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NCERT Solutions class 10 Maths Exercise 12.3 Ch 12 Area Related to Circles

NCERT Solutions for Class 10 Maths Exercise 12.3 Chapter 12 Area Related to Circles- FREE PDF Download

NCERT Class 10 Maths Ch 12 is one of the most important ones in the NCERT syllabus. Duly following NCERT Solutions for Class 10 Maths
Chapter 12 ensures you that there will be no hindrance when you opt for more advanced branches of Maths. This is where CoolGyan comes in. Our free Class 10 Area Related to Circles solutions will help you understand the chapter thoroughly.

NCERT Solutions for Class 10 Maths Chapter 12 – Area Related to Circles



Unless stated otherwise, take 

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.

Ans. In the given figure, RPQ = [Angle in semi-circle is 

 

 = 49 + 576 = 625

RQ = 25 cm

 Diameter of the circle = 25 cm

 Radius of the circle =  cm

Area of the semicircle = 

 = 

Area of right triangle RPQ = 1212x PQ x PR

 = 

Area of shaded region = Area of semicircle – Area of right triangle RPQ

 = 


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 = 

Ans. Area of shaded region = Area of sector OAC – Area of sector OBD 


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.

Ans. Area of shaded region 

= Area of square ABCD – (Area of semicircle APD + Area of semicircle BPC)

196[12×227×7×7+12×227×7×7]196−[12×227×7×7+12×227×7×7]

= 196 – 154 = 


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.

Ans. Area of shaded region 

= Area of circle + Area of equilateral triangle OAB – Area common to the circle and the triangle (Area of sector)

πr2+34(a)2θ360×πr2πr2+34(a)2−θ360∘×πr2

 = 


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 figure.

Ans. Area of remaining portion of the square 

= Area of square – [(4  Area of a quadrant + Area of a circle)]

16[227(4×14+1)]16−[227(4×14+1)]

 = 


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 (shaded region).

Ans. Area of design = Area of circular table cover – Area of the equilateral triangle ABC 

………(i)

 G is the centroid of the equilateral triangle.

 radius of the circumscribed circle =  cm

According to the question, 

 = 48 cm

Now, 

 

 

 

 

  = 3072

  cm

 Required area = [From eq. (i)]


7. In figure ABCD is a square of side 14 cm. With centers 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.

Ans. Area of shaded region = Area of square – 4  Area of sector 

 = 196 – 154 = 


8. Figure 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.

Ans. (i) Distance around the track along its inner edge 

=212+2×[12×2×227×602]=212+2×[12×2×227×602]

 =  =  m

(ii)Area of track = 

2120+227(40+30)(4030)2120+227(40+30)(40−30)

 = 


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.

Ans. Area of shaded region = Area of circle with diameter OD + Area of semicircle ACB – Area of ACB 

227×(72)2+12×227×(7)2(12×7×7+12×7×7)227×(72)2+12×227×(7)2−(12×7×7+12×7×7)

 =  = 


10. The area of an equilateral triangle ABC is . 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. 

Ans. Given: Area of equilateral triangle =  = 17320.5 

 

 

  cm

Area of shaded region = Area of ΔΔABC – Area of 3 sectors

= 17320.5  – 

= 17320.5 – 3[16×3.14×100×100]3[16×3.14×100×100]

= 17320.5 – 3 x 5233.33

= 17320.5 – 15700 = 


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.

Ans. Radius of each design = 7 cm, then Diamter = 7 x 2 = 14 cm 

Therefore, side of square = 14 + 14 + 14 = 42 cm

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

= 1764 – 1386 = 


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

Ans. (i) Area of quadrant OACB =  

 = 

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

77812×OB×OD778−12×OB×OD

 =  cm2


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

Ans. OB =      [Using Pythagoras theorem] 

OA =  cm

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


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 =  find the area of the shaded region.

Ans. Area of shaded region = Area of sector OAB – Area of sector OCD 

 = 693776693−776

 =  


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.

Ans. In right triangle BAC, [Pythagoras theorem] 

 = 

 BC =  cm

Radius of the semicircle =  cm

 Required area = Area BPCQB

= Area BCQB – Area BCPB

= Area BCQB – (Area BACPB – Area BAC)

= 154 – (154 – 98) = 


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

Ans. In right triangle ADC, [Pythagoras theorem]
 = 
 AC =  =  cm 

Draw BMAC.
Then AM = MC = AC =  =  cm
In right triangle AMB,
 [Pythagoras theorem]
 
 = 64 – 32 = 32
 BM =  cm
Area of ABC = 
 = 
 Half Area of shaded region

 = 
 Area of designed region
 =