1. The radii of the two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has a circumference equal to the sum of the circumferences of the two circles.
1. The radii of the two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has a circumference equal to the sum of the circumferences of the two circles.
Explanation:
If we assume that R is the radius of the circle whose circumference equals the total of the circumferences of the two circles, then the answer to the question is:
R = 19 + 9
R = 28 cm
2. The radii of the two circles are 8 cm and 6 cm respectively. Find the radius of the circle having an area equal to the sum of the areas of the two circles.
2. The radii of the two circles are 8 cm and 6 cm respectively. Find the radius of the circle having an area equal to the sum of the areas of the two circles.
Explanation:
Considering that the area of the circle with R as its radius equals the sum of the areas of the two circles
The answer to the query is
R2 = 64 + 36
R2 = 100
R = 10 cm
3. Figure depicts an archery target marked with its five scoring areas from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing the Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of the five scoring regions.
3. Figure depicts an archery target marked with its five scoring areas from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing the Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of the five scoring regions.
Explanation:
Gold: Diameter = 21 cm
Radius = cm
gold scoring region’s Area =
= = 346.5 cm2
Red: red scoring region’s Area =
=
=
= 1386 – 346.5 =
Blue: blue scoring region’s Area =
=
=
= 3118.5 – 1386 =
Black: black scoring region’s Area =
=
=
= 5544 – 3118.5 =
White: white scoring region’s Area =
=
=
= 8662.5 – 5544 =
4. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is traveling at a speed of 66 km per hour?
4. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is traveling at a speed of 66 km per hour?
Explanation:
The size of the wheel= 80 cm
Wheel's radius = 40 cm
Length covered by a wheel during one rotation= = = cm
The distance a wheel may travel in one hour= 66 km = 66000 m = 6600000 cm
Wheel distance traveled in 10 minutes= = 1100000 cm
Total No. of revolutions =
= = 4375
5. Tick the correct answer in the following and justify your choice: If the perimeter and area of a circle are numerically equal, then the radius of the circle is:
(A) 2 units
(B) units
(C) 4 units
(D) 7 units
5. Tick the correct answer in the following and justify your choice: If the perimeter and area of a circle are numerically equal, then the radius of the circle is:
(A) 2 units
(B) units
(C) 4 units
(D) 7 units
Explanation:
(A) Since the circle's area equals its perimeter,
Circumference = Area
= 2 units
As a result, option (A), which states that the circle's radius is 2 units, is right.
6. Find the area of a sector of a circle with a radius 6 cm, if the angle of the sector is
6. Find the area of a sector of a circle with a radius 6 cm, if the angle of the sector is
Explanation:
Consider, = 6 cm and
sector’s Area =
= = cm2
7. Find the area of a quadrant of a circle whose circumference is 22 cm.
7. Find the area of a quadrant of a circle whose circumference is 22 cm.
Explanation:
Presented, = 22
cm
Consider that for the quadrant of a circle,
quadrant’s Area =
= = cm2
8. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
8. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Explanation:
Consider, = 14 cm and
Area swept =
= =
9. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding: (i) minor segment, (ii) major segment.
9. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding: (i) minor segment, (ii) major segment.
Explanation:
(i) Consider, = 10 cm and
Minor sector’s Area =
= =
OAB’S Area =
= =
minor segment’s Area = minor sector’s Area – OAB’s Area
= 78.5 – 50 =
(ii) Major sector, radius = 10 cm and
Major sector’s Area =
= =
10. 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.
10. 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.
Explanation:
Presented, = 21 cm and
(i) arc’s Length =
= = 22 cm
(ii) Sector’s Area =
= =
(iii) Segment’s Area created by the appropriate chord
= OAB’s Area
segment’s Area = 231 – OAB’s Area……….(i)
By right angle triangles OMA and OMB,
OM = OB[The same circle's radii]
OM = OM [Common]
OMA OMB [ By RHS congruency]
AM = BM[CPCT]
M is the mid-point of AB and AOM = BOM
AOM = BOM
= AOB =
Hence, in the right angle triangle OMA,
OM = cm
AM = cm
AB = 2 AM = = 21 cm
OAB’s Area =
= =
Utilizing eq. (i),
Area of the section that the matching chord forms=
11. 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.
11. 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.
Explanation:
Consider, = 15 cm and
Minor sector’s Area=
=
=
By AOB’s Area,
To draw OMAB.
