To match the format of the options, we can factor out \(\frac{1}{4}\) or \(\frac{l^2}{4}\):
Area of major segment = \(\frac{l^2}{4} \times \frac{1}{2} (3\pi + 2)\)
Area of major segment = \(\frac{l^2}{4} \left(\frac{3\pi}{2} + \frac{2}{2}\right)\)
Area of major segment = \(\frac{l^2}{4} \left(\frac{3\pi}{2} + 1\right)\)
This result matches option 1.
| Geometric Shape | Formula | Calculated Area (in terms of \(l\)) |
|---|---|---|
| Circle | \(\pi r^2\) | \(\frac{\pi l^2}{2}\) |
| Sector (90°) | \(\frac{90}{360}\pi r^2\) | \(\frac{\pi l^2}{8}\) |
| Triangle (OAB, 90°) | \(\frac{1}{2}r^2\) | \(\frac{l^2}{4}\) |
| Minor Segment | Area of Sector - Area of Triangle | \(\frac{\pi l^2}{8} - \frac{l^2}{4} = l^2\left(\frac{\pi-2}{8}\right)\) |
| Major Segment | Area of Circle - Area of Minor Segment | \(\frac{\pi l^2}{2} - l^2\left(\frac{\pi-2}{8}\right) = \frac{l^2}{4}\left(\frac{3\pi}{2} + 1\right)\) |
Summary of Steps
- Use the chord length \(l\) and the 90° central angle to find the radius \(r\) squared: \(r^2 = \frac{l^2}{2}\).
- Calculate the area of the triangle formed by the chord and the radii: \(\frac{l^2}{4}\).
- Calculate the area of the sector formed by the 90° angle: \(\frac{\pi l^2}{8}\).
- Calculate the area of the minor segment (Sector Area - Triangle Area).
- Calculate the area of the circle: \(\frac{\pi l^2}{2}\).
- Calculate the area of the major segment (Circle Area - Minor Segment Area).
- Simplify the expression to match the given options.
Revision Table: Circle Segment Geometry
| Term | Definition/Concept |
|---|---|
| Chord | A line segment connecting two points on the circumference of a circle. |
| Sector | A region of a circle enclosed by two radii and an arc. |
| Segment | A region of a circle enclosed by a chord and an arc. Can be minor (smaller) or major (larger). |
| Central Angle | The angle formed by two radii at the center of the circle. |
| 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 (\(a^2 + b^2 = c^2\)). Useful for finding the radius when a chord forms a right angle at the center. |
Additional Information: Area Formulas in Circles
Understanding the basic area formulas for different parts of a circle is key to solving geometry problems like this.
- Area of Circle: \(\pi r^2\), where \(r\) is the radius.
- Area of Sector: \(\frac{\theta}{360^\circ} \times \pi r^2\), where \(\theta\) is the central angle in degrees.
- Area of Triangle within Sector: If the central angle is \(\theta\), the area of the triangle formed by the radii and the chord is \(\frac{1}{2}r^2 \sin(\theta)\). For a 90° angle, \(\sin(90^\circ) = 1\), so the area is \(\frac{1}{2}r^2\).
- Area of Minor Segment: Area of Sector - Area of Triangle.
- Area of Major Segment: Area of Circle - Area of Minor Segment.
In this specific problem, the 90° angle makes the triangle calculation straightforward using \(\frac{1}{2}r^2\) or by recognizing it as a right triangle with legs \(r\). The chord length \(l\) becoming the hypotenuse allows us to express \(r^2\) directly in terms of \(l^2\), simplifying the entire calculation process.
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