NCERT solution class 10 chapter 12 Areas Related to Circles exercise 12.3 mathematics ~ evidya

EXERCISE 12.3

Q 1, Ex 12.3 - Area Related to Circles - Chapter 12 - Mathematics Class 10th - NCERT

Page No 234:

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.

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

Radius of circle,

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

Area of ΔPQR

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

cm²

Q 2, Ex 12.3 - Area Related to Circles - Chapter 12 - Mathematics Class 10th - NCERT

Page No 235:

Question 2:

Find the area of the shaded region in the given figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Answer:

Radius of inner circle = 7 cm
Radius of outer circle = 14 cm
Area of shaded region = Area of sector OAFC − Area of sector OBED
=40°360°×π(14)2-40°360°×π(7)2=19×227×14×14-19×227×7×7=6169-1549=4629=1543cm2

Q 3, Ex 12.3 - Area Related to Circles - Chapter 12 - Mathematics Class 10th - NCERT

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.

Answer:

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

Area of each semi-circle =

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

Q 4, Ex 12.3 - Area Related to Circles - Chapter 12 - Mathematics Class 10th - NCERT

Question 4:

Find the area of the shaded region in the given figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.

Answer:

We know that each interior angle of an equilateral triangle is of measure 60°.

Area of sector OCDE

Area of

Area of circle = πr²

Area of shaded region = Area of ΔOAB + Area of circle − Area of sector OCDE

Q 5, Ex 12.3 - Area Related to Circles - Chapter 12 - Maths Class 10th - NCERT

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.

Answer:

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

Area of square = (Side)² = (4)² = 16 cm²
Area of circle = πr² = π (1)²

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

Q 6, Ex 12.3 - Area Related to Circles - Chapter 12 - Maths Class 10th - NCERT

Question 6:

In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the given figure. Find the area of the design (Shaded region).

Answer:

Radius (r) of circle = 32 cm
AD is the median of

ABC.

AD = 48 cm
In ΔABD,
AB² = AD² + BD²

Area of equilateral triangle,

Area of circle = πr²

Area of design = Area of circle − Area of ΔABC

Q 7, Ex 12.3 - Area Related to Circles - Chapter 12 - Maths Class 10th - NCERT

Page No 236:

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.

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 each sector

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.

Q 8, Ex 12.3 - Area Related to Circles - Chapter 12 - Maths Class 10th - NCERT

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

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.

Q 9, Ex 12.3 - Area Related to Circles - Chapter 12 - Maths Class 10th - NCERT

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.

Answer:

Radius (r₁) of larger circle = 7 cm
Radius (r₂) of smaller circle

Area of smaller circle

Area of semi-circle AECFB of larger circle

Area of

Area of the shaded region
= Area of smaller circle + Area of semi-circle AECFB − Area of ΔABC

Q 10, Ex 12.3 - Area Related to Circles - Chapter 12 - Maths Class 10th - NCERT

Question 10:

The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (See the given figure). Find the area of shaded region. [Use π = 3.14 and

]

Answer:

Let the side of the equilateral triangle be a.
Area of equilateral triangle = 17320.5 cm²

Each sector is of measure 60°.
Area of sector ADEF

Area of shaded region = Area of equilateral triangle − 3 × Area of each sector

Q 11, Ex 12.3 - Area Related to Circles - Chapter 12 - Maths Class 10th - NCERT

Page No 237:

Question 11:

On a square handkerchief, nine circular designs each of radius 7 cm are made (see the given figure). Find the area of the remaining portion of the handkerchief.

Answer:

From the figure, it can be observed that the side of the square is 42 cm.
Area of square = (Side)² = (42)² = 1764 cm²
Area of each circle = πr²

Area of 9 circles = 9 × 154 = 1386 cm²
Area of the remaining portion of the handkerchief = 1764 − 1386 = 378 cm²

Q 12, Ex 12.3 - Area Related to Circles - Chapter 12 - Maths Class 10th - NCERT

Question 12:

In the given figure, OACB is a quadrant of circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the
(i) Quadrant OACB
(ii) Shaded region

Answer:

(i) Since OACB is a quadrant, it will subtend 90° angle at O.
Area of quadrant OACB

(ii) Area of ΔOBD

Area of the shaded region = Area of quadrant OACB − Area of ΔOBD

Q 13, Ex 12.3 - Area Related to Circles - Chapter 12 - Maths Class 10th - NCERT

Question 13:

In the given figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. [Use π = 3.14]

Answer:

In ΔOAB,
OB² = OA² + AB²
= (20)² + (20)²

Radius (r) of circle

Area of quadrant OPBQ

Area of OABC = (Side)² = (20)² = 400 cm²
Area of shaded region = Area of quadrant OPBQ − Area of OABC
= (628 − 400) cm²
= 228 cm²

Q 14, Ex 12.3 - Area Related to Circles - Chapter 12 - Maths Class 10th - NCERT

Question 14:

AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see the given figure). If ∠AOB = 30°, find the area of the shaded region.