Areas Related to Circles Class 10 Mathematics

20 Questions | 76 Marks

A wire is in the shape of a circle of radius 21 cm. It is bent into a square. Find the difference between the area of the square and the area of the circle. (Use

π = \frac{22}{7}

) 5marks

From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in Fig. 11.12. Find the area of the remaining portion of the square. (Use

π = 3.14

) 3marks

A circular park is surrounded by a path 2.1 m wide. If the diameter of the park is 105 m, find the area of the path. (Use

π = \frac{22}{7}

) 3marks

A farmer has a field in the shape of a rhombus with diagonals 30 m and 24 m. He wants to divide the field into two equal parts for planting two types of crops. He uses a rope of length 12 m to fence off one part. Find the area of the part he can fence. 5marks

In Fig. 11.13, two circular action zones are marked with chalk. The larger one has radius 10 m and the smaller one has radius 6 m. Both the centres are same. Find the area of the region between the two action zones. (Use

π = 3.14

) 2marks

A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use

π = 3.14

and

\sqrt{3} = 1.73

) 5marks

The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 15 minutes. 3marks

A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding minor segment. [Use

π = 3.14

] 3marks

A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors. Find the total length of the silver wire required. 3marks

In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find the area of the segment formed by the corresponding chord. [Use

π = \frac{22}{7}

] 3marks

A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope. Find the area of that part of the field in which the horse can graze. (Use

π = 3.14

) 3marks

An umbrella has 8 ribs which are equally spaced. Assuming the umbrella to be a flat circle of radius 45 cm, find the area between two consecutive ribs of the umbrella. 3marks

A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades. 3marks

To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use

π = 3.14

) 3marks

A round table cover has six equal designs as shown in Fig. 11.11. If the radius of the cover is 28 cm, find the area of the area of one design. [Use

π = \frac{22}{7}

and

\sqrt{3} = 1.7

] 5marks

A circle of diameter 20 cm is equally divided into five sectors. Find the area of one of the sectors. 3marks

A sector of a circle has an area of 40

π

sq. units and a central angle of 72°. What is the radius of the circle? 3marks

A square ABCD is inscribed in a circle of radius 10 cm. Find the area of the shaded region. (Use

π = 3.14

) 3marks

The area of a semicircle of diameter 'd' is: (A)

π d^{2} / 16

(B)

π d^{2} / 4

(C)

π d^{2} / 8

(D)

π d^{2} / 2

1mark

The area of a sector of a circle of radius 12 cm is 60

π

cm². The central angle of this sector is: (A) 120° (B) 6° (C) 75° (D) 150° 1mark

Answers

Radius of the circle,

r = 21

cm.Area of the circle =

π r^{2} = \frac{22}{7} \times 21 \times 21 = 22 \times 3 \times 21 = 1386

^{2}

.The wire is bent into a square. So, the perimeter of the square is equal to the circumference of the circle.Circumference of the circle =

2 π r = 2 \times \frac{22}{7} \times 21 = 2 \times 22 \times 3 = 132

cm.Let the side of the square be

a

.Perimeter of the square =

4 a = 132

cm.

a = \frac{132}{4} = 33

cm.Area of the square =

a^{2} = 33^{2} = 1089

^{2}

.Difference between the area of the square and the area of the circle = Area of circle – Area of square =

1386 - 1089 = 297

^{2}

.5marks

Side of the square = 4 cm. Area of the square =

4 \times 4 = 16

^{2}

.Radius of each quadrant = 1 cm.Area of each quadrant =

\frac{1}{4} π r^{2} = \frac{1}{4} \times 3.14 \times 1^{2} = 0.785

^{2}

.There are 4 such quadrants, so total area of 4 quadrants =

4 \times 0.785 = 3.14

^{2}

.Diameter of the circle cut from the centre = 2 cm. Radius of the circle = 1 cm.Area of the circle =

π r^{2} = 3.14 \times 1^{2} = 3.14

^{2}

.Area of the remaining portion = Area of square – (Area of 4 quadrants + Area of the circle)=

16 - (3.14 + 3.14) = 16 - 6.28 = 9.72

^{2}

.3marks

Radius of the circular park,

r = \frac{105}{2}

m = 52.5 m.Width of the path = 2.1 m.Outer radius of the park including the path,

R = 52.5 + 2.1 = 54.6

m.Area of the path = Area of the outer circle – Area of the inner park

= π R^{2} - π r^{2} = π (R^{2} - r^{2}) = π (R - r) (R + r)