In right angle triangles OMA and OMB,
OA = OB[The same circle's radius]
OM = OM[Common]
OMA OMB [By RHS congruency]
AM = BM [CPCT]
AM = BM = AB and
AOM = BOM = AOB =
In right angle triangle OMA,
OM = cm
AM = cm
2 AM = = 15 cm
AB = 15 cm
AOB’s Area =
= =
= =
Minor segment’s Area = Minor sector’s Area – AOB’s Area
= 117.75 – 97.3125 = c
Major segment’s Area = Minor segment’s Area
= 706.5 – 20.4375 =
12. 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.
12. 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.
Explanation:
Consider, = 15 cm and
Corresponding sector’s Area=
=
=
ForAOB’s Area,
To Draw OMAB.
In right angle triangles OMA and OMB,
OA = OB[The same circle's radius]
OM = OM[Common]
OMA OMB [By RHS congruency]
AM = BM[ CPCT]
AM = BM = AB and
AOM = BOM = AOB =
In right angle triangle OMA,
OM = 6 cm
AM = cm
2 AM = = cm
AB = cm
AOB’s Area =
= =
= =
Corresponding segment’s Area = Corresponding sector’s Area – AOB’s Area
= 150.72 – 62.28 =
13. A horse is tied to a peg at one corner of a square-shaped grass field of side 15 m by means of a 5 m long rope (see figure). Find:
(i) the area of that part of the field in which the horse can graze.
(ii) the increase in the grazing area if the rope were 10 m long instead of 5 cm.
13. A horse is tied to a peg at one corner of a square-shaped grass field of side 15 m by means of a 5 m long rope (see figure). Find:
(i) the area of that part of the field in which the horse can graze.
(ii) the increase in the grazing area if the rope were 10 m long instead of 5 cm.
Explanation:
(i) Quadrant’s Area has 5m rope
=
= =
(ii) Quadrant’s Area having10 m rope
=
= =
The expansion of the grazing area
= 78.5 – 19.625
=
14. A brooch is made with silver wire in the form of a circle with a diameter of 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in the figure. Find:
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.
14. A brooch is made with silver wire in the form of a circle with a diameter of 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in the figure. Find:
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.
Explanation:
(i)Diameter = 35 mm
Radius = mm
Circumference = =
= 110 mm……….(i)
Size of five diameters= = 175 mm……….(ii)
The total amount of silver wire needed = 110 + 175 = 285 mm
(ii) mm and
The size of each brooch sector =
= =
15. An umbrella has 8 ribs that are equally spaced (see figure). Assuming the umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
15. An umbrella has 8 ribs that are equally spaced (see figure). Assuming the umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
Explanation:
Consider, = 45 cm and
The area between the umbrella's two successive ribs
=
=
=
16. A car has two wipers that do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of Find the total area cleaned at each sweep of the blades.
16. A car has two wipers that do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of Find the total area cleaned at each sweep of the blades.
Explanation:
Consider, = 25 cm and
The total area cleaned after each blade sweep
=
=
=
17. To warn ships of underwater rocks, a lighthouse spreads a red-coloured light over a sector of angle to a distance of 16.5 km. Find the area of the sea over which the ships are warned.
17. To warn ships of underwater rocks, a lighthouse spreads a red-coloured light over a sector of angle to a distance of 16.5 km. Find the area of the sea over which the ships are warned.
Explanation:
Consider, = 16.5 km and
The maritime region that ships are warned to avoid=
= =
18. A round table cover has six equal designs as shown in the figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Ts. 0.35 per cm2.
18. A round table cover has six equal designs as shown in the figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Ts. 0.35 per cm2.
Explanation:
= 28 cm and
Minor sector’s Area =
=
= =
For, AOB’s Area,
To Draw OMAB.
In right angle triangles OMA and OMB,
OA = OB[The same circle's radius]
OM = OM[Common]
OMA OMB [By RHS congruency]
AM = BM[CPCT]
AM = BM = AB and AOM = BOM = AOB =
In right angle triangle OMA,
OM = cm
AM = 14 cm
2 AM = = 28 cm
AB = 28 cm
AOB’s Area =
= =
= =
Minor segment’s Area = Minor sector’s Area – AOB’s Area
= 410.67 – 333.2 =
One design’s Area =
Six designs’ Area = =
Price of creating designs = = Rs. 162.68
19. Tick the correct answer in the following:
The area of a sector of angle (in degrees) of a circle with radius R is:
(A)
(B)
(C)
(D)
19. Tick the correct answer in the following:
The area of a sector of angle (in degrees) of a circle with radius R is:
(A)
(B)
(C)
(D)
Explanation:
