Class 10 Maths Chapter 11 Areas Related to Circles Important Questions with Answers in English

Q. Define a sector of a circle.

Answer: A sector is the portion of the circular region enclosed by two radii and the corresponding arc.

Q. Define a segment of a circle.

Answer: A segment is the portion of the circular region enclosed between a chord and the corresponding arc.

Q. The angle of a minor sector is $75°$. What is the angle of the corresponding major sector?

Answer: $360°-75°=285°$.

Q. Write the formula for the length of an arc of a sector of angle $\theta$ of a circle of radius $r$.

Answer: Length of arc $=\frac{\theta}{360}\times2\pi r$.

Q. Find the area of a quadrant of a circle of radius 14 cm. ($\pi=\frac{22}{7}$)

Answer: Quadrant $=\frac14\times\frac{22}{7}\times14\times14=\frac14\times616=154$ cm².

Q. Through what angle does the minute hand of a clock turn in 35 minutes?

Answer: $6°$ per minute, so $35\times6°=210°$.

Q. What is the area of a sector of angle $360°$ of a circle of radius $r$?

Answer: $\frac{360}{360}\times\pi r^2=\pi r^2$, the area of the whole circle.

Q. Chord AB of a circle with centre O and radius 8 cm subtends $90°$ at O. Find the area of $\triangle OAB$.

Answer: $\triangle OAB$ is right-angled at O: $\frac12\times8\times8=32$ cm².

Q. Find the length of the arc and the perimeter of a sector of a circle of radius 28 cm and angle $45°$. ($\pi=\frac{22}{7}$)

Answer: Arc $=\frac{45}{360}\times2\times\frac{22}{7}\times28=\frac18\times176=22$ cm.
Perimeter $=2r+\text{arc}=56+22=78$ cm.

Q. Find the area swept by the minute hand of a clock, 7 cm long, in 20 minutes. ($\pi=\frac{22}{7}$)

Answer: Angle $=20\times6°=120°$.
Area $=\frac{120}{360}\times\frac{22}{7}\times7\times7=\frac13\times154=\frac{154}{3}\approx51.33$ cm².

Q. The area of a sector of a circle of radius 10.5 cm is 57.75 cm². Find the angle of the sector. ($\pi=\frac{22}{7}$)

Answer: $\frac{\theta}{360}\times\frac{22}{7}\times10.5\times10.5=57.75$
$\frac{\theta}{360}\times346.5=57.75\Rightarrow\frac{\theta}{360}=\frac{57.75}{346.5}=\frac16\Rightarrow\theta=60°$.

Q. An arc of length 44 cm subtends an angle of $120°$ at the centre of a circle. Find the radius. ($\pi=\frac{22}{7}$)

Answer: $\frac{120}{360}\times2\times\frac{22}{7}\times r=44\Rightarrow\frac{44r}{21}=44\Rightarrow r=21$ cm.

Q. Find the area of a sector of a circle of radius 5 cm and angle $72°$. ($\pi=3.14$)

Answer: Area $=\frac{72}{360}\times3.14\times5\times5=\frac15\times78.5=15.7$ cm².

Q. The minor sector of a circle of radius 14 cm has angle $60°$. Find the area of the major sector by two methods. ($\pi=\frac{22}{7}$)

Answer: Method 1: angle of major sector $=360°-60°=300°$, so area $=\frac{300}{360}\times\frac{22}{7}\times14\times14=\frac56\times616=\frac{1540}{3}\approx513.33$ cm².
Method 2: minor sector $=\frac16\times616=\frac{308}{3}$ cm², so major sector $=616-\frac{308}{3}=\frac{1540}{3}\approx513.33$ cm².

Q. Find the ratio of the areas of two sectors of the same circle whose angles are $45°$ and $120°$.

Answer: Same radius, so the areas are in the ratio of the angles: $45:120=3:8$.

