NCERT Solutions for Class 10 Maths Chapter 11: Areas Related to Circles — Free PDF

Area of a circle $= \pi r^2$

Circumference $= 2\pi r$

Use $\pi = \dfrac{22}{7}$ when the radius is a multiple of $7$; use $\pi = 3.14$ otherwise (unless specified in the problem).

A sector is the region enclosed by two radii and an arc. For a sector with central angle $\theta$:

Remember: the perimeter of a sector includes the two straight radii plus the curved arc.

A segment is the region between a chord and its arc.

For a segment with central angle $\theta$:

For common angles:
- $\theta = 60^\circ$: triangle area $= \dfrac{\sqrt{3}}{4} r^2$ (equilateral triangle)
- $\theta = 90^\circ$: triangle area $= \dfrac{1}{2} r^2$
- $\theta = 120^\circ$: triangle area $= \dfrac{\sqrt{3}}{4} r^2$ (use $\dfrac{1}{2}r^2 \sin 120^\circ$)

For shaded-region problems:
1. Identify all the individual shapes (circles, semicircles, squares, triangles, sectors).
2. Determine whether the shaded region requires adding or subtracting areas.
3. Calculate each area separately.
4. Combine the results.

Always draw a clear diagram and label the dimensions before computing.

The radii of two circles are $19$ cm and $9$ cm. Find the radius of a circle whose circumference equals the sum of their circumferences.

Solution:

Answer: Radius $= 28$ cm.

The radii of two circles are $8$ cm and $6$ cm. Find the radius of a circle whose area equals the sum of their areas.

Solution:

Answer: Radius $= 10$ cm.

Find the perimeter of a semicircular region of radius $7$ cm.

Solution:
Perimeter of semicircle $= \pi r + 2r = r(\pi + 2)$

Answer: Perimeter $= 36$ cm.

Note: the perimeter includes the curved part ($\pi r$) and the straight diameter ($2r$).

Find the area of a sector with radius $6$ cm and angle $60^\circ$.

Solution:

Answer: Area $= \dfrac{132}{7}$ cm$^2 \approx 18.86$ cm$^2$.

Find the area of a sector with radius $4$ cm and arc length $5.5$ cm.

Solution:
Using the shortcut formula:

Answer: Area $= 11$ cm$^2$.

This shortcut works because Area $= \dfrac{1}{2} \times l \times r$ where $l$ is the arc length. It is analogous to the triangle area formula $\dfrac{1}{2} \times \text{base} \times \text{height}$.

Find the area of the minor segment if radius $= 21$ cm and angle $= 120^\circ$.

Solution:
Area of sector:

Area of triangle (with angle $120^\circ$ between two radii of $21$ cm):

Area of segment $= 462 - 190.96 = 271.04$ cm$^2$.

Answer: Area of segment $\approx 271$ cm$^2$.

A sector has angle $90^\circ$ and radius $14$ cm. Find the arc length and perimeter of the sector.

Solution:
Arc length $= \dfrac{90}{360} \times 2\pi r = \dfrac{1}{4} \times 2 \times \dfrac{22}{7} \times 14 = 22$ cm.

Perimeter of sector $= 2r + \text{arc length} = 2(14) + 22 = 50$ cm.

Answer: Arc length $= 22$ cm, Perimeter $= 50$ cm.

Find the area of the shaded region where ABCD is a square of side $14$ cm with four equal semicircles drawn on each side.

Solution:
Area of square $= 14^2 = 196$ cm$^2$.

Each semicircle has diameter $= 14$ cm, so radius $= 7$ cm.
Area of $4$ semicircles $= 4 \times \dfrac{1}{2} \pi r^2 = 2 \times \dfrac{22}{7} \times 49 = 308$ cm$^2$.

Shaded area $= 308 - 196 = 112$ cm$^2$.

Answer: Shaded area $= 112$ cm$^2$.

A round table cover has $6$ equal designs made by drawing arcs from each vertex at radius $28$ cm, each subtending $60^\circ$. Find the cost of making the designs at Rs $0.35$ per cm$^2$.

Solution:
Each of the $6$ segments has a $60^\circ$ arc with $r = 28$ cm.

Area of each segment $=$ Area of sector $-$ Area of equilateral triangle:

Total design area $= 6 \times 71.19 = 427.14$ cm$^2$.

Cost $= 427.14 \times 0.35 \approx$ Rs $149.50$.

Answer: Cost $\approx$ Rs $149.50$.

A circle is inscribed in a square of side $14$ cm. Find the area of the region between the square and the circle (the shaded corners).

Solution:
Side of square $= 14$ cm, so radius of inscribed circle $= 7$ cm.

Area of square $= 196$ cm$^2$.
Area of circle $= \dfrac{22}{7} \times 7^2 = 154$ cm$^2$.

Shaded area $= 196 - 154 = 42$ cm$^2$.

Answer: Shaded area $= 42$ cm$^2$.

