| Exercise 13.1 |
Exercise 13.1
Q1 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 1
The circumference of a circle exceeds its diameter by 16.8 cm. Find the circumference of the circle.
Sol :We know that, circumference of a circle, C = 2πr
and diameter, D = 2r
According to the question,
C = D + 16.8
⇒ 2πr = 2r + 16.8
⇒ 2πr – 2r = 16.8
⇒ 2r(π – 1)= 16.8
$\Rightarrow \mathrm{r}\left(\frac{22}{7}-1\right)=\frac{16.8}{2}$ $\left[\because \pi=\frac{22}{7}\right]$
$\Rightarrow \mathrm{r}\left(\frac{22-7}{7}\right)=8.4$
⇒ 15r = 8.4 × 7
⇒ r = 3.92 cm
$\Rightarrow \mathrm{r}\left(\frac{22-7}{7}\right)=8.4$
⇒ 15r = 8.4 × 7
⇒ r = 3.92 cm
∴ Circumference of circle = 2πr
$=2 \times \frac{22}{7} \times 3.92$
=24.64cm
Given: Radius of a circle, r = 42cm
Central Angle of the sector, θ = 150°
To find: Length of the arc i.e. AB
Now,
Length of an arc of a sector of angle θ$=\frac{\theta}{360} \times 2 \pi r$
⇒ Length of an arc of a sector of angle θ $=\frac{150}{360} \times 2 \times \frac{22}{7} \times 42$
$\left[\because \pi=\frac{22}{7}\right]$
⇒ Length of an arc of a sector of angle θ$=\frac{5}{12} \times 2 \times 22 \times 6$
⇒ Length of an arc of a sector of angle θ = 5×22 =110 cm
Given: Length of the arc = 8.8 cm
Central Angle of the sector, θ = 60°
To find: Radius of a circle
Now,
Length of an arc of a sector of angle $\theta=\frac{\theta}{360} \times 2 \pi r$
$=2 \times \frac{22}{7} \times 3.92$
=24.64cm
Q2 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 2
A sector is cut from a circle of radius 42cm. The central angle of the sector is 150°. Find the length of the arc.
Sol :Given: Radius of a circle, r = 42cm
Central Angle of the sector, θ = 150°
To find: Length of the arc i.e. AB
Now,
Length of an arc of a sector of angle θ$=\frac{\theta}{360} \times 2 \pi r$
⇒ Length of an arc of a sector of angle θ $=\frac{150}{360} \times 2 \times \frac{22}{7} \times 42$
$\left[\because \pi=\frac{22}{7}\right]$
⇒ Length of an arc of a sector of angle θ$=\frac{5}{12} \times 2 \times 22 \times 6$
⇒ Length of an arc of a sector of angle θ = 5×22 =110 cm
Q3 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 3
A pendulum swings through an angle 60° and describes an arc 8.8 cm in length. Find the length of the pendulum $\left[\right.$ Use $\left.\pi=\frac{22}{7}\right]$
Sol :Given: Length of the arc = 8.8 cm
Central Angle of the sector, θ = 60°
To find: Radius of a circle
Now,
Length of an arc of a sector of angle $\theta=\frac{\theta}{360} \times 2 \pi r$
$\Rightarrow 8.8=\frac{60}{360} \times 2 \times \frac{22}{7} \times 1$ $\left[\because \pi=\frac{22}{7}\right]$
$\Rightarrow 8.8=\frac{1}{6} \times 2 \times \frac{22}{7} \times \mathrm{r}$
$\Rightarrow 8.8=\frac{1}{3} \times \frac{22}{7} \times \mathrm{r}$
$\Rightarrow \mathrm{r}=\frac{8.8 \times 21}{22}$
⇒ r = 8.4cm
Given: Radius of the circle = 5.6cm
So, the circumference of the circle = 2πr
$=2 \times \frac{22}{7} \times 5.6$
= 35.2cm
Now,
The perimeter of square = Circumference of a circle
= 35.2cm
Therefore,
Side of a square$=\frac{\text { The perimeter of a square }}{4}$
$=\frac{35.2}{4}$
= 8.8cm
Hence, the side of a square = 8.8cm
Given: Length of the arc = 35.2 cm
Radius of a circle, r = 42cm
To find: Central Angle of the sector
Now,
Length of an arc of a sector of angle θ
$\Rightarrow 8.8=\frac{1}{6} \times 2 \times \frac{22}{7} \times \mathrm{r}$
$\Rightarrow 8.8=\frac{1}{3} \times \frac{22}{7} \times \mathrm{r}$
$\Rightarrow \mathrm{r}=\frac{8.8 \times 21}{22}$
⇒ r = 8.4cm
Q4 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 4
A wire made of silver is looped in the form of circular ear ring of radius 5.6 cm. It is rebent into a square form. Determine the length of the side of the square.
