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

EXERCISE 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.

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

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

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°.

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.

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

= 196 − 77 − 77 = 196 − 154 = 42 cm2

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.

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

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.

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

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).

Radius (r) of circle = 32 cm

AD is the median of ABC.

In ΔABD,

Area of equilateral triangle,

Area of circle = πr2

Area of design = Area of circle − Area of ΔABC

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.

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.

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

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.

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.

Radius (r1) of larger circle = 7 cm

Area of smaller circle

Area of semi-circle AECFB of larger circle

Area of

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

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]

Let the side of the equilateral triangle be a.

Area of equilateral triangle = 17320.5 cm2

Each sector is of measure 60°.

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

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.

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

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

(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]

In ΔOAB,

OB2 = OA2 + AB2

= (20)2 + (20)2

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

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.

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.

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

In ΔABC,

BC2 = AC2 + AB2

= (14)2 + (14)2

Radius (r1) of semi-circle drawn on

Area of

Area of sector

= 154 − (154 − 98)

= 98 cm2

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.

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)

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