(D) Presented, = R and
Sector’s Area =
=
=
=
20. Find the area of the shaded region in the figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.
20. Find the area of the shaded region in the figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.
Explanation:
RPQ = [Semi-circle’s Angle is ]
= = 49 + 576 = 625
RQ = 25 cm
Circle having Diameter = 25 cm
Circle having Radius = cm
Semicircle’s Area =
= =
Right triangle RPQ’S Area =
= =
Shaded region’s Area = Semicircle’s Area – Right triangle RPQ’s Area
= =
21. Find the area of the shaded region in the figure, if the radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and AOC =
21. Find the area of the shaded region in the figure, if the radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and AOC =
Explanation:
Shaded region’s Area = Sector OAC’S Area – Sector OBD’S Area
=
=
=
=
=
22. Find the area of the shaded region in the figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.
22. Find the area of the shaded region in the figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.
Explanation:
Area of the shaded region
= Area of square ABCD – (Area of semicircle APD + Area of semicircle BPC)
=
=
= 196 – 154 =
23. Find the area of the shaded region in the 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.
23. Find the area of the shaded region in the 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.
Explanation:
Shaded region’s Area
= circle’s Area + equilateral triangle’s Area OAB – Common Area to the circle and the triangle
=
=
= =
=
24. 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 the figure. Find the area of the remaining portion of the figure.
24. 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 the figure. Find the area of the remaining portion of the figure.
Explanation:
Area of the remaining portion of the square
= Square’s Area – (4 quadrant’s Area + circle’s Area)
=
= =
25. In a circular table covering of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the figure. Find the area of the design (shaded region).
25. In a circular table covering of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the figure. Find the area of the design (shaded region).
Explanation:
Design’s Area = Circular table cover’s Area – Equilateral triangle ABC’s Area
= ………(i)
triangle has G as the Centroid.
An equilaterala circumscribed circle having radius= cm
By the given question,
= 48 cm
= 3072
Area Required= [From eq. (i)]
=
=
26. 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.
26. 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.
Explanation:
Shaded region’s Area = Square’s Area – 4 sector’s Area
=
= = 196 – 154 =
27. 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.
27. 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.
Explanation:
(i)distance along the inner edge of the track around it.
=
= = = m
(ii) Track having Area =
=
= =
28. In the 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.
28. In the 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.
Explanation:
shaded region’s Area = Circle’s Area + semicircle ACB’S Area – ACB’s Area
=
= = =
29. The area of an equilateral triangle ABC is. With each vertex of the triangle as centre, a circle is drawn with a radius equal to half the length of the side of the triangle (see figure). Find the area of the shaded region.
29. The area of an equilateral triangle ABC is. With each vertex of the triangle as centre, a circle is drawn with a radius equal to half the length of the side of the triangle (see figure). Find the area of the shaded region.
Explanation:
Equilateral triangle’s Area = = 17320.5
cm
Shaded region’s Area = Area for
ABC –
= 17320.5 – 15700 =
30. 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.
30. 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.
Explanation:
Remaining portion of the handkerchief’s Area = Square ABCD’S Area – 9 circular designs Area
=
= 1764 – 1386 =
31. In the 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
31. In the 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
Explanation:
(i)Quadrant OACB’S Area =
= =
(ii) Shaded region’s Area = quadrant OACB’s Area – OBD’s Area
=
=
= =
32. In the figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region.
32. In the figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region.
Explanation:
OB =
=
= OA = cm
Shaded region’s Area = Quadrant OPBQ’s Area – Square OABC’s Area
=
=
=
=
33. 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.
33. 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.
Explanation:
shaded region’s Area = Sector OAB’s Area – Sector OCD’s Area
=
=
= =
= =
34. In the 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.
34. In the 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.
Explanation:
In right angle triangle BAC, [By Pythagoras theorem]
=
BC = cm
Semicircle having Radius = cm
Area Required= BPCQB’s Area
= BCQB’s Area – BCPB’s Area
= BCQB’s Area – (BACPB’s Area – BAC’s Area)
=
=
= 154 – (154 – 98) =
35. Calculate the area of the designed region in the figure common between the two quadrants of circles of radius 8 cm each.
35. Calculate the area of the designed region in the figure common between the two quadrants of circles of radius 8 cm each.
Explanation:
In right angle triangle ADC, [By Pythagoras theorem]
=
AC = = cm
To Draw BMAC.
AM = MC = AC = = cm
In right angle triangle AMB,
[By Pythagoras theorem]
= 64 – 32 = 32
BM = cm
ABC’s Area =
= =
½ Area of the shaded region
=
= =
Designed the region’s Area
= =
Chapter-12, Area related to circle