NCERT Solution Class X Mathematics Areas Related to Circles Question 9 (Ex 12.3)

Question 9:

In the given figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region. [Use π = 22/7]

Answer:

Radius (r1) of larger circle = 7 cm

Radius (r2) of smaller circle = 7/2 cm

Area of the shaded region

= Area of smaller circle + Area of semi-circle AECFB − Area of ΔABC

NCERT Solution Class X Mathematics Areas Related to Circles Question 8 (Ex 12.3)

Question 8:

The given figure depicts a racing track whose left and right ends are semicircular.

The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find:

(i) The distance around the track along its inner edge

(ii) The area of the track

 [Use π = 22/7]

Answer:

Distance around the track along its inner edge = AB + arc BEC + CD + arc DFA

Area of the track = (Area of GHIJ − Area of ABCD) + (Area of semi-circle HKI − Area of semi-circle BEC) + (Area of semi-circle GLJ − Area of semi-circle AFD)

Therefore, the area of the track is 4320 m2.

NCERT Solution Class X Mathematics Areas Related to Circles Question 7 (Ex 12.3)

Question 7:

In the given figure, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touches externally two of the remaining three circles. Find the area of the shaded region.  [Use π = 22/7]

Answer:

Area of each of the 4 sectors is equal to each other and is a sector of 90° in a circle of 7 cm radius.

Area of square ABCD = (Side)2 = (14)2 = 196 cm2

Area of shaded portion = Area of square ABCD − 4 × Area of each sector

Therefore, the area of shaded portion is 42 cm2.

NCERT Solution Class X Mathematics Areas Related to Circles Question 5 (Ex 12.3)

Question 5:

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 the given figure. Find the area of the remaining portion of the square.  [Use π = 22/7]

Answer:

 

Each quadrant is a sector of 90° in a circle of 1 cm radius.

Area of square = (Side)2 = (4)2 = 16 cm2

Area of circle = πr2 = π (1)2 = 22/7 cm2

Area of the shaded region = Area of square − Area of circle − 4 × Area of quadrant

NCERT Solution Class X Mathematics Areas Related to Circles Question 3 (Ex 12.3)

Question 3:

Find the area of the shaded region in the given figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.   [Use π = 22/7]

Answer:

It can be observed from the figure that the radius of each semi-circle is 7 cm.

Area of square ABCD = (Side)2 = (14)2 = 196 cm2

Area of the shaded region

= Area of square ABCD − Area of semi-circle APD − Area of semi-circle BPC

= 196 − 77 − 77 = 196 − 154 = 42 cm2

NCERT Solution Class X Mathematics Areas Related to Circles Question 1 (Ex 12.3)

Question 1:

Find the area of the shaded region in the given figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle. [Use π = 22/7]

Answer:

It can be observed that RQ is the diameter of the circle. Therefore, ∠RPQ will be 90º.

By applying Pythagoras theorem in ΔPQR,

RP2 + PQ2 = RQ2

(7)2 + (24)2 = RQ2

RQ = √625 = 25

Radius of circle, OR = RQ/2 = 25/2

Since RQ is the diameter of the circle, it divides the circle in two equal parts.

Area of shaded region = Area of semi-circle RPQOR − Area of ΔPQR

NCERT Solution Class X Mathematics Areas Related to Circles Question 13 (Ex 12.2)

Question 13:

A round table cover has six equal designs as shown in figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs.0.35 per cm2. [Use √3 = 1.7]

Answer:

It can be observed that these designs are segments of the circle.

Consider segment APB. Chord AB is a side of the hexagon. Each chord will substitute 360º/6 = 60º at the centre of the circle.

In ΔOAB,

∠OAB = ∠OBA (As OA = OB)

∠AOB = 60°

∠OAB + ∠OBA + ∠AOB = 180°

2∠OAB = 180° − 60° = 120°

∠OAB = 60°

Therefore, ΔOAB is an equilateral triangle.