= \frac{22}{7} \times (54.6 - 52.5) \times (54.6 + 52.5)

= \frac{22}{7} \times 2.1 \times 107.1

= 22 \times 0.3 \times 107.1 = 6.6 \times 107.1 = 706.86

^{2}

.3marks

Diagonals of the rhombus are

d_{1} = 30

m and

d_{2} = 24

m.Area of the rhombus =

\frac{1}{2} d_{1} d_{2} = \frac{1}{2} \times 30 \times 24 = 15 \times 24 = 360

^{2}

.The field is divided into two equal parts, so the area of each part is

\frac{360}{2} = 180

^{2}

.The farmer uses a rope of length 12 m to fence off one part. This means the area to be fenced is a sector of a circle (assuming the peg is at the center of the division line, or any point within the field which allows maximum area coverage for a given rope length). However, the question implies that the rope is used to define the boundary of the fenced area. If the rope is used as a radius to fence off an area from a point, it would be a sector. If the rope is used to form a boundary, it's about perimeter. Given the context of fencing off a part, and the mention of the area of the part, it's likely the rope is used as a radius from a point within the field to define a circular area. But without a clear anchor point, this interpretation is ambiguous.

Let's re-interpret: If the farmer divides the field into two equal parts, the dividing line is one of the diagonals (or a line joining midpoints of opposite sides). If the rope is used to fence off one part, and it's a rhombus, perhaps the rope is used to mark a boundary within the rhombus. If the rope is of length 12m, and it's used from a corner, it would sweep a quadrant.

Let's assume the question means the rope is used to enclose an area of 180 m $^{2}$ . If the rope is of length 12 m, and it's used as a radius from a corner. The field is a rhombus. Let's assume it's anchored at a vertex. The internal angle of a rhombus with diagonals 30 and 24 is not directly given, but we can find the side length. Side length $s = \sqrt{(\frac{d_{1}}{2})^{2} + (\frac{d_{2}}{2})^{2}} = \sqrt{15^{2} + 12^{2}} = \sqrt{225 + 144} = \sqrt{369} = 3 \sqrt{41}$ m.
The area of the rhombus can also be given by $s^{2} \sin α$ , where $α$ is an internal angle.

This question seems to have a disconnect between the rhombus and the rope fencing. Let's assume the question implies that part of the field is fenced using the rope of 12m. If the rope is used from a vertex, it can sweep a sector. We need an angle.

Let's consider a scenario where the rope is used to mark the boundary of the 180 m $^{2}$ area. If the rope is 12m long and used as a radius from a point. The problem statement is not clear.

Let's assume a simpler interpretation: the rope is used to form a boundary of length 12m. This doesn't make sense for fencing an area.

Let's consider the possibility that the rope is used to enclose a sector. If anchored at a vertex, we need the angle.
Alternatively, if the rope is used to form a circular boundary.

Let's assume the question implies the rope forms a circular arc of radius 12m. If the rope is anchored at one corner of the rhombus, and it's used to fence an area. This implies a sector.
Let's assume the farmer divides the field along one diagonal. The field is divided into two triangles. Area of each triangle is 180 m $^{2}$ . If the rope is used from a vertex of the rhombus.

Given the typical context of such problems, it's likely that the rope is used from one of the vertices of the rhombus. We need an angle.

Let's look at the area of the rhombus again. Area = 360 m $^{2}$ . Side = $3 \sqrt{41}$ .
If the rope is anchored at a vertex, and used as a radius. This suggests a sector area.
Area of a sector = $\frac{θ}{360} π r^{2}$ . If $r = 12$ , and Area=180.
$180 = \frac{θ}{360} π (12^{2}) ⟹ θ = \frac{180 \times 360}{π \times 144} = \frac{5 \times 360}{π \times 4} = \frac{450}{π}$ degrees. This is a very large angle.

Let's consider another interpretation: The question intends to ask for the area of a sector formed by a rope of 12m. However, the context of the rhombus is confusing.

Let's assume the question implies that the rope is used to enclose the maximum possible area within the rhombus field. If the rope is 12m, it can form a circle of radius 12m. Area would be $π (12^{2}) = 144 π \approx 452$ m $^{2}$ . This is larger than the field.

This question is ill-posed or requires a specific interpretation not explicitly stated. Assuming the farmer divides the field into two equal parts and then uses the rope to fence off part of one of these parts.

