Important Questions for CBSE Class 10 Maths Chapter 12 - Areas Related to Circles 3 Mark Question -CoolGyan

CBSE Class 10 Maths Chapter-12 Areas Related to Circles – Free PDF Download

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CBSE Class 10 Maths Chapter-12 Areas Related to Circles Important Questions

CBSE Class 10 Maths Important Questions Chapter 12 – Areas Related to Circles

3 Mark Questions

1. The cost of fencing a circular field at the rate of Rs. 24 per metre is Rs. 5280. The field is to be ploughed at the rate of Rs. 0.50 Per m². Find the cost of ploughing the field.
Ans. Since for Rs. 24, the length of fencing = 1metre
for Rs 5280, the length of fencing
Perimeter i.e. circumference of the field = 220m
Let r be radius of the field.

Area of the field =
Rate = Rs 0.50 Per m²
Total cost of ploughing the field =

2. The length of the minute hand of a clock is 14cm. Find the area swept by the minute hand in 5 minutes.
Ans. Angle covered by minute hand in 60 minutes
Angle covered in 1 minute
Angle covered in 5 minutes =
We know that area swept by the minute hand during this period
=Area of sector with sector angle

3. An umbrella has 8 ribs which are equally spaced as given in the figure. Assuming umbrella to be a flat circle of radius 45cm. Find the area between two consecutive ribs of the umbrella.

Ans. Area between two consecutive ribs of the umbrella
= Area of the sector with angle made at centre

4. Find the area of the shaded region if PQ = 24cm, PR = 7cm and O is the centre of the circle.

Ans. Area of the shaded region
= Area of the semi-circle with O as centre and OQ as radius – area of
Since QR is a diameter passing through the centre O of the circle
[Angle of semi-circle]

Þ
Diameter of the circle = 25cm
And
Area of the semi circle

Also, area of

Hence, area of the shaded region

5. The perimeter of a sector of a circle of radius 5.7m is 27.2m.Calculate:
(i) The length of arc of the sector in cm.
(ii) The area of the sector in cm² correct to the nearest cm²

Ans. Let O be the centre of a circle of radius 5.7m and OACB be the given sector with perimeter 27.2m.
Then OA=OB=5.7m
Now,

(i) Length of arc AB = 15.8m
(ii) Area of sector

Area of sector correct to nearest

6. Find the area of shaded region in the given figure where ABCD is a square of side 10cm and semi-circles are drawn with each side of square as diameter.

Ans. Area of region I + II = area of ABCD – area of 2 semicircles of each radius 5cm

Similarly, area of III + area of IV =
Area of region I, II, III, and IV
Thus, area of shaded region = Area ABCD – Area of (I, II, III, IV)

7. Find the area of the shaded region where a circular arc of radius 6cm has been drawn with vertex O of an equilateral triangle OAB of side 12cm as centre.

Ans. Area of shaded region
= Area of major sector OPLQO + Area of equilateral

8. In the given figure is an equilateral triangle inscribed in a circle of radius 4 cm and centre O. Show that the area of the shaded region is
Ans. In

Area of shaded region = Area of sector OBPC – Area of

9. The radii of two circles are 8cm and 6cm respectively. Find the radius of the circle having its area equal to the sum of the areas of the two circles.
Ans. A₁ = Area of the first circle =
A₂ = Area of the second circle =
A₁ + A₂ = Total area =
Let R be the radius of the circle with area A₁ + A₂

Hence, required radius = 10cm

10. A chord of a circle of radius 10cm subtends a right angle at the centre. Find the area of the corresponding: (Use )
(i) minor sector
(ii) major sector
(iii) minor segment

(iv) major segment
Ans. (i) Area of minor sector =

(ii) Area of major sector = Area of circle – Area of minor sector

(iii) We know that area of minor segment
= Area of minor sector OAB – Area of

(iv) Area of major segment = Area of the circle – Area of minor segment

11. Find the area of the shaded region in the given figure where ABCD is a square of side 10cm and semi-circle are drawn with each side of the square as diameter.

Ans. Let us mark the four unshaved regions as I, II, III and IV
Thus, area of I + Area of III
= Area of ABCD – Area of two semi circles of each of radius 5 cm

Similarly, area of II + area of IV = 21.5cm²
Hence, area of the unshaded region i.e.,
I, II, III, IV
= 2(21.5) cm²=45cm²
Thus, area of the shaded region
= Area ABCD – Area of (I,II,III and IV) = 100 – 43= 57cm²

12. Find the area of the shaded region where a circular arc of radius 6cm has been drawn with vertex O of an equilateral triangle OAB of side 12cm as centre.

