NCERT Solutions Maths Class 10 Exercise 12.3

# NCERT Solutions Maths Class 10 Exercise 12.3

## Maths Class 10 Exercise 12.3

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

Solution: RPQ = 90°               [Angle in semicircle is 90°]

In right-angled ∆PQR, we have

RQ2 = PR2 + PQ2                                      [By Pythagoras theorem]

= (7)2 + (24)2

= 49 + 576

= 625

RQ = 25 cm

Diameter of the circle = 25 cm

Radius of the circle = 25/2 cm

Area of the semicircle = ½ × Ï€r2

=

= 6875/28 cm2

Area of right-angled ∆PQR = ½ × PQ × PR

= ½ × 24 × 7

= 84 cm2

Area of the shaded region = Area of semicircle – Area of right-angled ∆PQR

=

= 4523/28 cm2

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

Solution: Area of the shaded region = Area of sector OAC – Area of sector OBD

=

=

=

= 154/3 cm2

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

Solution: Area of the shaded region = Area of square ABCD – (Area of semicircle APD + Area of semicircle BPC)

=

= 196 – 22/7 × 7 × 7

= 196 – 154 = 42 cm2

4. Find the area of the shaded region in 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.

Solution: Area of the shaded region = Area of circle + Area of equilateral triangle OAB – Area of common part to the circle and the triangle (i.e., a sector)

=

= 36Ï€ + 36√3 – 6Ï€

= 30Ï€ + 36√3

= 30 × 22/7 + 36√3

=

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

Solution: Area of the remaining portion of the square = Area of square – (4 × Area of a quadrant + Area of a circle)

= 16 – 2 × 22/7

= 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 figure. Find the area of the design.

Solution: Area of the design = Area of circular table cover – Area of the equilateral triangle ABC

=               ……… (i)

Since G is the centroid of the equilateral triangle.

Therefore, radius of the circumscribed circle = 2h/3 cm

According to the question, 2h/3 = 32

h = 48 cm

Now, a2 = h2 + (a/2)2

a2 = h2 + a2/4

a2 – a2/4 = h2

3a2/4 = h2

a2 = 4h2/3

a2 = 4(48)2/3

a2 = 3072

Hence, the required area =            [From equation (i)]

=

=

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.

Solution: Area of the shaded region = Area of square – 4 × Area of a sector

=

=

= 196 – 154

= 42 cm2

8. Figure given below 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.

Solution: (i) Distance around the track along its inner edge

= 106 + 106 + 2 × ½ × 2Ï€r

= 212 + 2 × 22/7 × 30

= 212 + 1320/7

= 2804/7 m

(ii) Area of the track =

=

= 2120 + 22/7 × 700

= 4320 m2

9. In 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.

Solution: Area of the shaded region = Area of the smaller circle + Area of semicircle ACB – Area of ∆ACB

=

= 77/2 + 187 – 49

= 133/2

= 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 figure). Find the area of the shaded region. (Use Ï€ = 3.14 and √3 = 1.73205)

Solution: Area of equilateral triangle = √3a2/4 = 17320.5

a = 200 cm

Area of the shaded region = Area of ABC –

= 17320.5 – 15700

= 1620.5 cm2

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

Solution: Area of the remaining portion of the handkerchief = Area of square ABCD – Area of 9 circular designs

=

= 1764 – 1386

= 378 cm2

12. In figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the:

Solution: (i) Area of the quadrant OACB =

=

= 77/8 cm2

(ii) Area of the shaded region = Area of the quadrant OACB – Area of OBD

= 77/8 – ½ × OB × OD

= 77/8 – ½ × 3.5 × 2

= 77/8 – 35/10

= 49/8 cm2

13. In figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region.  (Use Ï€ = 3.14)

Solution: Using Pythagoras theorem, OB =

=

√2 OA = √2 × 20 = 20√2 cm

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

=

= 1/4 × 3.14 × 800 – 400

= 200 × 3.14 – 400

= 228 cm2

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

Solution: Area of the shaded region = Area of sector OAB – Area of sector OCD

=

=

= 231/2 – 77/6

= (692 – 77)/6

= 616/6

= 308/3 cm2

15. In 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.

Solution: In right-angled triangle ABC, BC2 = AB2 + AC2    [By Pythagoras theorem]

BC2 = (14)2 + (14)2 = 2(14)2

BC = 14√2 cm

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

Area of the shaded region = Area of the semicircle BCQB – Area of segment BCPB

= Area of the semicircle BCQB – (Area of the quadrant BACPB – Area of ABC)

=

=

= 154 – (154 – 98)

= 98 cm2

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

Solution: In right-angled triangle ADC, AC2 = AD2 + CD2        [By Pythagoras theorem]

AC2 = (8)2 + (8)2 = 2(8)2

AC = 8√2 cm

Draw BM perpendicular to AC.

Then AM = MC = ½ AC = ½ × 8√2 = 4√2 cm

In right-angled triangle AMB,

AB2 = AM2 + BM2                 [By Pythagoras theorem]

(8)2 = (4√2)2 + BM2

BM2 = 64 – 32 = 32

BM = 4√2 cm

Area of ABC = ½ × AC × BM

= 32 cm2

Half area of shaded region =

= 16 × 22/7 – 32

= 128/7 cm2

Area of the designed region = 2 × 128/7 = 256/7 cm2