Q. Show that the area of a sector of radius $r$ whose arc has length $l$ is $\frac12lr$.

Answer: Area $=\frac{\theta}{360}\times\pi r^2=\frac12\times r\times\left(\frac{\theta}{360}\times2\pi r\right)=\frac12\times r\times l=\frac12lr$.

Q. A chord of a circle of radius 14 cm subtends a right angle at the centre. Find the areas of the minor segment and the major segment. ($\pi=\frac{22}{7}$)

Answer: Area of circle $=\frac{22}{7}\times14\times14=616$ cm².
Sector ($90°$) $=\frac14\times616=154$ cm².
$\triangle OAB$ (right-angled at O) $=\frac12\times14\times14=98$ cm².
Minor segment $=154-98=56$ cm².
Major segment $=616-56=560$ cm².

Q. A chord of a circle of radius 10 cm subtends an angle of $60°$ at the centre. Find the area of the minor segment. ($\pi=3.14$, $\sqrt3=1.73$)

Answer: Sector $=\frac{60}{360}\times3.14\times100=\frac{314}{6}\approx52.33$ cm².
$OA=OB$ and $\angle O=60°$, so $\triangle OAB$ is equilateral with side 10 cm: area $=\frac{\sqrt3}{4}\times100=\frac{1.73\times100}{4}=43.25$ cm².
Minor segment $=52.33-43.25=9.08$ cm² (approx.).

Q. A chord AB of a circle with centre O and radius 14 cm subtends $120°$ at O. Find the area of the minor segment. ($\pi=\frac{22}{7}$, $\sqrt3=1.73$)

Answer: Sector $=\frac{120}{360}\times\frac{22}{7}\times14\times14=\frac13\times616=\frac{616}{3}\approx205.33$ cm².
Draw $OM\perp AB$. By RHS congruence M is the mid-point of AB and $\angle AOM=60°$.
$OM=14\cos60°=7$ cm; $AM=14\sin60°=7\sqrt3$ cm; $AB=14\sqrt3$ cm.
$\triangle OAB=\frac12\times14\sqrt3\times7=49\sqrt3=49\times1.73=84.77$ cm².
Minor segment $=205.33-84.77=120.56$ cm² (approx.).

Q. A circular pizza of radius 14 cm is cut into 8 equal slices from the centre. Find (i) the area of the top of each slice, (ii) the length of the curved crust on each slice. ($\pi=\frac{22}{7}$)

Answer: Each slice is a sector of angle $360°\div8=45°$.
(i) Area $=\frac18\times\frac{22}{7}\times14\times14=\frac18\times616=77$ cm².
(ii) Crust (arc) $=\frac18\times2\times\frac{22}{7}\times14=\frac18\times88=11$ cm.

Q. ABCD is a square of side 21 cm. With A as centre, a quadrant of radius 21 cm is drawn inside the square, passing through B and D. Find the area of the part of the square outside the quadrant. ($\pi=\frac{22}{7}$)

Answer: Area of square $=21\times21=441$ cm².
Quadrant $=\frac14\times\frac{22}{7}\times21\times21=\frac14\times1386=346.5$ cm².
Required area $=441-346.5=94.5$ cm².

Q. A car has two wipers which do not overlap. Each blade is 28 cm long and sweeps through $120°$. Find the total area cleaned at each sweep of the blades. ($\pi=\frac{22}{7}$)

Answer: Area cleaned by one blade $=\frac{120}{360}\times\frac{22}{7}\times28\times28=\frac13\times2464=\frac{2464}{3}$ cm².
Two blades: $2\times\frac{2464}{3}=\frac{4928}{3}\approx1642.67$ cm².

Q. A sector of angle $60°$ is cut from a circle of radius 42 cm. Find the area and the perimeter of the sector. ($\pi=\frac{22}{7}$)

Answer: Area $=\frac{60}{360}\times\frac{22}{7}\times42\times42=\frac16\times5544=924$ cm².
Arc $=\frac16\times2\times\frac{22}{7}\times42=\frac16\times264=44$ cm.
Perimeter $=2\times42+44=128$ cm.

