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

Ans: (d)

Perimeter of sector

Ans:
(A) Area of square shaped field
= 20 × 20
= 400 sq. m.
(B) Area of 4 quadrant = area of a. circle of radius 7m = πr²

OR
New radius = 10 m
So, area grazed by one horse =

(C) Area of ungrazed portion = Area of square field – Area of circle with radius 7 m
 

Ans: (c)
Given diameter of semi circle = d

∴ Radius, r = d/2
Area of semi circle
=

Ans: (i) Length of play ground . AB = 14 units, Breadth of play ground. AD = 7 units
Radius of semi - circular part is 7/2 units
Total perimeter of parking area = πr + 2r

=
= 11 + 7 = 18 Units
(ii) (a): Area of parking = πr² / 2
=
= 19.25 sq. units
Area of two quadrants (I) a n d [II) =1/2 x 1/4 x πr²
=
= 6.29 sq. units
Total area of parking and two quadrant
= 19.25 + 6.29
= 25.54 sq. units

Ans: Here, radius t(r) = 14 cm and  Sector angle (θ) = 60°
∴ Area of the sector

= 102.67 cm²

Since ∠O = 60° and OA = OB = 14 cm
∴ AOB is an equilateral triangle.
⇒ AS = 14 cm and ∠A = 60°
Draw OM ⊥ AB.
In ΔAMO

Now,

Now, area of the minor segment= (Area of minor sector) - (ar ΔAOB)
= 102.67 - 84.87 cm²
= 17.8 cm²
Area of the major segment
= Area of the circle - Area of the minor segment  

= (616 - 17.8) cm² = 598.2 cm²

Ans: (d)
Angle formed by minute hand of a clock in 60 minutes = 360°
∴ Angle formed by minute hand of a clock in 10 minutes = 10/60 x 360° = 60°
Length of minute hand of a dock = radius = 7 cm
∴ Required area
=
=

Ans: (d)

Let O be the centre of the circle. So. OA = OB = AB = 1 cm
So ΔOAB is an equilateral triangle.
∴ ∠AOB = 60°

∴  Required area = 8 x area of one segment with r = 1 cm,θ = 60°

= 8 x  (area of sector - area of ΔAOB)

Ans: Let AB be the wire of length 22 cm in the form of an art of a circle so blending an ∠AOB - 60° at centre O.

∵ Length of arc =
⇒
= 21 cm
Hence, radius of the circle is 21cm.

Ans: 

Given: Width of road = 21 m 
Radius of park, r₁ = 105 m  
⇒ Radius of the whole circular portion (park + road) 
r₂ = 105 + 21 = 126 m 
So, Area of road = Area of park and road – Area of park 

Hence, the area of the road is 15246 m².

Ans: Here radius (r) = 21 cm
5ector angle (θ) = 120°
∴ Area cleaned by each sweep of the blades
[∵ there are 2 blades]
=
= 22 x 7 x 3 x 2 cm²
= 924 cm²

Ans: Given. O is the centre of the circle of radius 21cm and AB is the chord that subtends an angle of 120° at the centre.

Draw OM ⊥ AB,
Area of the minor segment AMBP = Area of sector OAPB - Area of ΔAOB
Now, area of sector OAPB
=
= 462 cm²
Since, OM ⊥ AB.
[∵ Perpendicular from the centre to the chord bisects the angle subtended by the chord at the centre.]
In ΔAOM, sin60° = AM/AO, cos60° = OM/OA
⇒
⇒

∵
Area of ΔAOB =

Hence, Required Area =
= 462 - 381.92= 80.08 cm² 

Ans: Radius (r) of circle = 7 cm
Area of shaded region =
 
=
=
= 77 cm²

Ans:  Radius of the circle = 10 cm
Central angle subtended by chord AB = 60°
Area of minor sector OACB

Area of equilateral triangle OAB formed by radii and chord

∴ Area of minor segment ACBD
= Area of sector OACB - Area of triangle OAB
= (52.38 - 43.30) cm² = 9.08 cm²
Area of circle = πr² 

∴ Area of major segment ADBE
= Area circle - Area of minor segment
= (314.28 - 9.08) cm² = 305.20 cm²

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


Hence area of the shaded region = 6 cm²

Ans: In right angle triangle ABC



Area of shaded region = area of semicircle - area of ΔCAB + area of quadrant BOD

Ans: Let ‘r’ and ‘R’ be the radii of the smaller and bigger circles, respectively. 
Then, OO’ = R – r = 6 cm [Given] ...(i) 
Also, sum of their areas = 116π cm²
i.e., πR² + πr² = 116π

⇒ R² + r² = 116 ...(ii)
We know, (R – r)² = R² + r² – 2Rr 
⇒ (6)² = 116 – 2Rr 
⇒ 2Rr = 116 – 36 = 80 
⇒ Rr = 40 ...(iii) 
Also, (R + r)² = R² + r² + 2Rr = 116 + 2 × 40 [Using (ii) and (iii)] 
⇒ (R + r)² = 196 
⇒ R + r = 14 ...(iv) 
Now, adding equations (i) and (iv), we get 
2R = 20 
⇒ R = 10 cm 
Putting the value of R in equation (i), we get 
r = 4 cm 
Hence, the radii of the bigger and smaller circles are 10 cm and 4 cm, respectively.

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

Ans: Since OP = PQ = QO
⇒ APOQ is an equilateral triangle
∴ ∠POQ = 60°
Area of segment PAQM


Area of semicircle with M as centre
Area of shaded region

Ans: Given AD = 14 cm, AB = CD = 3.5 cm
∴ BC = 7 cmArea of shaded region

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

Now, area of the equilateral triangle OAB

∴ Area of shaded region = area of circular portion + area of equilateral triangle OAB

Ans:




PA = 5√S cm = BP (Tangents from an external point are equal)


Shaded area = 43 .25 - 26 .17 = 17.08 cm²

Ans:
In right ΔAOB

Perimeter of shaded region

Ans: Given, A circle of radius (r) = 10 cm in which ∠AOB = 90º. 
Area of the minor segment AQBP = Area of sector OAPB – Area of ΔAOB 
= θ / 360° × πr² – 1 /2 × OA × OB 
= 90 / 360 °  × 3.14 × 10 × 10 – 1/ 2 × 10 × 10 
= 3.14 × 5 × 5 – 5 × 10 
= 78.5 – 50 
= 28.5 cm² 

So, Area of the minor segment AQBP = 28.5 cm²
Area of the major segment ALBQA = Area of circle – Area of minor segment AQBP 
= 3.14 × (10)² – 28.5 
= 314 – 28.5 
= 285.5 cm²
Area of major segment ALBQA = 285.5 cm². Hence, the area of the minor segment AQBP is 28.5 cm² and the area of the major segment ALBQA is 285.5 cm².

Ans: Area of shaded region = area of trapezium - (area of 4 sectors)


Total area of all sector

[Sum of angles of a quadrilateral is 360°]
From (i) and (ii),
Area of shaded region = 350 - 154 = 196 cm²