CLASS 10 MATH NCERT SOLUTIONS FOR CHAPTER – 12 AREAS RELATED TO CIRCLES EX – 12.3

Areas Related to Circles

Page No: 234

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.

Ncert solution class 10 chapter 12-12

Solution:

Here, P is in the semi-circle and so,

P = 90°

So, it can be concluded that QR is hypotenuse of the circle and is equal to the diameter of the circle.

∴ QR = D

Using Pythagorean theorem,

QR= PR2+PQ2

Or, QR= 72+242

QR= 25 cm = Diameter

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

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

= (22/7)×(25/2)×(25/2)/2 cm2

= 13750/56 cm= 245.54 cm2

Also, area of the ΔPQR = ½×PR×PQ

=(½)×7×24 cm2

= 84 cm2

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

= 161.54 cm2

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°.

Solution:

Ncert solution class 10 chapter 12-13

Given,

Angle made by sector = 40°,

Radius the inner circle = r = 7 cm, and

Radius of the outer circle = R = 14 cm

We know,

Area of the sector = (θ/360°)×πr2

So, Area of OAC = (40°/360°)×πrcm2

= 68.44 cm2

Area of the sector OBD = (40°/360°)×πrcm2

= (1/9)×(22/7)×7= 17.11 cm2

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

= 68.44 cm2 – 17.11 cm= 51.33 cm2

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.

Ncert solution class 10 chapter 12-14

Solution:

Side of the square ABCD (as given) = 14 cm

So, Area of ABCD = a2

= 14×14 cm2 = 196 cm2

We know that the side of the square = diameter of the circle = 14 cm

So, side of the square = diameter of the semicircle = 14 cm

∴ Radius of the semicircle = 7 cm

Now, area of the semicircle = (πR2)/2

= (22/7×7×7)/2 cm=

= 77 cm2

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

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

= 196 cm-154 cm2

= 42 cm2

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.

Ncert solution class 10 chapter 12-15

Solution:

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= (√3/40×12= 36√3 cm2

Area of the circle = πR2 = (22/7)×6= 792/7 cm2

Area of the sector making angle 60° = (60°/360°) ×πrcm2

= (1/6)×(22/7)× 6cm= 132/7 cm2

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

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 circle = 1 cm

Four quadrant of a circle are cut from corner and one circle of radius are cut from middle.

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

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

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

Area of the circle = πRcm2 = (22/7×12) = 22/7 cm2

Area of the shaded region = Area of square – (Area of the 4 quadrants + Area of the circle)

= 16 cm2-(22/7) cm2+(22/7) cm2

= 68/7 cm2

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.

Ncert solution class 10 chapter 12-16

Solution:

Radius of the circle = 32 cm

Draw a median AD of the triangle passing through the centre of the circle.

⇒ BD = AB/2

Since, AD is the median of the triangle

∴ AO = Radius of the circle = (2/3) AD

⇒ (2/3)AD = 32 cm

⇒ AD = 48 cm

In ΔADB,

Ncert solution class 10 chapter 12-17

By Pythagoras theorem,

AB= AD+BD2

⇒ AB= 482+(AB/2)2

⇒ AB= 2304+AB2/4

⇒ 3/4 (AB2)= 2304

⇒ AB= 3072

⇒ AB= 32√3 cm

Area of ΔADB = √3/4 ×(32√3)cm= 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

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.

Ncert solution class 10 chapter 12-18

Solution:

Side of square = 14 cm

Four quadrants are included in the four sides of the square.

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

Area of the square ABCD = 14= 196 cm2

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

= 77/2 cm2

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

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

= 196 cm– 154 cm2

= 42 cm2

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.

Ncert solution class 10 chapter 12-19

Solution:

Width of the track = 10 m

Distance between two parallel lines = 60 m

Length of parallel tracks = 106 m

Ncert solution class 10 chapter 12-20

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

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.

Ncert solution class 10 chapter 12-21

Solution:

Radius of larger circle, R = 7 cm

Radius of smaller circle, r = 7/2 cm

Height of ΔBCA = OC = 7 cm

Base of ΔBCA = AB = 14 cm

Area of ΔBCA = 1/2 × AB × OC = (½)×7×14 = 49 cm2

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

Area of larger semicircle = 154/2 cm= 77 cm2

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

Area of the shaded region = Area of larger circle – Area of triangle – Area of larger semicircle + Area of smaller circle

Area of the shaded region = (154-49-77+77/2) cm2

= 133/2 cm2 = 66.5 cm2

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)

Ncert solution class 10 chapter 12-22

Solution:

ABC is an equilateral triangle.

∴ ∠ A = ∠ B = ∠ C = 60°

There are three sectors each making 60°.

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

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

Area of the sector = (60°/360°)×π rcm2

= 1/6×3.14×(100)cm2

= 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 cm= 1620.5 cm2

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.

Ncert solution class 10 chapter 12-23

Solution:

Number of circular designs = 9

Radius of the circular design = 7 cm

There are three circles in one side of square handkerchief.

∴ Side 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 r= (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

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.

Ncert solution class 10 chapter 12-24

Solution:

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) cm= 49/8 cm2

= 6.125 cm2

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)

Ncert solution class 10 chapter 12-25

Solution:

Side of square = OA = AB = 20 cm

Radius of the quadrant = OB

OAB is right angled triangle

By Pythagoras theorem in ΔOAB,

OB= AB2+OA2

⇒ OB= 20+202

⇒ OB= 400+400

⇒ OB= 800

⇒ OB= 20√2 cm

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

Area of the square = 20×20 = 400 cm2

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

= 628-400 cm= 228cm2

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.

Ncert solution class 10 chapter 12-26

Solution:

Radius of the larger circle, R = 21 cm

Radius of the smaller circle, r = 7 cm

Angle made by sectors of both concentric circles = 30°

Area of the larger sector = (30°/360°)×πRcm2

= (1/12)×(22/7)×21cm2

= 231/2cm2

Area of the smaller circle = (30°/360°)×πrcm2

= 1/12×22/7×7cm2

=77/6 cm2

Area of the shaded region = (231/2) – (77/6) cm2

= 616/6 cm2 = 308/3cm2

15. In Fig. 12.33, 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.

Ncert solution class 10 chapter 12-27

Solution:

Radius of the quadrant ABC of circle = 14 cm

AB = AC = 14 cm

BC is diameter of semicircle.

ABC is right angled triangle.

By Pythagoras theorem in ΔABC,

BC= AB+AC2

⇒ BC= 14+142

⇒ BC = 14√2 cm

Radius of semicircle = 14√2/2 cm = 7√2 cm

Area of ΔABC =( ½)×14×14 = 98 cm2

Area of quadrant = (¼)×(22/7)×(14×14) = 154 cm2

Area of the semicircle = (½)×(22/7)×7√2×7√2 = 154 cm2

Area of the shaded region =Area of the semicircle + Area of ΔABC – Area of quadrant

= 154 +98-154 cm= 98cm2

16. Calculate the area of the designed region in Fig. 12.34 common between the two quadrants of circles of radius 8 cm each.

Solution:

Ncert solution class 10 chapter 12-28

AB = BC = CD = AD = 8 cm

Area of ΔABC = Area of ΔADC = (½)×8×8 = 32 cm2

Area of quadrant AECB = Area of quadrant AFCD = (¼)×22/7×82

= 352/7 cm2

Area of shaded region = (Area of quadrant AECB – Area of ΔABC) = (Area of quadrant AFCD – Area of ΔADC)

= (352/7 -32)+(352/7- 32) cm2

= 2×(352/7-32) cm2

= 256/7 cm2

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