Q. In a circle of radius 12 cm, a chord AB subtends $60°$ at the centre. Find (i) the length of chord AB, (ii) the length of arc AB, (iii) the perimeter of the minor segment. ($\pi=3.14$)

Answer: (i) $OA=OB=12$ cm and $\angle O=60°$, so $\triangle OAB$ is equilateral: $AB=12$ cm.
(ii) Arc $=\frac{60}{360}\times2\times3.14\times12=\frac{75.36}{6}=12.56$ cm.
(iii) Perimeter of segment $=$ arc $+$ chord $=12.56+12=24.56$ cm.

Q. A chord AB of a circle of radius 28 cm subtends an angle of $60°$ at the centre O. Find (i) the area of the minor sector, (ii) the area of $\triangle OAB$, (iii) the area of the minor segment, (iv) the area of the major segment. ($\pi=\frac{22}{7}$, $\sqrt3=1.73$)

Answer: Area of circle $=\frac{22}{7}\times28\times28=2464$ cm².
(i) Minor sector $=\frac16\times2464=\frac{1232}{3}\approx410.67$ cm².
(ii) $\triangle OAB$ is equilateral with side 28 cm: $\frac{\sqrt3}{4}\times784=196\sqrt3=196\times1.73=339.08$ cm².
(iii) Minor segment $=410.67-339.08=71.59$ cm² (approx.).
(iv) Major segment $=2464-71.59=2392.41$ cm² (approx.).

Q. Using the Unitary Method, derive the formula for the area of a sector of angle $\theta$ of a circle of radius $r$, and the formula for the length of its arc. Hence find the area and the arc length of a sector of radius 35 cm and angle $72°$. ($\pi=\frac{22}{7}$)

Answer: Area: a sector of $360°$ is the whole circle, area $\pi r^2$. So $1°$ gives $\frac{\pi r^2}{360}$, and $\theta$ gives $\frac{\theta}{360}\times\pi r^2$.
Arc: the whole circumference $2\pi r$ belongs to $360°$, so the arc of $\theta$ is $\frac{\theta}{360}\times2\pi r$.
Numerical: $\frac{72}{360}=\frac15$.
Area $=\frac15\times\frac{22}{7}\times35\times35=\frac15\times3850=770$ cm².
Arc $=\frac15\times2\times\frac{22}{7}\times35=\frac15\times220=44$ cm.

Q. A round table cover of radius 14 cm has six equal designs, each a segment cut off by a side of a regular hexagon drawn inside the circle. Find the cost of making the designs at ₹0.50 per cm². ($\pi=\frac{22}{7}$, $\sqrt3=1.73$)

Answer: Each side of the hexagon subtends $360°\div6=60°$ at the centre, so each design is a $60°$ segment and each triangle OAB is equilateral with side 14 cm.
Area of circle $=\frac{22}{7}\times14\times14=616$ cm².
Area of one triangle $=\frac{\sqrt3}{4}\times196=49\sqrt3=49\times1.73=84.77$ cm²; hexagon $=6\times84.77=508.62$ cm².
Area of the six designs $=616-508.62=107.38$ cm².
Cost $=107.38\times0.50=$ ₹53.69.

Q. A cow is tied with a 14 m rope to a peg at one corner of a rectangular field 20 m long and 16 m wide. Find (i) the area of the field the cow can graze, (ii) the area it cannot graze, (iii) the grazing area lost if the rope were 7 m shorter. ($\pi=\frac{22}{7}$)

Answer: The corner angle is $90°$ and the rope (14 m) is shorter than both sides, so the grazed region is a quadrant.
(i) $\frac14\times\frac{22}{7}\times14\times14=154$ m².
(ii) Field $=20\times16=320$ m²; not grazed $=320-154=166$ m².
(iii) With a 7 m rope: $\frac14\times\frac{22}{7}\times7\times7=38.5$ m². Area lost $=154-38.5=115.5$ m².