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 minor segment ?
Understanding the Problem: Finding the Area of a Minor Segment
The question asks for the area of the minor segment of a circle. We are given the length of a chord, denoted by \(l\), and the fact that this chord creates a 90° angle at the center of the circle. To find the area of the minor segment, we need to calculate the area of the sector formed by the central angle and subtract the area of the triangle formed by the chord and the two radii.
Given Information
- Length of the chord = \(l\)
- Central angle subtended by the chord, \(\theta = 90^\circ\)
Relationship Between Chord Length and Radius
When a chord subtends a 90° angle at the center of a circle, the triangle formed by the two radii connecting the endpoints of the chord to the center and the chord itself is a right-angled triangle. Let the radius of the circle be \(r\). The two radii forming the 90° angle are the legs of the right triangle, and the chord is the hypotenuse.
Using the Pythagorean theorem in this right-angled triangle:
\(r^2 + r^2 = l^2\)
\(2r^2 = l^2\)
\(r^2 = \frac{l^2}{2}\)
This gives us the relationship between the radius squared (\(r^2\)) and the chord length squared (\(l^2\)). We will use \(r^2\) in the area calculations.
Calculating the Area of the Sector
The area of a sector with central angle \(\theta\) (in degrees) and radius \(r\) is given by the formula:
\(A_{sector} = \frac{\theta}{360^\circ} \times \pi r^2\)
In this case, \(\theta = 90^\circ\), so the area of the sector is:
\(A_{sector} = \frac{90^\circ}{360^\circ} \times \pi r^2 = \frac{1}{4} \pi r^2\)
Now, substitute the value of \(r^2\) in terms of \(l\):
\(A_{sector} = \frac{1}{4} \pi \left(\frac{l^2}{2}\right) = \frac{\pi l^2}{8}\)
Calculating the Area of the Triangle
The triangle formed by the chord and the two radii is a right-angled triangle with legs of length \(r\). The area of a right triangle is given by \(\frac{1}{2} \times base \times height\).
In this case, base = \(r\) and height = \(r\), so the area of the triangle is:
\(A_{triangle} = \frac{1}{2} \times r \times r = \frac{1}{2} r^2\)
Now, substitute the value of \(r^2\) in terms of \(l\):
\(A_{triangle} = \frac{1}{2} \left(\frac{l^2}{2}\right) = \frac{l^2}{4}\)
Calculating the Area of the Minor Segment
The area of the minor segment is the difference between the area of the sector and the area of the triangle:
\(A_{segment} = A_{sector} - A_{triangle}\)
\(A_{segment} = \frac{\pi l^2}{8} - \frac{l^2}{4}\)
To match the format of the options, we can factor out a common term. Let's factor out \(\frac{l^2}{4}\):
\(A_{segment} = \frac{l^2}{4} \left(\frac{\pi l^2 / 8}{l^2 / 4} - \frac{l^2 / 4}{l^2 / 4}\right)\)
\(A_{segment} = \frac{l^2}{4} \left(\frac{\pi l^2}{8} \times \frac{4}{l^2} - 1\right)\)
\(A_{segment} = \frac{l^2}{4} \left(\frac{4\pi}{8} - 1\right)\)
\(A_{segment} = \frac{l^2}{4} \left(\frac{\pi}{2} - 1\right)\)
This result matches one of the given options.
Checking the Options
Let's compare our derived area of the minor segment, \(\frac{l^2}{4} \left(\frac{\pi}{2} - 1\right)\), with the given options:
- Option 1: \(\frac{l^2}{2}\left(\pi-\frac{1}{2}\right)\)
- Option 2: \(\frac{l^2}{4}\left(\pi-\frac{1}{2}\right)\)
- Option 3: \(\frac{l^2}{4}\left(\frac{\pi}{2}-1\right)\)
- Option 4: \(\frac{l^2}{2}\left(\frac{\pi}{2}-\frac{1}{2}\right)\)
Our calculated area matches Option 3.
| Concept | Formula Used |
|---|---|
| Pythagorean Theorem | \(a^2 + b^2 = c^2\) |
| Area of Sector | \(\frac{\theta}{360^\circ} \times \pi r^2\) |
| Area of Right Triangle | \(\frac{1}{2} \times base \times height\) |
| Area of Minor Segment | Area of Sector - Area of Triangle |
Revision Table: Key Concepts in Circle Geometry
| Term | Definition | Related Formula (if any) |
|---|---|---|
| Chord | A straight line segment connecting two points on the circumference of a circle. | |
| Radius (r) | A line segment from the center of the circle to any point on its circumference. | |
| Central Angle (\(\theta\)) | The angle formed by two radii at the center of the circle. | |
| Sector | A region of a circle bounded by two radii and the intercepted arc. | \(A = \frac{\theta}{360^\circ} \times \pi r^2\) (for angle in degrees) |
| Segment | A region of a circle bounded by a chord and the arc subtended by the chord. Minor segment is the smaller area, major segment is the larger area. | \(A_{segment} = A_{sector} - A_{triangle}\) (for minor segment) |
| Area of Triangle | The space enclosed by three straight sides. | \(A = \frac{1}{2} \times base \times height\) or \(A = \frac{1}{2}ab\sin(C)\) |
Additional Information: Special Cases and Relationships
Understanding the relationship between the chord length, radius, and central angle is crucial in circle geometry problems involving segments and sectors.
- For a central angle of 90°, the triangle formed is a right-angled isosceles triangle. The chord length \(l\) is related to the radius \(r\) by \(l = r\sqrt{2}\), or \(r = \frac{l}{\sqrt{2}}\). Squaring the radius gives \(r^2 = \frac{l^2}{2}\), which was used in our calculations.
- For a central angle of 180°, the chord is a diameter (\(l=2r\)). The segment is a semicircle. The triangle area is 0 (a degenerate triangle). The sector area is \(\frac{180}{360}\pi r^2 = \frac{1}{2}\pi r^2\). Area of segment = \(\frac{1}{2}\pi r^2\).
- For a central angle of 60°, the triangle formed is an equilateral triangle with side length equal to the radius, so \(l=r\). The area of the triangle is \(\frac{\sqrt{3}}{4}r^2 = \frac{\sqrt{3}}{4}l^2\). The area of the sector is \(\frac{60}{360}\pi r^2 = \frac{1}{6}\pi r^2\). Area of segment = \(\frac{1}{6}\pi r^2 - \frac{\sqrt{3}}{4}r^2\).
These special cases highlight how the geometry of the inscribed triangle changes with the central angle, affecting the segment area calculation.
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