Two equal circles of radius $7$ cm are placed inside a rectangle such that each circle touches two sides and the other circle. Find the area of the shaded region (inside the rectangle but outside both circles).

Solution:
The rectangle has length $= 2 \times 14 = 28$ cm and width $= 14$ cm.

Area of rectangle $= 28 \times 14 = 392$ cm$^2$.
Area of $2$ circles $= 2 \times \dfrac{22}{7} \times 49 = 308$ cm$^2$.

Shaded area $= 392 - 308 = 84$ cm$^2$.

Answer: Shaded area $= 84$ cm$^2$.

The length of the minute hand of a clock is $14$ cm. Find the area swept by it in $30$ minutes.

Solution:
In $30$ minutes, the minute hand sweeps $180^\circ$.

Answer: Area swept $= 308$ cm$^2$.

Find the area of the minor segment of a circle of radius $14$ cm where the central angle is $90^\circ$.

Solution:
Area of sector $= \dfrac{90}{360} \times \dfrac{22}{7} \times 14^2 = \dfrac{1}{4} \times 616 = 154$ cm$^2$.

Area of right triangle $= \dfrac{1}{2} \times 14 \times 14 = 98$ cm$^2$.

Area of segment $= 154 - 98 = 56$ cm$^2$.

Answer: Area of minor segment $= 56$ cm$^2$.

A flower bed is in the shape of a ring between two concentric circles with radii $21$ m and $14$ m. Find the area of the flower bed.

Solution:

Answer: Area of flower bed $= 770$ m$^2$.

A figure consists of a rectangle $20$ cm $\times$ $14$ cm with semicircles drawn on each of the shorter sides. Find the perimeter and area of the figure.

Solution:
Radius of each semicircle $= 7$ cm.

Perimeter $= 2 \times 20 + 2 \times (\pi \times 7) = 40 + 2 \times \dfrac{22}{7} \times 7 = 40 + 44 = 84$ cm.

Area $=$ Area of rectangle $+ 2 \times$ Area of semicircle
$= 20 \times 14 + 2 \times \dfrac{1}{2} \times \dfrac{22}{7} \times 49 = 280 + 154 = 434$ cm$^2$.

Answer: Perimeter $= 84$ cm, Area $= 434$ cm$^2$.

Always identify the central angle $\theta$ of the sector before applying formulas. If the problem mentions a semicircle, $\theta = 180^\circ$. If it mentions a quadrant, $\theta = 90^\circ$. Read carefully to find the angle.

A sector includes the triangle formed by the two radii; a segment does not. Segment area $=$ Sector area $-$ Triangle area. Many students forget to subtract the triangle and lose marks.

Use $\pi = \dfrac{22}{7}$ when the radius is a multiple of $7$ (like $7$, $14$, $21$, $28$) as it simplifies calculations. Use $\pi = 3.14$ for other values. Always use the value specified in the problem if one is given.

The perimeter of a sector is $2r + \text{arc length}$, not just the arc length. Students often forget the two straight edges (the radii) when computing the perimeter.

In combined figure problems, carefully determine whether you need to add or subtract areas. Draw the figure, shade the required region, and decide which standard shapes make it up. A common error is subtracting when you should add, or vice versa.

Find the area of a sector with central angle $72^\circ$ and radius $10$ cm. (Use $\pi = 3.14$)

Answer: Area $= \dfrac{72}{360} \times 3.14 \times 100 = \dfrac{1}{5} \times 314 = 62.8$ cm$^2$.

Find the length of the arc of a sector with angle $150^\circ$ and radius $21$ cm.

Answer: Arc length $= \dfrac{150}{360} \times 2 \times \dfrac{22}{7} \times 21 = \dfrac{5}{12} \times 132 = 55$ cm.

Find the area of the minor segment of a circle with radius $7$ cm and central angle $60^\circ$.

Answer: Sector area $= \dfrac{60}{360} \times \dfrac{22}{7} \times 49 = \dfrac{154}{6} = 25.67$ cm$^2$. Triangle area $= \dfrac{\sqrt{3}}{4} \times 49 \approx 21.22$ cm$^2$. Segment area $= 25.67 - 21.22 = 4.45$ cm$^2$.

A square of side $28$ cm has a circle inscribed in it. Find the area of the region outside the circle but inside the square.

Answer: Radius $= 14$ cm. Area of square $= 784$ cm$^2$. Area of circle $= \dfrac{22}{7} \times 196 = 616$ cm$^2$. Shaded area $= 784 - 616 = 168$ cm$^2$.

A circular flower bed of radius $14$ m has a path of width $3.5$ m around it. Find the area of the path.

Answer: Outer radius $= 17.5$ m. Area of path $= \dfrac{22}{7}(17.5^2 - 14^2) = \dfrac{22}{7}(306.25 - 196) = \dfrac{22}{7} \times 110.25 = 346.5$ m$^2$.