Ans. Area of the shaded region
= Area of major sector OPLQO + Area of equilateral triangle OAB

13. In Akshita’s house, there is a flower pot. The sum of radii of circular top and bottom of a flowerpot is 140 cm and the difference of their circumference is 88cm, find the diameter of the circular top and bottom.
Ans. Sum of radii of circular top and bottom = 140cm
Let radius of top = r cm
rRadius of bottom =
Circumference of top =
Circumference of bottom =
Difference of circumference =
By the given condition,

Radius of top = 77cm
Diameter of top =
Radius of bottom = 140 – r=140 – 77=63cm
Diameter of bottom = = 126cm

14. A chord of a circle of radius 15cm subtends an angle of 60at the centre. Find the area of the corresponding minor and major segments of the circle. (use and )

Ans. We know that area of minor segment
= Area of minor sector OAB – area of

Again, area of major segment = Area of the circle – area of minor segment

15. Find the area of the shaded region of the two concentric circles with centre O and radii 7cm and 14 cm respectively and .

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

16. From each corner of a square of side 4cm, a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2cm is cut as shown in the figure. Find the area of the remaining portion of the square.

Ans. Area of the shaded region = Area of square ABCD -4 (area of quadrant of acircle with radius 1) –

(Radius of the circle = 1cm)

17. The length of the minute hand of clock is 14cm. Find the area swept by the minute hand is 5 minutes.
Ans. Angle covered by minute hand in 60 minutes =
Angle covered in 1 minute =
Angle covered in 5 minutes =
We know that area swept by the minute hand during this period
= Area of the sector with sector angle

18. A round table cover has six equal designs as shown in the figure. If the radius of the cover is 28cm, find the cost of making the design at the rate of Rs.0.35 per . (Use )

Ans. Area of one design = – Area of
– Area of equilateral

[Area of equilateral triangle with side a = ]

Total cost =

=

19. Find the area of the shaded region where ABCD is a square of side 14cm.

Ans. Area of square ABCD =
Diameter of each circle =
Radius of each circle =
Area of one circle =
Area of the four circles =

20. ABCD is a flower bed. If OA = 21m and DC = 14m. Find the area of the bed.

Ans. Here, OA = R = 21m and OC = r = 14m
Area of the flower bed (i.e., shaded portion)
= area of quadrant of a circle of radius R of the quadrant of a circle of radius r

21. In a circle of radius 21 cm, an arc subtends an angle of  at the centre. Find:
(i) the length of the arc.
(ii) area of the sector formed by the arc.
(iii) area of the segment formed by the corresponding chord

Ans. Given,  = 21 cm and
(i) Length of arc =  =  = 22 cm
(ii) Area of the sector =  =  = 231 cm²
(iii) Area of segment formed by corresponding chord
=  Area of OAB
Area of segment = 231 – Area of OAB ……….(i)
In right angled triangle OMA and OMB,
OM = OB [Radii of the same circle]
OM = OM [Common]
OMAOMB [RHS congruency]
AM = BM [By CPCT]
M is the mid-point of AB and AOM = BOM
AOM = BOM = AOB =
Therefore, in right angled triangle OMA,

OM = cm
Also,

AM =  cm
AB = 2 AM =  = 21 cm
Area of OAB =  x AB x OM
=  = cm²
Using eq. (i),
Area of segment formed by corresponding chord = cm²

22. A chord of a circle of radius 15 cm subtends an angle of  at the centre. Find the area of the corresponding segment of the circle.

Ans. Here,  = 15 cm and
Area of minor sector =
=
= 117.75 cm²
For, Area of AOB,
Draw OMAB.
In right triangles OMA and OMB,
OA = OB [Radii of same circle]
OM = OM [Common]
OMAOMB [RHS congruency]
AM = BM [By CPCT]
AM = BM =  AB and AOM
= BOM = AOB =
In right angled triangle OMA,

OM =  cm
Also,

AM =  cm
2 AM =  = 15 cm
AB = 15 cm
Area of AOB =  x AB x OM
=  =
=  = 97.3125 cm²
Area of minor segment = Area of minor sector – Area of AOB
= 117.75 – 97.3125 = 20.4375 cm²
And, Area of major segment =  Area of minor segment
= 706.5 – 20.4375 = 686.0625 cm²

23. A chord of a circle of radius 12 cm subtends an angle of  at the centre. Find the area of the corresponding segment of the circle.

Ans. Here,  = 15 cm and
Area of corresponding sector =
=
= 150.72 cm²
For, Area of AOB,
Draw OMAB.
In right triangles OMA and OMB,
OA = OB [Radii of same circle]
OM = OM [Common]
OMAOMB [RHS congruency]
AM = BM [By CPCT]
AM = BM =  AB and AOM = BOM
= AOB =
In right angled triangle OMA,

OM = 6 cm
Also,

AM =  cm
2 AM =  =  cm
AB =  cm
Area of AOB =  x AB x OM
=  =
= 36 x 1.73 = 62.28 cm²
Area of corresponding segment = Area of corresponding sector – Area of AOB
= 150.72 – 62.28 = 88.44 cm²

Unless stated otherwise, take
24. 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. RPQ =  [Angle in semi circle is ]
RQ² = PR² + PQ² =  = 49 + 576 = 625
RQ = 25 cm
Diameter of the circle = 25 cm
Radius of the circle =  cm
Area of the semicircle =  =
= cm²
Area of right triangle RPQ =  x PQ x PR
= = 84 cm²
Area of shaded region = Area of semicircle – Area of right triangle RPQ
=
= cm²