Sol :Given: Radius of the circle = 5.6cm
So, the circumference of the circle = 2πr
$=2 \times \frac{22}{7} \times 5.6$
= 35.2cm
Now,
The perimeter of square = Circumference of a circle
= 35.2cm
Therefore,
Side of a square$=\frac{\text { The perimeter of a square }}{4}$
$=\frac{35.2}{4}$
= 8.8cm
Hence, the side of a square = 8.8cm
Q5 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 5
An arc of a circle of radius 42cm has a length 35.2 cm. Find the angle subtended by the arc at the centre of the circle.
Sol :Given: Length of the arc = 35.2 cm
Radius of a circle, r = 42cm
To find: Central Angle of the sector
Now,
Length of an arc of a sector of angle θ
$=\frac{\theta}{360} \times 2 \pi r$
$\Rightarrow 35.2=\frac{\theta}{360} \times 2 \times \frac{22}{7} \times 42$ $\left[\because \pi=\frac{22}{7}\right]$
$\Rightarrow 35.2=\frac{\theta}{360} \times 2 \times 22 \times 6$
$\Rightarrow \theta=\frac{35.2 \times 360}{12 \times 22}$
⇒ θ = 48°
Hence, the angle subtended by the arc at the centre of the circle is 48°
Diameter of a car wheel = 80cm
∴ radius = $\mathrm{r}=\frac{80}{2}=40 \mathrm{cm}$
Distance covered in 1 revolution = Circumference of wheel
= 2πr
= 2 × π × 40
= 80π cm
Now, we find total distance covered
$\Rightarrow 35.2=\frac{\theta}{360} \times 2 \times \frac{22}{7} \times 42$ $\left[\because \pi=\frac{22}{7}\right]$
$\Rightarrow 35.2=\frac{\theta}{360} \times 2 \times 22 \times 6$
$\Rightarrow \theta=\frac{35.2 \times 360}{12 \times 22}$
⇒ θ = 48°
Hence, the angle subtended by the arc at the centre of the circle is 48°
Q6 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 6
A car has wheels which are 80 cm in diameter. How many complete revolutions does each wheel make in 10 minutes when the car is moving at a speed of 80 km per hour?
Sol :Diameter of a car wheel = 80cm
∴ radius = $\mathrm{r}=\frac{80}{2}=40 \mathrm{cm}$
Distance covered in 1 revolution = Circumference of wheel
= 2πr
= 2 × π × 40
= 80π cm
Now, we find total distance covered
We know that,
Speed $=\frac{\text { Distance }}{\text { Time }}$
Here, Speed = 80 km/hr
Time = 10 minutes
$=\frac{10}{60}$ hour $=\frac{1}{6}$ hour
Speed $=\frac{\text { Distance }}{\text { Time }}$
Here, Speed = 80 km/hr
Time = 10 minutes
$=\frac{10}{60}$ hour $=\frac{1}{6}$ hour
Putting a value in the formula, we get
$80=\frac{\text { Distance }}{\frac{1}{6}}$
$\Rightarrow$ Distance $=\frac{80}{6}$
⇒ Distance = 13.33 km
= 13.33 × 1000 m
= 1333.3 m
= 1333.3 × 100 cm
= 1333333.3 cm
Now,
Number of revolutions $=\frac{\text { Total distance }}{\text { Distance covered in } 1 \text { revolution }}$
$=\frac{1333333.3}{80 \pi}$
$=\frac{1333333.3 \times 7}{80 \times 22}$
= 5303.02
= 5303 (approx.)
