NCERT Solutions class 10 maths chapter 12

Area Related to Circles

NCERT Solutions class 10 maths chapter 12 Exercise 12.3

Exercise 12.1 Exercise 12.2

Question 1

Find the area of the shaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

Answer

we know that any triangle drawn from diameter RQ to the circle is 90°

∠P = 90°

Using Pythagorean theorem,

QR²= PR²+PQ²

QR²= 7²+24²

QR= 25 cm

Hence, the radius of the circle = 25/2 cm

Also, area of the ΔPQR = 1/2 × PR × PQ

=1/2 × 7 × 24 cm²

= 84 cm²

Now, the area of the semicircle = (πR²)/2

= ((22/7)×(25/2)×(25/2))/2 cm²

= 13750/56 cm²= 245.54 cm²

the area of the shaded region = 245.54 cm²-84 cm²

= 161.54 cm²

Question 2

Find the area of the shaded region in Fig. 12.20, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and AOC = 40°

Answer

Given,

Angle made by sector = 40°,

the inner Radius = r = 7 cm, and

the outer Radius = R = 14 cm

We know,

Area of the sector = (θ/360°)×πr²

So, Area of OAC = (40°/360°)×πr²cm²

= 68.44 cm²

Area of the sector OBD = (40°/360°)×πr²cm²

= (1/9)×(22/7)×7²= 17.11 cm²

Now, area of the shaded region ABDC = Area of OAC – Area of the OBD

= 68.44 cm² – 17.11 cm²= 51.33 cm²

Question 3

Find the area of the shaded region in Fig. 12.21, if ABCD is a square of side 14 cm and APD and BPC are semicircles

Answer

Side of the square ABCD = 14 cm

So, Area of square= a²

= 14×14 cm² = 196 cm²

here the side of the square = diameter of the circle = 14 cm

∴ Radius of the semicircle = 7 cm

area of the semicircle = (πR²)/2

= (22/7×7×7)/2 cm²=

= 77 cm²

∴ Area of two semicircles = 2×77 cm²= 154 cm²

area of the shaded region = Area of the Square – Area of two semicircles

= 196 cm²-154 cm²

= 42 cm²

Question 4

Find the area of the shaded region in Fig. 12.22, 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

It is given that OAB is an equilateral triangle having each angle as 60°

Area of the sector is common in both.

Radius of the circle = 6 cm.

Side of the triangle = 12 cm.

Area of the equilateral triangle = (√3/4) (OA)²= (√3/4×12²= 36√3 cm²

Area of the sector making angle 60° = (60°/360°) ×πr²cm²

= (1/6)×(22/7)× 6²cm²= 132/7 cm²

Area of the circle = πR² = (22/7)×6²= 792/7 cm²

tota Area of the shaded region = Area of the equilateral triangle + Area of the circle – Area of the sector

= 36√3 cm² +792/7 cm²-132/7 cm²

= (36√3+660/7) cm²

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 Fig. 12.23. Find the area of the remaining portion of the square.

Solution:

Side of the square = 4 cm

Radius of the quadrant = 1 cm

Area of the quadrant = (60°/360°) x πR² cm² = (22/7)×(1²)/4 = 11/14 cm²

∴So Total area of the 4 quadrants = 4 ×(11/14) cm² = 22/7 cm²

Area of square = (side)²= 4²= 16 cm²

Area of the circle = πR²cm² = (22/7×1²) = 22/7 cm²

Area of the shaded region = Area of square – (Area of the all quadrants + Area of the circle)

= 16 cm²-(22/7) cm²+(22/7) cm²

= 68/7 cm²

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 Fig. 12.24. Find the area of the design.

Answer

Radius of the circle = 32 cm

Draw a median AD of the triangle

∴ AO = (2/3) AD

⇒ (2/3)AD = 32 cm

⇒ AD = 48 cm

In ΔADB,

By Pythagoras theorem,

AB²= AD²+BD²

⇒ AB²= 48²+(AB/2)²

⇒ AB²= 2304+AB²/4

⇒ 3/4 (AB²)= 2304

⇒ AB²= 3072

⇒ AB= 32√3 cm

Area of ΔADB = √3/4 ×(32√3)²cm²= 768√3 cm²

Area of circle = πR² = (22/7)×32×32 = 22528/7 cm²

Area of the design = Area of circle – Area of ΔADB

= (22528/7 – 768√3) cm²

Question 7

In Fig. 12.25, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.

