Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev

Mathematics (Maths) Class 10

Class 10 : Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev

The document Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.
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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]
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev   
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
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev

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 Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev    [CBSE (Delhi) 2016]
Ans.
Since OP = PQ = QO
⇒ APOQ is an equilateral triangle
∴ ∠POQ = 60°
Area of segment PAQM
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
Area of semicircle with M as centre Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
Area of shaded region Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev

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. Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev     [Foreign 2016]
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
Ans. 
Given AD = 14 cm, AB = CD = 3.5 cm
∴ BC = 7 cm
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
Area of shaded region
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev

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 ]
 Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev   
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
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
Now, area of the equilateral triangle OAB
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
∴ Area of shaded region = area of circular portion + area of equilateral triangle OAB
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev

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
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
Area of equilateral triangle OAB formed by radii and chord
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
∴ 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
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
∴ 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]
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
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 Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
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.
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev    [CBSE (AI) 2017]
Ans.
In right angle triangle ABC
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
Area of shaded region = area of semicircle - area of ΔCAB + area of quadrant BOD
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev

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]
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev   
Ans.
 Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
PA = 5√S cm = BP (Tangents from an external point are equal)
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
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 Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev    [CBSE (AI) 2016]
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev   
Ans.
 Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
In right ΔAOB
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
Perimeter of shaded region Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev

Q.6. Find the area of the shaded region in Fig. 12.48, where Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev are semicircles of diameter 14 cm, 3.5 cm, 7 cm and 3.5 cm respectively. Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev    [CBSE (E) 2016]
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev   
Ans.
Area of shaded region = Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev

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)
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
Total area of all sector
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
[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]
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
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)
Previous Year Questions - Areas Related to Circles (part-2) Class 10 Notes | EduRev
∴ 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

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