Hence, number of revolutions = 5303
Area of circular park = 88704 m2
and Area of circle = πr2 = 88704
$\Rightarrow \frac{22}{7} \mathrm{r}^{2}=88704$
$\Rightarrow \mathrm{r}^{2}=\frac{88704 \times 7}{22}$
⇒ r2 = 28224
⇒ r = √28224
⇒ r = 168m
[taking positive root, because radius can’t be negative]
Perimeter of circle = 2πr
$=2 \times \frac{22}{7} \times 168$
= 1056 m
= 1.056 km
So, total distance after 10 rounds = 1.056 × 10 = 10.56km
$80=\frac{\text { Distance }}{\frac{1}{6}}$
$\Rightarrow$ Distance $=\frac{80}{6}$
⇒ Distance = 13.33 km
= 13.33 × 1000 m
= 1333.3 m
= 1333.3 × 100 cm
= 1333333.3 cm
Now,
Number of revolutions $=\frac{\text { Total distance }}{\text { Distance covered in } 1 \text { revolution }}$
$=\frac{1333333.3}{80 \pi}$
$=\frac{1333333.3 \times 7}{80 \times 22}$
= 5303.02
= 5303 (approx.)
Hence, number of revolutions = 5303
Q7 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 7
Rajeev walks around a circular park of area 88704 sq. m How long will he take to walk 10 rounds at the speed of 4.5 km per hour?
Sol :Area of circular park = 88704 m2
and Area of circle = πr2 = 88704
$\Rightarrow \frac{22}{7} \mathrm{r}^{2}=88704$
$\Rightarrow \mathrm{r}^{2}=\frac{88704 \times 7}{22}$
⇒ r2 = 28224
⇒ r = √28224
⇒ r = 168m
[taking positive root, because radius can’t be negative]
Perimeter of circle = 2πr
$=2 \times \frac{22}{7} \times 168$
= 1056 m
= 1.056 km
So, total distance after 10 rounds = 1.056 × 10 = 10.56km
Now, we know that
speed $=\frac{\text { Distance }}{\text { Time }}$
Here, Speed = 4.5 km/hr
Distance = 10.56 km
Putting value in formula, we get
$4.5=\frac{10.56}{\text { Time }}$
$\Rightarrow$ Time $=\frac{10.56}{4.5}$
= 2.34 hour
= 2 hours 20 minutes 24 seconds
Hence, Rajeev will take 2 hours 20 minutes and 24 seconds to walk 10 rounds.
Diameter of a bus wheel = 140cm
∴ radius = $\mathrm{r}=\frac{140}{2}=70 \mathrm{cm}$
Distance covered in 1 revolution = Circumference of wheel
= 2πr
= 2 × π × 70
= 140π cm
Now, we find total distance covered
We know that,
Speed $=\frac{\text { Distance }}{\text { Time }}$
speed $=\frac{\text { Distance }}{\text { Time }}$
Here, Speed = 4.5 km/hr
Distance = 10.56 km
Putting value in formula, we get
$4.5=\frac{10.56}{\text { Time }}$
$\Rightarrow$ Time $=\frac{10.56}{4.5}$
= 2.34 hour
= 2 hours 20 minutes 24 seconds
Hence, Rajeev will take 2 hours 20 minutes and 24 seconds to walk 10 rounds.
Q8 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 8
The diameter of the wheels of a bus is 140cm. How many revolutions per minute must a wheel make to move at a speed of 66km per hour?
