Question

In the given figure, chord AB subtends an angle of 120\(^{\circ}\) at the centre of the circle with radius 7 cm. Find (i) perimeter of major sector OACB, and (ii) area of the shaded segment, if area of \(\Delta\) OAB = 21.2 cm\(^2\).

Hint:

Remember that "perimeter of a sector" is not just the arc length.
It includes the two straight boundaries (radii) as well: \( \text{Perimeter} = L + 2r \).
For segment areas, subtracting the given triangle area directly from the sector area is extremely straightforward, so keep track of units and rounding decimals.

Answer 1

Step 1: Find the length of the minor arc first.
The circle has radius $r = 7$ cm, and chord $AB$ subtends an angle of $120^{\circ}$ at the centre $O$, marking off the minor arc $AB$ and the major arc $ACB$.
The length of the minor arc, matching the $120^{\circ}$ angle, is:
\[ \text{Minor arc length} = \frac{120^{\circ}}{360^{\circ}} \times 2\pi r = \frac{1}{3} \times 2 \times \frac{22}{7} \times 7 = \frac{1}{3} \times 44 = \frac{44}{3}\text{ cm} \]

Step 2: Get the major arc length by subtracting from the full circumference, instead of computing it directly.
The full circumference of the circle is:
\[ 2\pi r = 2 \times \frac{22}{7} \times 7 = 44\text{ cm} \]
Since the minor arc and the major arc together make up the whole circle:
\[ \text{Major arc length} = 44 - \frac{44}{3} = \frac{132 - 44}{3} = \frac{88}{3}\text{ cm} \approx 29.33\text{ cm} \]
So the perimeter of the major sector $OACB$ is this major arc length plus the two bounding radii $OA$ and $OB$:
\[ \text{Perimeter} = \frac{88}{3} + 2(7) = \frac{88}{3} + 14 = \frac{88 + 42}{3} = \frac{130}{3} \approx 43.33\text{ cm} \]

Step 3: Find the area of the minor sector by subtracting the major sector area from the full circle, instead of computing it directly.
The area of the full circle is:
\[ \pi r^2 = \frac{22}{7} \times 7 \times 7 = 154\text{ cm}^2 \]
The major sector corresponds to the angle $360^{\circ} - 120^{\circ} = 240^{\circ}$, so its area is:
\[ \text{Major sector area} = \frac{240^{\circ}}{360^{\circ}} \times 154 = \frac{2}{3} \times 154 = \frac{308}{3} \approx 102.67\text{ cm}^2 \]
Since the major sector and the minor sector together make up the whole circle:
\[ \text{Minor sector area} = 154 - \frac{308}{3} = \frac{462 - 308}{3} = \frac{154}{3} \approx 51.33\text{ cm}^2 \]

Step 4: Subtract the triangle area to get the shaded segment.
We are given that the area of triangle $OAB$ is $21.2$ cm$^2$. The shaded minor segment is the region inside the minor sector but outside triangle $OAB$, so:
\[ \text{Area of segment} = \text{Minor sector area} - \text{Area of } \triangle OAB = 51.33 - 21.2 = 30.13\text{ cm}^2 \]

Final Answer:
\[ \boxed{\text{Perimeter of major sector} \approx 43.33\text{ cm}, \quad \text{Area of shaded segment} \approx 30.13\text{ cm}^2} \]