Answer

Side of square = 14 cm

∴ Radius of the circles = 14/2 cm = 7 cm

Area of the square ABCD = 14²= 196 cm²

Area of the quadrant of one circle = (πR²)/4 cm² = (22/7) ×7²/4 cm²

= 77/2 cm²

Total area of the quadrant = 4×77/2 cm²= 154cm²

Area of the shaded region = Area of the square ABCD – Area of the quadrant

= 196 cm²– 154 cm²

= 42 cm²

Question 8

Fig. 12.26 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

Width of the track = 10 m

Length of tracks = 106 m

DE = CF = 60 m

Radius of inner semicircle, r = OD = O’C

= 60/2 m = 30 m

Radius of outer semicircle, R = OA = O’B

= 30+10 m = 40 m

Also, AB = CD = EF = GH = 106 m

Distance around the track along its inner edge = CD+EF+2×(Circumference of inner semicircle)

= 106+106+(2×πr) m = 212+(2×22/7×30) m

= 212+1320/7 m = 2804/7 m

Area of the track = Area of ABCD + Area EFGH + 2 × (area of outer semicircle) – 2 × (area of inner semicircle)

= (AB×CD)+(EF×GH)+2×(πr²/2) -2×(πR²/2) m²

= (106×10)+(106×10)+2×π/2(r²-R²) m²

= 2120+22/7×70×10 m²

= 4320 m²

Question 9

In Fig. 12.27, 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 of larger circle, R = 7 cm

Radius of smaller circle, r = 7/2 cm

Height of ΔBCA = OC = 7 cm

AB = 14 cm

Area of ΔBCA = 1/2 × AB × OC = (1/2) × 7 × 14 = 49 cm²

Area of larger circle = πR²= (22/7)×7² = 154 cm²

Area of larger semi-circle = 154/2 cm²= 77 cm²

Area of smaller circle = πr² = (22/7)×(7/2)×(7/2) = 77/2 cm²

area of segment BC and AC
area of semi-circle - Area of ΔABC =77 - 49 =28 cm²

Area of the shaded region = Area of larger semi-circle + area of segment BC and AC

Area of the shaded region = (77/2)+ 28 cm²

= 133/2 cm² = 66.5 cm²

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 Fig. 12.28). Find the area of the shaded region. (Use π = 3.14 and √3 = 1.73205)

Answer

Area of ΔABC = 17320.5 cm²

⇒ √3/4 ×(side)² = 17320.5

⇒ (side)² =17320.5×4/1.73205

⇒ (side)² = 4×10⁴

⇒ side = 200 cm

ABC is an equilateral triangle.

∴ So, ∠ A = ∠ B = ∠ C = 60°
AB=a=200cm

Radius of the circles = 200/2 cm = 100 cm

Area of the sector of 1st = (60°/360°)×π r²cm²

= 1/6×3.14×(100)²cm²

= 15700/3cm²

Area of 3 sectors = 3×15700/3 = 15700 cm²

Thus, area of the shaded region = Area of equilateral triangle ABC – Area of 3 sectors

= 17320.5-15700 cm²= 1620.5 cm²

Question 11

On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig. 12.29). Find the area of the remaining portion of the handkerchief.

Answer

Radius of the each circular = 7 cm

There are three circles in one side of square handkerchief.

∴ Length of the square = 3 × diameter of circle = 3 × 14 = 42 cm

Area of the square = 42×42 cm² = 1764 cm²

Area of the circle = π r²= (22/7)×7×7 = 154 cm²

Total area of the design = 9 × 154 = 1386 cm²

Area of the remaining portion of the handkerchief = Area of the square – Total area of the design = 1764 – 1386 = 378 cm²

Question 12

In Fig. 12.30, OACB is a quadrant of a 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

here radius of the quadrant = 3.5 cm = 7/2 cm

(i) Area of quadrant OACB = (πR²)/4 cm²

= (22/7)×(7/2)×(7/2)/4 cm²

= 77/8 cm²

(ii) Area of triangle BOD = (½)×(7/2)×2 cm²

= 7/2 cm²

Area of shaded region = Area of quadrant – Area of triangle BOD

= (77/8)-(7/2) cm²= 49/8 cm²

= 6.125 cm²

Question 13

In Fig. 12.31, 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

here length of square =20 cm

Radius of the quadrant = OB

OAB is right angled triangle

By Pythagoras theorem in ΔOAB,

OB²= AB²+OA²

⇒ OB²= 20²+20²

⇒ OB²= 400+400

⇒ OB²= 800

⇒ OB= 20√2 cm

Area of the quadrant = (πR²)/4 cm²= (3.14/4)×(20√2)²cm²= 628cm²

Area of the square = 20×20 = 400 cm²

Area of the shaded region = Area of the quadrant – Area of the square

= 628-400 cm²= 228cm²

Question 14

AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see Fig. 12.32). If ∠AOB = 30°, find the area of the shaded region.