Sol :Diameter of a bus wheel = 140cm
∴ radius = $\mathrm{r}=\frac{140}{2}=70 \mathrm{cm}$
Distance covered in 1 revolution = Circumference of wheel
= 2πr
= 2 × π × 70
= 140π cm
Now, we find total distance covered
We know that,
Speed $=\frac{\text { Distance }}{\text { Time }}$
Here, Speed = 66 km/hr
Time = 1 minutes
$=\frac{1}{60}$ hour $=\frac{1}{60}$ hour
Putting a value in the formula, we get
$66=\frac{\text { Distance }}{\frac{1}{60}}$
$\Rightarrow$ Distance $=\frac{66}{60}$
⇒ Distance = 1.1 km
= 1.1 × 1000 m
= 1100 m
= 1100 × 100 cm
= 110000 cm
Now,
Number of revolutions $=\frac{\text { Total distance }}{\text { Distance covered in } 1 \text { revolution }}$
$=\frac{110000}{140 \pi}$
$=\frac{110000 \times 7}{140 \times 22}$
Number of revolutions = 250
Given: Radius of circle = 4cm
And Central angle, θ = 30°
To find: Area of the sector
Now, Area of the sector of angle θ$=\frac{\theta}{360}\times \pi \times \mathrm{r}^{2}$
⇒ Area of the sector of angle θ$=\frac{30}{360} \times 3.14 \times(4)^{2}$ [∵ π = 3.14]
⇒ Area of the sector of angle θ$=\frac{1}{12} \times 3.14 \times 4 \times 4$
$=\frac{3.14 \times 4}{3}$
=4.19cm2
Now, we have to find the area of major sector (unshaded region)
= Area of circle – Area of sector OAPBO
= πr2 – 4.19
= {3.14 × (4)2} – 4.19
= {3.14 × 16} – 4.19
= 50.24 – 4.19
= 46.05 cm2
Hence, the area of sector = 4.19cm2 and area of the major sector = 46.05cm2
Given: Circumference of a circle = 22cm
⇒ 2πr = 22
$\Rightarrow 2 \times \frac{22}{7} \times \mathrm{r}=22$
⇒ 2r = 7
$\Rightarrow \mathrm{r}=\frac{7}{2} \mathrm{cm}$
Now, we have to find the area of Quadrant
Area of Quadrant $=\frac{1}{4}\times \pi \mathrm{r}^{2}$
Given: Length of minute hand = 12cm
Let us look at a clock:
So, minute hand covers a total of 60 minutes in one round of the clock.
So, 60 minutes = 360˚
1 minute $=\frac{360^{\circ}}{60}=6^{\circ}$
35 minutes = (35 × 6)˚ = 210˚
So in 35 minutes, minute hand subtends an angle of 210˚.
Now Area of segment $=\frac{\theta}{360^{\circ}} \pi r^{2}$
Where θ is the angle subtended.
Therefore, the area covered by minute hand$=\frac{210^{\circ}}{360^{\circ}} \pi r^{2}$
The radius of the circle = Length of the minute hand = 12cm
The area covered by minute hand $=\frac{7}{12} \times \frac{22}{7} \times 12^{2}$
The area covered by minute hand = (22 × 12) cm2
Area = 264 cm2
Hence, The area covered by minute hand in 35 minutes is 264cm2.
Sol :
Let OAB be a the given sector with perimeter 27.2cm
Let arc AB = l
Perimeter of sector OAB = 27.2cm
⇒ OA + AB + OB = 27.2
⇒ 5.6 + l + 5.6 = 27.2
⇒ l = 27.2 – 11.2
⇒ l = 16cm
Now, we know that
Length of the $\operatorname{arc}=\frac{\theta}{360} \times 2 \pi r$
$\Rightarrow 16=\frac{\theta}{360} \times 2 \pi r$
$\Rightarrow \frac{16}{2 \pi r}=\frac{\theta}{360}$ …(i)
Area of sector OAB$=\frac{\theta}{360} \times \pi r^{2}$
$=\frac{16}{2 \pi r} \times \pi r^{2}$ [from (i)]
= 8r
= 8 × 5.6
= 44.8cm2
$66=\frac{\text { Distance }}{\frac{1}{60}}$
$\Rightarrow$ Distance $=\frac{66}{60}$
⇒ Distance = 1.1 km
= 1.1 × 1000 m
= 1100 m
= 1100 × 100 cm
= 110000 cm
Now,
Number of revolutions $=\frac{\text { Total distance }}{\text { Distance covered in } 1 \text { revolution }}$
$=\frac{110000}{140 \pi}$
$=\frac{110000 \times 7}{140 \times 22}$
Number of revolutions = 250
Q9 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 9
Find the area of the sector of a circle with radius 4cm and angle 30°. Also, find the area of the corresponding major sector
[use π=3.14].
Sol :Given: Radius of circle = 4cm
And Central angle, θ = 30°
To find: Area of the sector
Now, Area of the sector of angle θ$=\frac{\theta}{360}\times \pi \times \mathrm{r}^{2}$
⇒ Area of the sector of angle θ$=\frac{30}{360} \times 3.14 \times(4)^{2}$ [∵ π = 3.14]
⇒ Area of the sector of angle θ$=\frac{1}{12} \times 3.14 \times 4 \times 4$
$=\frac{3.14 \times 4}{3}$
=4.19cm2
Now, we have to find the area of major sector (unshaded region)
= Area of circle – Area of sector OAPBO
= πr2 – 4.19
= {3.14 × (4)2} – 4.19
= {3.14 × 16} – 4.19
= 50.24 – 4.19
= 46.05 cm2
Hence, the area of sector = 4.19cm2 and area of the major sector = 46.05cm2
Q10 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 10
Find the area of a quadrant of a circle whose circumference is 22 cm.