Cost of making 1 cm2 designs = Rs 0.35

Cost of making 464.76 cm2 designs = 464.8 × 0.35 = = Rs 162.68

Therefore, the cost of making such designs is Rs 162.68.

NCERT Solution Class X Mathematics Areas Related to Circles Question 10 (Ex 12.2)

Question 10:

An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.   [Use π = 22/7]

Answer:

There are 8 ribs in an umbrella. The area between two consecutive ribs is subtending 360°/8 = 45° at the centre of the assumed flat circle.

Area between two consecutive ribs of circle

NCERT Solution Class X Mathematics Areas Related to Circles Question 9 (Ex 12.2)

Question 9:

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 as shown in figure. Find

(i) The total length of the silver wire required.

(ii) The area of each sector of the brooch     [Use π = 22/7]

Answer:

Total length of wire required will be the length of 5 diameters and the circumference of the brooch.

Radius of circle = 35/2 mm

Circumference of brooch = 2πr

= 110 mm

Length of wire required = 110 + 5 × 35

= 110 + 175 = 285 mm

It can be observed from the figure that each of 10 sectors of the circle is subtending 36° at the centre of the circle.

Therefore, area of each sector

NCERT Solution Class X Mathematics Areas Related to Circles Question 8 (Ex 12.2)

Question 8:

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 (see the given figure). Find

(i) The area of that part of the field in which the horse can graze.

(ii) The increase in the grazing area of the rope were 10 m long instead of 5 m. [Use π = 3.14]

Answer:

From the figure, it can be observed that the horse can graze a sector of 90° in a circle of 5 m radius.

Area that can be grazed by horse = Area of sector OACB

Increase in grazing area = (78.5 − 19.625) m2 = = 58.875 m2

NCERT Solution Class X Mathematics Areas Related to Circles Question 7 (Ex 12.2)

Question 7:

A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. [Use π = 3.14 and √3 = 1.73]

Answer:

Let us draw a perpendicular OV on chord ST. It will bisect the chord ST.

SV = VT

In ΔOVS,

Area of segment SUT = Area of sector OSUT − Area of ΔOST

= 150.72 − 62.28 = 88.44 cm2

NCERT Solution Class X Mathematics Areas Related to Circles Question 6 (Ex 12.2)

Question 6:

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 √3 = 1.73]

Answer:

Radius (r) of circle = 15 cm

Area of sector OPRQ

In ΔOPQ,

∠OPQ = ∠OQP (As OP = OQ)

∠OPQ + ∠OQP + ∠POQ = 180°

2 ∠OPQ = 120°

∠OPQ = 60°

 ΔOPQ is an equilateral triangle.

Area of segment PRQ = Area of sector OPRQ − Area of ΔOPQ

= 117.75 − 97.3125

= 20.4375 cm2

Area of major segment PSQ = Area of circle − Area of segment PRQ

NCERT Solution Class X Mathematics Areas Related to Circles Question 5 (Ex 12.2)

Question 5:

In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:

(i) The length of the arc

(ii) Area of the sector formed by the arc

(iii) Area of the segment forced by the corresponding chord        [Use π = 22/7]

 Answer:

Radius (r) of circle = 21 cm

Angle subtended by the given arc = 60°

Length of an arc of a sector of angle θ

 

In ΔOAB,

∠OAB = ∠OBA (As OA = OB)

∠OAB + ∠AOB + ∠OBA = 180°

2∠OAB + 60° = 180°

∠OAB = 60°

Therefore, ΔOAB is an equilateral triangle.

NCERT Solution Class X Mathematics Areas Related to Circles Question 4 (Ex 12.2)

Question 4:

A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:

(i) Minor segment         (ii) Major sector     [Use π = 3.14]

Answer:

Let AB be the chord of the circle subtending 90° angle at centre O of the circle.

Area of major sector OADB

Area of minor segment ACB = Area of minor sector OACB − Area of ΔOAB = 78.5 − 50 = 28.5 cm2

© Copyright Entrance India - Engineering and Medical Entrance Exams in India | Website Maintained by Firewall Firm - IT Monteur