## EXERCISE 12.3

#### 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,

RP^{2} + PQ^{2} = RQ^{2}

(7)^{2} + (24)^{2} = RQ^{2}

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

=

=

= cm^{2}

#### 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

#### 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 cm^{2}

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 cm^{2}

#### 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 = π*r*^{2}

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

#### 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 cm^{2}

Area of circle = π*r*^{2} = π (1)^{2}

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

#### 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,

AB^{2} = AD^{2} + BD^{2}

Area of equilateral triangle,

Area of circle = π*r*^{2}

Area of design = Area of circle − Area of ΔABC

#### 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 cm^{2}

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

Therefore, the area of shaded portion is 42 cm^{2}.

#### Question 8:

Thegivenfigure 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 m^{2}.

#### 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 (*r*_{1}) of larger circle = 7 cm

Radius (*r*_{2}) 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

#### Question 10:

The area of an equilateral triangle ABC is 17320.5 cm^{2}. 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 cm^{2}

Each sector is of measure 60°.

Area of sector ADEF

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

#### 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 cm^{2}

Area of each circle = π*r*^{2}

Area of 9 circles = 9 × 154 = 1386 cm^{2}

Area of the remaining portion of the handkerchief = 1764 − 1386 = 378 cm^{2}

#### 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

#### 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,

OB^{2} = OA^{2} + AB^{2}

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

Radius (*r*) of circle

Area of quadrant OPBQ

Area of OABC = (Side)^{2} = (20)^{2} = 400 cm^{2}

Area of shaded region = Area of quadrant OPBQ − Area of OABC

= (628 − 400) cm^{2}

= 228 cm^{2}

#### 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.

#### Answer:

Area of the shaded region = Area of sector OAEB − Area of sector OCFD

#### Question 15:

In the given figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

#### Answer:

As ABC is a quadrant of the circle, ∠BAC will be of measure 90º.

In ΔABC,

BC^{2} = AC^{2} + AB^{2}

= (14)^{2} + (14)^{2}

Radius (*r*_{1}) of semi-circle drawn on

Area of

Area of sector

= 154 − (154 − 98)

= 98 cm^{2}

#### Question 16:

Calculate the area of the designed region in the given figure common between the two quadrants of circles of radius 8 cm each.

#### Answer:

The designed area is the common region between two sectors BAEC and DAFC.

Area of sector

Area of ΔBAC

Area of the designed portion = 2 × (Area of segment AEC)

= 2 × (Area of sector BAEC − Area of ΔBAC)