NCERT Solutions For Class 10 Math Chapter – 12 Exercise – 12.3

NCERT Solutions For Class 10 Math Chapter – 12 Exercise – 12.3

 

Q1. Find  the area of the shaded region If PQ = 24 cm , PR = 7 cm and O is the center of the circle.

Solution:-

QR² = PR² + PQ²

QR² = 7² + 24²

QR² = 49 + 576

QR² = 625

QR = 25 cm

Radius of the circle = 25/2 cm

Area  of the semicircle = (∏r²)/2

= (22/7 x 25/2 x 25/2)/2

= 245.54 cm²

Area of the ∆PQR = Q/2 X PQ / PR

= 1/2 X 24 X 7

= 84 cm²

Area  of the shade region = 245.54 cm² – 84 cm²

= 161.54 cm²

 

Q2. Find the area of the shaded region If radii of the two concentric circles with center O are 7 cm and 14 cm respectively and ∟AOC = 40⁰

Solution:-

 

 

 

 

 

 

 

 

Area of sector OAC = (40⁰/360⁰) X ∏r²

= 1/9 x 22/7  x 14² = 68.44 cm²

Area  of sector OBD = (40⁰/360⁰) x ∏r²

= 1/9 x 22/7 x 14²

= 68.44 cm²

 

Area of the sector OBD  = (40⁰/360⁰) X ∏R²

= 1/9 X 22/7 X 7²

= 17.11 cm²

 

Area of the shaded region ABDC = Area of the sector OAC – Area of the sector

= 6.44 cm² – 17.11 cm²

= 51.33 cm²

Q3. Find the area of the shaded region If ABCD is a square of side 14 cm and APD and BPC are semicircles.

Solution:-

 

Area of the square = 14 x 14 = 196 cm²

Area of the semicircle = (∏r²)/2

= (22/7 x 7 x 7)/2

= 77 cm²

 

Area of two semicircles = 2 x 77 cm² = 154 cm²

Area of the shaded region = 196  cm² – 154 cm²

= 42 cm²

 

Q4. Find the area of the shaded region where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as center.

Solution:-

Area of  the equilateral triangle = √3/4 x (OA)²

= √3/4 X 12²

= 36√3 cm²

Area of the circle = ∏r² = 22/7 x 6² = 792/7 cm²

Area of sector = (60⁰/360⁰) x ∏r²

= 1/6 x 22/7 x 6²

= 132/7 cm²

 

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

= 36√3 + 792/7 – 132/7

= (36√3 + 660√7) cm²

Q5. 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 Find the area of the remaining  portion of the square.

Solution:-

 

 

 

 

 

 

 

 

 

Area of square = (side)² = 4² = 16 cm²

Area of quadrant = (∏r²)/4 cm² = (22/7 x 1²)/4 = 11/4 cm²

Total area of 4 quadrant = 4 x (11/14) cm² = 22/7 cm²

Area of the circle = ∏r²

= (22/7 x 1²) = 22/7 cm²

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

= 16 – (22/7 + 22/7)

= 68/7 cm²

 

Q6. In a circular table cover of radius 32 cm , a design  is formed leaving an equilateral triangle ABC in the middle Find the area of the design.

Solution:-

 

 

 

 

 

 

 

 

 

AO = Radius of the circle = 2/3 AD

2/3 AD = 32 cm

AD = 48 cm

In ∆ADB,

 

 

 

 

 

 

 

By Pythagoras theorem,

AB² = AD² + BD²

AB² = 48² + (AB/2)²

AB²  = 2304 + AB²/4

AB =32√3 cm.

Area of ∆ADB = √3/4 X (32√3)² cm² = 768√3 cm²

Area of circle = ∏r² = 22/7 x 32 x 32

= 22528/7 cm²

Area of the design = Area of circle – Area of ∆ADB

= (22528/7 – 768√3) cm²

 

Q7. ABCD is a square of side 14 cm. With centers A , B , C and D, four circles are drawn such that each circle touch external two of the remaining three circles. Find  the area of the shaded region.

Solution:-

 

 

 

 

 

 

 

 

Area of the square ABCD  =14² = 196 cm²

Area of the quadrant = (∏r²)/4 cm² 

  = (22/7 x 7²)/4

= 77/2 cm²

 

Total area of the quadrant = 4 x 77/2 cm²

= 154 cm²

 

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

= 196 – 154

= 42 cm²

 

Q8. Fig 12.26 depicts racking 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  wide Find:

• The distance around the track along its inner edge
• the area of the track

Solution:-

 

Radius of inner semicircle = 60/2 = 30 m

Radius of outer semicircle = 30 + 10 = 40 m

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

= 106 + 106 + (2 x ∏r) m

= 212 + 1320/7 m

= 2804/7 m

 

Area of a track = Area of ABCD + Area EFGH + 2 X (Area of outer semicircle) – 2 x (  area of inner semicircle)

= (AB X CD) + (EF X GH) + 2 (∏R²/2) – 2 X (∏r²/2)

= (106 x 10) + (106 x 10) + 2 x ∏/2 (r² – R²) 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.

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 = πR2 = (22/7)×72 = 154 cm2

Area of larger semicircle = 154/2 cm2 = 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)

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 r2 cm2

= 1/6×3.14×(100)2 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 cm2 = 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.

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 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

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.

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) cm2 = 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)

Solution:

Side of square = OA = AB = 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)2 cm2 = 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

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.

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°)×πR2 cm2

= (1/12)×(22/7)×212 cm2

= 231/2cm2

Area of the smaller circle = (30°/360°)×πr2 cm2

= 1/12×22/7×72 cm2

=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.

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,

BC2 = AB2 +AC2

⇒ BC2 = 142 +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 cm2 = 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:

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