Let's consider the possibility of a typo in the question or a missing piece of information.
If we ignore the rhombus shape and just consider a farmer fencing off an area of 180 m $^{2}$ using a 12m rope. The question asks "Find the area of the part he can fence." This implies the rope defines the area.

Re-evaluating the problem: "A farmer has a field in the shape of a rhombus... He wants to divide the field into two equal parts... He uses a rope of length 12 m to fence off one part."
This implies the rope is used to demarcate the boundary of the 180 m $^{2}$ part.
Let's assume the farmer is anchoring the rope at the center of the rhombus. The distance from the center to the vertices are half the diagonals: 15m and 12m. The distance to the sides would be the height of the triangle formed by the center and a side.
If the rope is 12m and anchored at the center, it can define a circle of radius 12m.
Area of this circle = $π (12^{2}) = 144 π \approx 452.16$ m $^{2}$ .
The rhombus has area 360 m $^{2}$ . So, this circle is larger than the field.

Consider anchoring the rope at one of the vertices. The distances to other vertices are the side lengths.
Side length $s = \sqrt{15^{2} + 12^{2}} = \sqrt{225 + 144} = \sqrt{369} \approx 19.2$ m.
If anchored at a vertex, and the rope is 12m. It can sweep a sector.
We need the internal angle of the rhombus at that vertex.
Let the angle be $α$ . Area of rhombus = $s^{2} \sin α$ .
$360 = (3 \sqrt{41})^{2} \sin α = 369 \sin α$ .
$\sin α = \frac{360}{369} = \frac{40}{41}$ .
$α = \arcsin (\frac{40}{41})$ . This is an obtuse angle. Let's find the acute angle.
Let the acute angle be $β$ . $\tan (β / 2) = \frac{12}{15} = \frac{4}{5}$ .
The other angle $α = 180 - β$ .
The two angles of the rhombus are $2 \arctan (4 / 5)$ and $180 - 2 \arctan (4 / 5)$ .
Let $θ = 2 \arctan (4 / 5)$ . $\tan (θ / 2) = 4 / 5$ .
$\tan θ = \frac{2 \tan (θ / 2)}{1 - \tan^{2} (θ / 2)} = \frac{2 (4 / 5)}{1 - (4 / 5)^{2}} = \frac{8 / 5}{1 - 16 / 25} = \frac{8 / 5}{9 / 25} = \frac{8}{5} \times \frac{25}{9} = \frac{40}{9}$ .
So, $\tan θ = 40 / 9$ . This is the acute angle at the vertex where the shorter diagonal is opposite. The angle where the longer diagonal is opposite is obtuse.
The question says "fence off one part". If the farmer divides it along the longer diagonal (30m), the two parts are isosceles triangles with sides $3 \sqrt{41}$ , $3 \sqrt{41}$ and 30.
If the farmer divides it along the shorter diagonal (24m), the two parts are isosceles triangles with sides $3 \sqrt{41}$ , $3 \sqrt{41}$ and 24.

This question is highly ambiguous and likely flawed. Without clarification on how the rope is used or anchored, it's impossible to provide a definitive answer. Given the context of "areas related to circles", it might be implied that the rope sweeps a sector. If anchored at a vertex with the acute angle $θ$ , the area of the sector would be $\frac{θ}{360} π (12^{2})$ .

Let's assume a different interpretation based on common problems: The field is divided into two equal parts, each of area 180 m $^{2}$ . The rope of 12m is used to enclose an area from a vertex. The question asks for the area of the part he can fence. If the part to be fenced is part of the 180 m $^{2}$ area, and the rope is 12m.

Let's consider the possibility that the question implies the rope is used to create a boundary around the 180 m $^{2}$ area. This doesn't fit.

Given the mark allocation of 5 marks, it usually involves a multi-step calculation or a concept like sector/segment area.

Let's assume the rope is used from the center of the rhombus and sweeps an angle.
The area of the rhombus is 360 m $^{2}$ .
If the farmer divides it into two equal parts, area of each part is 180 m $^{2}$ .
If the rope is 12m and is used from the center. The distance to the midpoints of sides is the height of the triangle formed by center and a side.
Let's assume the farmer fences off an area using the rope as a radius from a point within the 180m $^{2}$ part.

This question cannot be solved as stated without significant assumptions about the fencing method and anchor point. I will provide a placeholder answer based on a common interpretation of "rope tied to a peg" problems, which usually imply a sector. However, the rhombus context is problematic.

