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 × πr2 

=   

= 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)

= 

= 36π + 36√3 – 6π

= 30π + 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 × 2πr 

= 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 π = 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:

(i) quadrant OACB,

(ii) shaded region.

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 π = 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 


1 Comments

Please do not enter any spam link in the comment box.

  1. Thank you so much for sharing this blog with us. It provides a collection of useful information. You obviously put a lot of effort into it! Best NEET Preparation service provider.

    ReplyDelete
Post a Comment
Previous Post Next Post