Class 10 Exam  >  Class 10 Notes  >  Mathematics (Maths) Class 10  >  Previous Year Questions: Areas Related to Circles - 2

Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

Q.13. In Fig. 12.34, O is the centre of a circle such that diameter AB =13 cm and AC = 12 cm. ISC is joined. Find the area of the shaded region. (Take k = 3.14)    [CBSE (AI) 2016]
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles   
Ans.
  In ΔABC, ∠ACB = 90° (Angle in the semicircle)
∴ BC2 + AC2 = AB2
∴ BC2 = AB2 -AC2
= 169 - 144 = 25
∴ BC = 5 cm
Area of the shaded region = area of semicircle - area of right AABC
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

Q.14. In Fig. 12.35, are shown two arcs PAQ and PBQ. Arc PAQ is a part of circle with centre 0 and radius OP while arc PBQ is a semi-circle drawn on PQ as diameter with centre M.
If OP = PQ = 10 cm show that area of shaded region is 25 Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles    [CBSE (Delhi) 2016]
Ans.
Since OP = PQ = QO
⇒ APOQ is an equilateral triangle
∴ ∠POQ = 60°
Area of segment PAQM
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Area of semicircle with M as centre Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Area of shaded region Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

Q.15. In figure, the boundary of shaded region consists of four semicircular arcs, two smallest being equal. If diameter of the largest is 14 cm and that of the smallest is 3.5 cm, calculate the area of the shaded region. Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles     [Foreign 2016]
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Ans. 
Given AD = 14 cm, AB = CD = 3.5 cm
∴ BC = 7 cm
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Area of shaded region
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

Q.16. Find the area of the shaded region in Fig. 12.23, where a circular arc of radius 6 cm has been drawn with vertex 0 of an equilateral triangle ΔOAB of side 12 cm as centre.    [NCERT, CBSE (F) 2016 ]
 Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles   
Ans.
We have, radius of circular region = 6 cm and each side of ΔOAB = 12 cm.
∴ Area of the circular portion
= area of circle - area of the sector
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Now, area of the equilateral triangle OAB
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
∴ Area of shaded region = area of circular portion + area of equilateral triangle OAB
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

Long Answer Type Questions

Q.1. A chord PQ of a circle of radius 10 cm subtends an angle of 60° at the centre of circle. Find the area of major and minor segments of the circle.    [Delhi 2017]
Ans.
  Radius of the circle = 10 cm
Central angle subtended by chord AB = 60°
Area of minor sector OACB
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Area of equilateral triangle OAB formed by radii and chord
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
∴ Area of minor segment ACBD
= Area of sector OACB - Area of triangle OAB
= (52.38 - 43.30) cm2 = 9.08 cm2 
Area of circle = πr
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
∴ Area of major segment ADBE
= Area circle - Area of minor segment
= (314.28 - 9.08) cm2 = 305.20 cm2

Q.2. In the given figure, ΔABC is a right-angled triangle in which ∠A is 90°. Semicircles are drawn on AB, AC and BC as diameters. Find the area of the shaded region.     [Al 2017]
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Ans. 
In right triangle ABC.
AB2 + AC2 = BC2
⇒ (3)2 + (4)2 = BC2 ⇒ 9 + 16 = BC2  ⇒ 25 = BC2 
∴ BC = 5 cm
Now, Area of shaded region = Area of semicircle on side AB + area of semicircle on side AC - area of semicircle on side BC + area of ΔABC
Now, Area o f semicircle on side AB Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Hence area of the shaded region = 6 cm2

Q.3. In Fig. 12.51, O is the centre of the circle with AC = 24 cm , AB = 7 cm and ∠BOD = 90°. Find the area of the shaded region.
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles    [CBSE (AI) 2017]
Ans.
In right angle triangle ABC
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Area of shaded region = area of semicircle - area of ΔCAB + area of quadrant BOD
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

Q.4. An elastic belt is placed around the rim of a pulley of radius 5 cm. (Fig. 12.46). From one point C on the belt, the elastic belt is pulled directly away from the centre O of the pulley until it is at P, 10 cm from the point O. Find the length o f the belt that is still in contact with the pulley. Also find the shaded area.  (Use π = 3.14 and √3 = 1.73)    [CBSE Delhi 2016]
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles   
Ans.
 Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
PA = 5√S cm = BP (Tangents from an external point are equal)
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Shaded area = 43 .25 - 26 .17 = 17.08 cm2

Q.5. In Fig. 12.47, a sector OAP of a circle with centre O, containing angle 0. AB is perpendicular to the radius OA and meets OP produced at B. Prove that the perimeter of shaded region is Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles    [CBSE (AI) 2016]
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles   
Ans.
 Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
In right ΔAOB
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Perimeter of shaded region Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

Q.6. Find the area of the shaded region in Fig. 12.48, where Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles are semicircles of diameter 14 cm, 3.5 cm, 7 cm and 3.5 cm respectively. Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles    [CBSE (E) 2016]
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles   
Ans.
Area of shaded region = Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

Q.7.  In figure, ABCD is a trapezium with AB || DC, AB = 18 cm, DC = 32 cm and distance between AB and DC = 14 cm. If arcs of equal radii 7 cm with centres A, B, C and D have been drawn, then find the area of the shaded region.     [Foreign 2015]
Ans.
Area of shaded region = area of trapezium - (area of 4 sectors)
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Total area of all sector
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
[Sum of angles of a quadrilateral is 360°]
From (i) and (ii),
Area of shaded region = 350 - 154 = 196 cm2

Q.8. Find the area of the shaded region given in Fig.    [Delhi 2015]
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Ans. 
Side of the square = 14 cm
Area of the square = a2 
= 142 - 196 cm2 
Let radius of a semicircle=x
radius of two semicircles = 2x
side of inner square = diameter of semicircle = 2x
According to figure 2x + 2x = 8
4x = 8 ⇒ x = 2 cm
⇒ Side of inner square = 4 cm
Area of unshaded region = area of inner square + 4 (Area of a semicircle)
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
∴ Area of shaded region = area of square - area of unshaded region
= (196-16-871) cm2 = (180-8π) cm2 
= 180-8 x 3.14 = 180-25.12 = 154.88 cm2

The document Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles is a part of the Class 10 Course Mathematics (Maths) Class 10.
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FAQs on Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

1. What is the formula to find the area of a circle?
Ans. The formula to find the area of a circle is A = πr², where A represents the area and r is the radius of the circle.
2. How do you find the circumference of a circle?
Ans. The circumference of a circle can be found using the formula C = 2πr, where C represents the circumference and r is the radius of the circle.
3. Can you calculate the area of a circle if only the diameter is given?
Ans. Yes, the area of a circle can be calculated if the diameter is given. The formula to find the area using the diameter is A = π(d/2)², where A represents the area and d is the diameter of the circle.
4. Is the value of π constant for all circles?
Ans. Yes, the value of π (pi) is a mathematical constant and is the same for all circles. Its approximate value is 3.14159.
5. How can the area of a sector be calculated?
Ans. The area of a sector can be calculated using the formula A = (θ/360)πr², where A represents the area, θ is the central angle of the sector, and r is the radius of the circle.
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