Let's assume the farmer divides the field into two equal triangular parts using the shorter diagonal (24m). Each triangle has base 24m and height 15m. Area = 180 m $^{2}$ .
If the rope is used from a vertex of this triangle. Which vertex?
If the rope is anchored at the vertex where the two sides of length $3 \sqrt{41}$ meet, and the diagonal is 24. The angles at this vertex would be $2 \arctan (12 / 15) = 2 \arctan (4 / 5)$ .
Let $α = 2 \arctan (4 / 5)$ . $\tan (α / 2) = 4 / 5$ .
$\tan α = \frac{2 (4 / 5)}{1 - (4 / 5)^{2}} = \frac{8 / 5}{9 / 25} = 40 / 9$ .
We need the angle in degrees to calculate the sector area.
$α = \arctan (40 / 9) \approx {77.32}^{\circ}$ .
Area of sector = $\frac{77.32}{360} π (12^{2}) = \frac{77.32}{360} \times 3.14 \times 144 \approx 97.1$ m $^{2}$ . This is less than 180 m $^{2}$ .

If the rope is anchored at a vertex of the longer diagonal (30m). The angle would be $180 - α$ .
$180 - 77.32 = {102.68}^{\circ}$ .
Area of sector = $\frac{102.68}{360} π (12^{2}) \approx 128.7$ m $^{2}$ . This is also less than 180 m $^{2}$ .

The problem is likely intended to be simpler or more direct. Given the difficulty in interpretation, I cannot provide a confident answer without assumptions. I will skip this question as it is ill-defined for a standard exam.

Radius of the larger circular action zone,

R = 10

m.Radius of the smaller circular action zone,

r = 6

m.Both circles have the same centre.Area of the region between the two zones = Area of the larger circle – Area of the smaller circle

= π R^{2} - π r^{2} = π (R^{2} - r^{2})

= 3.14 \times (10^{2} - 6^{2})

= 3.14 \times (100 - 36)

= 3.14 \times 64

= 200.96

^{2}

.2marks

Radius of the circle,

r = 15

cm.Angle subtended by the chord at the centre,

θ = 60^{\circ}

.Area of the sector OAB =

\frac{θ}{360} π r^{2} = \frac{60}{360} \times 3.14 \times 15^{2}

= \frac{1}{6} \times 3.14 \times 225 = 3.14 \times 37.5 = 117.75

^{2}

.Since

θ = 60^{\circ}

and OA = OB (radii), triangle OAB is an equilateral triangle.Side of the equilateral triangle = 15 cm.Area of triangle OAB =

\frac{\sqrt{3}}{4} \times s i d e^{2} = \frac{1.73}{4} \times 15^{2}

= \frac{1.73}{4} \times 225 = 1.73 \times 56.25 = 97.3125

^{2}

.Area of the minor segment = Area of sector OAB – Area of triangle OAB

= 117.75 - 97.3125 = 20.4375

^{2}

.Area of the major sector = Area of circle – Area of minor sectorArea of circle =

π r^{2} = 3.14 \times 15^{2} = 3.14 \times 225 = 706.5

^{2}

.Area of major sector =

706.5 - 117.75 = 588.75

^{2}

.Alternatively, angle of major sector =

360^{\circ} - 60^{\circ} = 300^{\circ}

.Area of major sector =

\frac{300}{360} π r^{2} = \frac{5}{6} \times 3.14 \times 15^{2} = \frac{5}{6} \times 706.5 = 5 \times 117.75 = 588.75

^{2}

.5marks

Length of the minute hand = radius,

r = 14

cm.In 60 minutes, the minute hand completes one full revolution, i.e.,

360^{\circ}

.In 1 minute, the angle swept by the minute hand =

\frac{360^{\circ}}{60} = 6^{\circ}

.In 15 minutes, the angle swept by the minute hand =

15 \times 6^{\circ} = 90^{\circ}

.Area swept by the minute hand in 15 minutes = Area of the sector with angle

90^{\circ}

.Area =

\frac{θ}{360} π r^{2} = \frac{90}{360} \times \frac{22}{7} \times 14^{2}

= \frac{1}{4} \times \frac{22}{7} \times 196

= \frac{1}{4} \times 22 \times 28 = 22 \times 7 = 154

^{2}

.3marks

Radius of the circle,

r = 10

cm.The chord subtends a right angle (

90^{\circ}

) at the centre. So,

θ = 90^{\circ}

.Area of the sector formed by the chord =

\frac{θ}{360} π r^{2} = \frac{90}{360} \times 3.14 \times 10^{2}

= \frac{1}{4} \times 3.14 \times 100 = 3.14 \times 25 = 78.5

^{2}

.Since the angle is

90^{\circ}

and the sides are radii (equal), the triangle formed is a right-angled isosceles triangle.Area of the triangle =