Sol :Given: Circumference of a circle = 22cm
⇒ 2πr = 22
$\Rightarrow 2 \times \frac{22}{7} \times \mathrm{r}=22$
⇒ 2r = 7
$\Rightarrow \mathrm{r}=\frac{7}{2} \mathrm{cm}$
Now, we have to find the area of Quadrant
Area of Quadrant $=\frac{1}{4}\times \pi \mathrm{r}^{2}$
$=\frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}$
$=\frac{11 \times 7}{8}$
= 9.625cm2Q11 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 11
The minute hand of a clock is 12cm long. Find the area of the face of the clock described by the minute hand in 35 minutes.
Sol :Given: Length of minute hand = 12cm
Let us look at a clock:
So, minute hand covers a total of 60 minutes in one round of the clock.
So, 60 minutes = 360˚
1 minute $=\frac{360^{\circ}}{60}=6^{\circ}$
35 minutes = (35 × 6)˚ = 210˚
So in 35 minutes, minute hand subtends an angle of 210˚.
Now Area of segment $=\frac{\theta}{360^{\circ}} \pi r^{2}$
Where θ is the angle subtended.
Therefore, the area covered by minute hand$=\frac{210^{\circ}}{360^{\circ}} \pi r^{2}$
The radius of the circle = Length of the minute hand = 12cm
The area covered by minute hand $=\frac{7}{12} \times \frac{22}{7} \times 12^{2}$
The area covered by minute hand = (22 × 12) cm2
Area = 264 cm2
Hence, The area covered by minute hand in 35 minutes is 264cm2.
Q12 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 12
The perimeter of a sector of a circle of radius 5.6 cm is 27.2 cm. Find the area of the sector.Sol :
Let OAB be a the given sector with perimeter 27.2cm
Let arc AB = l
Perimeter of sector OAB = 27.2cm
⇒ OA + AB + OB = 27.2
⇒ 5.6 + l + 5.6 = 27.2
⇒ l = 27.2 – 11.2
⇒ l = 16cm
Now, we know that
Length of the $\operatorname{arc}=\frac{\theta}{360} \times 2 \pi r$
$\Rightarrow 16=\frac{\theta}{360} \times 2 \pi r$
$\Rightarrow \frac{16}{2 \pi r}=\frac{\theta}{360}$ …(i)
Area of sector OAB$=\frac{\theta}{360} \times \pi r^{2}$
$=\frac{16}{2 \pi r} \times \pi r^{2}$ [from (i)]
= 8r
= 8 × 5.6
= 44.8cm2
Q13 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 13
A chord of a circle of radius 14cm makes a right angle at the centre. Find the areas of the minor and the major segments of the circle.
Sol :
Given: Radius of circle = 14cm
$\therefore$ Area of minor sector $=\frac{\theta}{360} \times \pi r^{2}$
$=\frac{90}{360} \times \frac{22}{7} \times 14 \times 14$
$=\frac{1}{4} \times 22 \times 2 \times 14$
=154 cm2
$=\frac{90}{360} \times \frac{22}{7} \times 14 \times 14$
$=\frac{1}{4} \times 22 \times 2 \times 14$
=154 cm2
Area of triangle$=\frac{1}{2}\times \text{base}\times \text{height}$
$=\frac{1}{2}\times 14\times 14$
=98 cm2
Area of minor segment=Area of minor sector - Area of triangle
=154-98
=56 cm2
Area of major segment=Area of circle - Area of minor segment
= πr2 - 56 cm2
=$\frac{22}{7}\times 14^2$ - 56 cm2
=(22×14×2) cm2 - 56 cm2
=616-56
=560 cm2
Q14 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 14
The area of a circle is 78.5 sq.cm. Calculate the circumference of the circle[Taken π=3.14].
Sol :
Given: Area of circle = 78.5cm2
πr2 = 78.5
⇒ 3.14 r2 = 78.5
$\Rightarrow \mathrm{r}^{2}=\frac{78.5}{3.14}$
⇒ r2 = 25
⇒ r = ± 5
⇒ r = 5cm
[taking positive root, because radius can’t be negative]
Now, circumference of circle = 2πr
= 2 × 3.14 × 5
= 31.4cm
Q15 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 15
Find the area of the shaded region in the given figure if radii of the two concentric circles with centre O are 7cm and 14 cm respectively and ∠AOC=40°.
