Class 10 — Areas Related to Circles

Chapter Overview & Weightage

Areas Related to Circles is one of the easiest chapters to score in Class 10 boards. CBSE allocates $5$ – $8$ marks across short-answer and long-answer questions. Students who memorise three formulas and practise composite-figure problems regularly score full marks.

Year	Marks	Pattern
2024	7	Sector + segment area
2023	5	Composite figure
2022	6	Length of arc + area of sector
2021	5	Area of segment
2020	6	Combined geometry

The board examiner picks composite figures involving circle + square or circle + triangle. Drill these. Take $\pi = 22/7$ unless told otherwise.

Key Concepts You Must Know

Circumference and area of a circle — $2\pi r$ , $\pi r^2$ .
Length of an arc — angle/360 × circumference.
Area of a sector — angle/360 × area of circle.
Area of a segment — area of sector minus area of triangle.
Composite figures — combinations of squares, rectangles, triangles, semicircles.
Inscribed and circumscribed figures — relationships between sides/radii.

Important Formulas

For a sector with radius $r$ and central angle $\theta$ (in degrees):

$\text{Length of arc} = \frac{\theta}{360}\times 2\pi r$

$\text{Area of sector} = \frac{\theta}{360}\times \pi r^2$

When to use: any “pie slice” problem.

$\text{Area of segment} = \text{Area of sector} - \text{Area of triangle}$

For minor segment with central angle $\theta$ :

$\text{Area} = \frac{\theta}{360}\pi r^2 - \tfrac{1}{2}r^2\sin\theta$

When to use: chord-bounded region problems.

$\pi \approx 22/7$ or $3.14$ .
$\sin 60° = \tfrac{\sqrt{3}}{2}$ , $\cos 60° = 1/2$ .
Area of equilateral triangle with side $a$ : $\tfrac{\sqrt{3}}{4}a^2$ .

Solved Previous Year Questions

PYQ 1 (CBSE 2024, 7 marks)

A chord of a circle of radius $14\,\text{cm}$ subtends a $60°$ angle at the centre. Find the area of the corresponding minor segment. Use $\pi = 22/7$ .

Solution. Sector area $= (60/360) \times (22/7) \times 14^2 = (1/6) \times (22/7) \times 196 = 102.67\,\text{cm}^2$ .

Triangle area $= \tfrac{1}{2}r^2\sin 60° = \tfrac{1}{2}\times 196 \times \tfrac{\sqrt{3}}{2} = 49\sqrt{3} \approx 84.87\,\text{cm}^2$ .

Segment area $= 102.67 - 84.87 \approx 17.80\,\text{cm}^2$ .

PYQ 2 (CBSE 2023, 5 marks)

A square is inscribed in a circle of radius $7\,\text{cm}$ . Find the area of the region between the circle and the square.

Solution. Diagonal of square = diameter of circle $= 14\,\text{cm}$ .

Side of square $= 14/\sqrt{2} = 7\sqrt{2}\,\text{cm}$ .

Area of square $= (7\sqrt{2})^2 = 98\,\text{cm}^2$ .

Area of circle $= (22/7)\times 49 = 154\,\text{cm}^2$ .

Required area $= 154 - 98 = 56\,\text{cm}^2$ .

PYQ 3 (CBSE 2022, 6 marks)

A pendulum swings through an angle of $30°$ and describes an arc of $8.8\,\text{cm}$ . Find the length of the pendulum. Use $\pi = 22/7$ .

Solution. Length of arc $= (\theta/360) \times 2\pi r$ .

$8.8 = \frac{30}{360}\times 2 \times \frac{22}{7}\times r$

$8.8 = \frac{1}{12}\times \frac{44}{7}\times r = \frac{44 r}{84}$

$r = \frac{8.8 \times 84}{44} = 16.8\,\text{cm}$

The pendulum is $16.8\,\text{cm}$ long.

Difficulty Distribution

Easy ( $\sim 40\%$ ): direct sector/arc formula application.
Medium ( $\sim 40\%$ ): segment area calculations.
Hard ( $\sim 20\%$ ): composite figures, inscribed/circumscribed combinations.

Expert Strategy

For composite-figure problems, always identify the regions in your sketch. Subtract overlapping areas, add complementary regions. Sketches save marks.

Three habits that guarantee full marks:

Always state the formula before substituting.
Carry units throughout (cm, cm $^2$ ).
For irrational answers, leave them in surd form unless decimals are specified.

Common Traps

Trap 1: Forgetting to convert angle to radians or to degrees consistently. CBSE uses degrees with the $\theta/360$ factor.

Trap 2: Confusing sector with segment. Sector includes radii (pie slice). Segment is bounded by chord and arc.

Trap 3: Using $\pi r^2$ when $\pi r$ (circumference) is needed. Always check what the question asks.

Trap 4: Forgetting that the chord-and-arc figure has two segments — minor and major. Read the question carefully.

Trap 5: Mixing up “area of triangle” formulas. For sector triangle, $\tfrac{1}{2}r^2\sin\theta$ where $\theta$ is the central angle.

Quick Revision Notes

Sector $\to$ pie slice. Segment $\to$ chord-bounded.
Use $\pi = 22/7$ when radii are multiples of $7$ . Use $\pi = 3.14$ otherwise.
Inscribed square in circle: diagonal $=$ diameter.
Circumscribed square around circle: side $=$ diameter.
Equilateral triangle inscribed in circle of radius $r$ : side $= r\sqrt{3}$ .

A high-confidence chapter for boards. Practice $20$ composite-figure problems and you’ll never lose marks here.