\frac{1}{2} \times b a s e \times h e i g h t = \frac{1}{2} \times r \times r = \frac{1}{2} \times 10 \times 10 = 50

^{2}

.Area of the corresponding minor segment = Area of sector – Area of triangle

= 78.5 - 50 = 28.5

^{2}

.3marks

Diameter of the circle = 35 mm. Radius,

r = \frac{35}{2} = 17.5

mm.The brooch is made of a circle and 5 diameters.Length of the wire for the circle = Circumference =

2 π r = 2 \times \frac{22}{7} \times 17.5

= 2 \times 22 \times 2.5 = 44 \times 2.5 = 110

mm.Length of the wire for 5 diameters =

5 \times diameter = 5 \times 35 = 175

mm.Total length of the silver wire required = Length for circle + Length for diameters

= 110 + 175 = 285

mm.3marks

Radius of the circle,

r = 21

cm.Angle subtended by the arc at the centre,

θ = 60^{\circ}

.Area of the sector formed by the arc =

\frac{θ}{360} π r^{2} = \frac{60}{360} \times \frac{22}{7} \times 21^{2}

= \frac{1}{6} \times \frac{22}{7} \times 441 = \frac{1}{6} \times 22 \times 63 = 11 \times 21 = 231

^{2}

.Since

θ = 60^{\circ}

and the triangle formed by the radii and the chord is isosceles, it is an equilateral triangle with side 21 cm.Area of the triangle =

\frac{\sqrt{3}}{4} \times s i d e^{2} = \frac{\sqrt{3}}{4} \times 21^{2} = \frac{\sqrt{3}}{4} \times 441

.Using

\sqrt{3} \approx 1.732

: Area

\approx \frac{1.732}{4} \times 441 = 0.433 \times 441 \approx 190.953

^{2}

.Area of the segment = Area of sector – Area of triangle

= 231 - 190.953 = 40.047

^{2}

.3marks

The horse is tied to a peg at one corner of a square field of side 15 m by means of a 5 m long rope. This means the horse can graze in a sector of a circle with radius equal to the length of the rope.Radius of the sector,

r = 5

m.Since the peg is at a corner of a square, the angle of the sector is

90^{\circ}

.Area of the field in which the horse can graze = Area of the sector.Area =

\frac{θ}{360} π r^{2} = \frac{90}{360} \times 3.14 \times 5^{2}

= \frac{1}{4} \times 3.14 \times 25 = \frac{78.5}{4} = 19.625

^{2}

.3marks

The umbrella has 8 ribs, equally spaced. Assuming it to be a flat circle of radius 45 cm.The angle between two consecutive ribs =

\frac{360^{\circ}}{8} = 45^{\circ}

.The area between two consecutive ribs is the area of a sector with angle

45^{\circ}

and radius 45 cm.Area =

\frac{θ}{360} π r^{2} = \frac{45}{360} \times \frac{22}{7} \times 45^{2}

= \frac{1}{8} \times \frac{22}{7} \times 2025 = \frac{11}{4 \times 7} \times 2025 = \frac{22275}{28}

= 795.5357... \approx 795.54

^{2}

.3marks

Each wiper has a blade of length 25 cm (radius,

r = 25

cm) and sweeps through an angle of 115°.Area swept by one wiper = Area of a sector with angle

115^{\circ}

and radius 25 cm.Area =

\frac{θ}{360} π r^{2} = \frac{115}{360} \times \frac{22}{7} \times 25^{2}

= \frac{23}{72} \times \frac{22}{7} \times 625 = \frac{23 \times 11 \times 625}{36 \times 7} = \frac{158125}{252}

\approx 627.48

^{2}

.Since there are two wipers and they do not overlap, the total area cleaned is the sum of the areas swept by each wiper.Total area =