Sol :
Given: Radius of the small circle, OB = 7cm
Radius of second circle, OA = 14cm
and ∠AOC=40°
$\therefore$ Area of minor sector $\mathrm{OBD}=\frac{\theta}{360} \times \pi r^{2}$
$=\frac{40}{360} \times \frac{22}{7} \times 7 \times 7$
$=\frac{1}{9} \times 22 \times 7$
= 17.11 cm2$\therefore$ Area of minor sector $\mathrm{OAC}=\frac{\theta}{360} \times \pi \mathrm{r}^{2}$
$=\frac{40}{360} \times \frac{22}{7} \times 14 \times 14$
$=\frac{1}{9} \times 22 \times 2 \times 14$
= 68.4 cm2
Area of the shaded region = Area of sector OAC – Area of sector OBD
= 68.4 – 17.1
= 51.3 cm2
Q16 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 16
A circular park, 42m is a diameter, has a path 3.5 wide running round it on the outside. Find the cost of gravelling the path at Rs. 4 per m2
Sol :
The diameter of a circular park = 42m
⇒ the radius of a circular park, $r=\frac{42}{2}=21 \mathrm{m}$
Width of path = 3.5m
⇒ radius of the park with path, R = 21 + 3.5 = 24.5m
∴ Area of path = Area of outer circle – Area of the inner circle
= πR2 – πr2
= π (R2 – r2)
= π (R – r)(R + r)
$=\frac{22}{7}(24.5-21)(24.5+21)$
$=\frac{22}{7}(3.5)(45.5)$
= 22 × 0.5 × 45.5= 500.5 m2
Now, the cost of gravelling the path of 1m2 = Rs 4
the cost of gravelling the path of 500.5m2 = Rs 4×500.5
= Rs 2002
Q17 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 17
The diameter of a circular pond is 17.5m. It is surrounded by a path of width 3.5 m. Find the area of the path.
Sol :
The diameter of a circular park = 17.5m
⇒ the radius of a circular park, $r=\frac{17.5}{2}$
Width of path = 3.5m
⇒ radius of the park with path,
$\mathrm{R}=\frac{17.5}{2}+3.5=\frac{17.5+7}{2}=12.25 \mathrm{m}$
∴ Area of path = Area of outer circle – Area of inner circle
= πR2 – πr2
= π (R2 – r2)
= π (R – r)(R + r)
Given: Let the radius of the inner circle = r
Area enclosed between two concentric circles = 770cm2
and Radius of the outer circle, R = 21cm
∴ The area enclosed between two concentric circles
= Area of the Outer circle – Area of the inner circle
770 = πR2 – πr2
770 = π(212 – r2)
⇒ r2 = 441 – 245
⇒ r2 = 196
⇒ r = √196
⇒ r = ±14
⇒ r = 14cm [taking positive square root, because radius can’t be negative]
Hence, the radius of the inner circle is 14cm.
Given: Difference between circumference and diameter of circular plot = 105m
We know that,
The circumference of a circle = 2πr
and diameter of circle = 2r
According to the question
2πr – 2r = 105
⇒ 2r(π – 1) = 105
$\Rightarrow \mathrm{r}=\frac{105}{2(\pi-1)}$
Now, Area of circular path = πr2
Total cost of fencing a circular field = Rs 5280
Cost of fencing per meter = Rs 24
So,
The total length of the field= $\frac{\text { The total cost of fencing }}{\text { cost of fencing per meter }}$
$=\frac{5280}{24}$
= 220m
Here, the total length of the field would be the circumference of the circular field.
∴ Total length = Circumference of circular field
⇒ 220 = 2πr
Now, Area of the field = πr2
Now,
Cost of ploughing 1m2 of the field = Rs 0.50
Cost of ploughing 3850m2 of the field = Rs 0.50 × 3850
= Rs 1925
Hence, the cost of ploughing the field is Rs 1925
Cost of ploughing 1m2 of the field = Rs1.50
The total cost of ploughing the field = Rs 5775
So,
Area of the field $=\frac{\text { The total cost of ploughing the field }}{\text { cost of ploughing per square. meter }}$
$=\frac{5775}{1.50}$
=3850m2
Given: A field is in the form of a circle.