2 \times 627.48 = 1254.96

^{2}

.3marks

The lighthouse spreads red coloured light over a sector of angle 80° to a distance of 16.5 km.Radius of the sector,

r = 16.5

km.Angle of the sector,

θ = 80^{\circ}

.Area of the sea over which the ships are warned = Area of the sector.Area =

\frac{θ}{360} π r^{2} = \frac{80}{360} \times 3.14 \times (16.5)^{2}

= \frac{2}{9} \times 3.14 \times 272.25

= \frac{2 \times 3.14 \times 272.25}{9} = \frac{1709.13}{9} = 189.9033... \approx 189.90

^{2}

.3marks

Radius of the round table cover,

r = 28

cm.There are six equal designs, which means the circle is divided into 6 equal sectors by the designs.The angle of each sector =

\frac{360^{\circ}}{6} = 60^{\circ}

.The design area is the area of the segment formed by the chord connecting the endpoints of the arc of the sector.Area of one sector =

\frac{60}{360} π r^{2} = \frac{1}{6} \times \frac{22}{7} \times 28^{2}

= \frac{1}{6} \times \frac{22}{7} \times 784 = \frac{1}{6} \times 22 \times 112 = \frac{2464}{6} = \frac{1232}{3}

^{2}

.The triangle formed by the radii and the chord is an equilateral triangle because the angle at the centre is

60^{\circ}

and the other two sides are radii.Side of the equilateral triangle = radius = 28 cm.Area of the equilateral triangle =

\frac{\sqrt{3}}{4} \times s i d e^{2} = \frac{1.7}{4} \times 28^{2}

= \frac{1.7}{4} \times 784 = 1.7 \times 196 = 333.2

^{2}

.Area of one design (segment) = Area of sector – Area of triangle

= \frac{1232}{3} - 333.2 = 410.666... - 333.2 = 77.466... \approx 77.47

^{2}

.5marks

Diameter of the circle = 20 cm. Radius,

r = \frac{20}{2} = 10

cm.The circle is equally divided into five sectors.Angle of each sector =

\frac{360^{\circ}}{5} = 72^{\circ}

.Area of one sector =

\frac{θ}{360} π r^{2} = \frac{72}{360} \times \frac{22}{7} \times 10^{2}

= \frac{1}{5} \times \frac{22}{7} \times 100 = \frac{2200}{35} = \frac{440}{7} \approx 62.86

^{2}

.3marks

Area of a sector = 40

π

sq. units.Central angle = 72°.Let the radius of the circle be

r

.Area of sector =

\frac{θ}{360} π r^{2}

40 π = \frac{72}{360} π r^{2}

40 π = \frac{1}{5} π r^{2}

Divide both sides by $π$ :

40 = \frac{1}{5} r^{2}

r^{2} = 40 \times 5 = 200

r = \sqrt{200} = \sqrt{100 \times 2} = 10 \sqrt{2}

units.3marks

Radius of the circle,

r = 10

cm.A square ABCD is inscribed in the circle. The diagonal of the square is equal to the diameter of the circle.Diagonal of the square =

2 r = 2 \times 10 = 20

cm.Let the side of the square be

a

.Using Pythagoras theorem,

a^{2} + a^{2} = (20)^{2}

2 a^{2} = 400

a^{2} = 200

^{2}

.Area of the square =

a^{2} = 200

^{2}

.Area of the circle =

π r^{2} = 3.14 \times 10^{2} = 3.14 \times 100 = 314

^{2}

.The shaded region is the area of the circle minus the area of the square.Area of shaded region = Area of circle – Area of square

= 314 - 200 = 114

^{2}

.3marks

Diameter of the semicircle,

d

.Radius of the semicircle,

r = d / 2

.Area of a circle =

π r^{2} = π (d / 2)^{2} = \frac{π d^{2}}{4}

.Area of a semicircle is half the area of the circle.Area of semicircle =

\frac{1}{2} \times \frac{π d^{2}}{4} = \frac{π d^{2}}{8}

.The correct option is (C).1mark

Radius of the circle,

r = 12

cm.Area of the sector =

60 π

^{2}

.Let the central angle be

θ

.Area of sector =

\frac{θ}{360} π r^{2}

60 π = \frac{θ}{360} π (12)^{2}

60 π = \frac{θ}{360} π (144)

Divide both sides by $π$ :

60 = \frac{θ \times 144}{360}

60 = \frac{θ \times 144}{360} = θ \times \frac{144}{360} = θ \times \frac{2}{5}

$θ = 60 \times \frac{5}{2} = 30 \times 5 = 150^{\circ}$ .
The correct option is (D).
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