Let the radius of the field = r
∴ Area of the field = πr2
⇒ 3850 = πr2
⇒ r = √1225
⇒ r = √(5×5×7×7)
⇒ r = 35m
Now,
Circumference = 2πr
$=2 \times \frac{22}{7} \times 35$
= 2 × 22 × 5
= 220 m
Cost of fencing 1m of the field = Rs 8.50
Cost of fencing 220 of the field = Rs 8.50 × 220
= Rs 1870
= πR2 – πr2
= π (R2 – r2)
= π (R – r)(R + r)
$=\frac{22}{7}\left(\frac{24.5}{2}-\frac{17.5}{2}\right)\left(\frac{24.5}{2}+\frac{17.5}{2}\right)$
$=\frac{22}{7}\left(\frac{24.5-17.5}{2}\right)\left(\frac{24.5+17.5}{2}\right)$
$=\frac{22}{7} \times \frac{7}{2} \times \frac{42}{2}$
= 231 m2Q18 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 18
The area enclosed between two concentric circles is 770 cm2. If the radius of the outer circle is 21 cm, find the radius of the inner circle.
Sol :Given: Let the radius of the inner circle = r
Area enclosed between two concentric circles = 770cm2
and Radius of the outer circle, R = 21cm
∴ The area enclosed between two concentric circles
= Area of the Outer circle – Area of the inner circle
770 = πR2 – πr2
770 = π(212 – r2)
$770=\frac{22}{7}\left(441-\mathrm{r}^{2}\right)$
$\frac{770 \times 7}{22}=441-\mathrm{r}^{2}$
245 = 441 – r2⇒ r2 = 441 – 245
⇒ r2 = 196
⇒ r = √196
⇒ r = ±14
⇒ r = 14cm [taking positive square root, because radius can’t be negative]
Hence, the radius of the inner circle is 14cm.
Q19 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 19
The difference between circumference and diameter of circular plot is 105m. Find the area of the circular plot.
Sol :Given: Difference between circumference and diameter of circular plot = 105m
We know that,
The circumference of a circle = 2πr
and diameter of circle = 2r
According to the question
2πr – 2r = 105
⇒ 2r(π – 1) = 105
$\Rightarrow \mathrm{r}=\frac{105}{2(\pi-1)}$
Now, Area of circular path = πr2
$=\pi\left(\frac{105}{2(\pi-1)}\right)^{2}$
$=\frac{22 \times 105 \times 105}{7 \times 4\left(\frac{22}{7}-1\right)^{2}}$
$=\frac{11 \times 15 \times 105}{2 \times\left(\frac{22-7}{7}\right)^{2}}$
$=\frac{11 \times 15 \times 105 \times 49}{2 \times 15 \times 15}$
$=\frac{11 \times 7 \times 49}{2}$
= 1886.5m2Q20 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 20
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
$\left[\right.$ Take $\left.\pi=\frac{22}{7}\right]$
Sol :Total cost of fencing a circular field = Rs 5280
Cost of fencing per meter = Rs 24
So,
The total length of the field= $\frac{\text { The total cost of fencing }}{\text { cost of fencing per meter }}$
$=\frac{5280}{24}$
= 220m
Here, the total length of the field would be the circumference of the circular field.
∴ Total length = Circumference of circular field
⇒ 220 = 2πr
$\Rightarrow \frac{220}{2 \pi}=\mathrm{r}$
$\Rightarrow \mathrm{r}=\frac{110}{\pi}$
Now, Area of the field = πr2
$=\pi\left(\frac{110}{\pi}\right)^{2}$
$=\frac{110 \times 110 \times 7}{22}$
= 3850m2Now,
Cost of ploughing 1m2 of the field = Rs 0.50
Cost of ploughing 3850m2 of the field = Rs 0.50 × 3850
= Rs 1925
Hence, the cost of ploughing the field is Rs 1925
Q21 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 21
A field is in the form of a circle. The cost of ploughing the field at Rs. 1.50 per m2 is Rs. 5775. Find the cost of fencing the field at Rs. 8.50 per metre.
Sol :Cost of ploughing 1m2 of the field = Rs1.50
The total cost of ploughing the field = Rs 5775
So,
Area of the field $=\frac{\text { The total cost of ploughing the field }}{\text { cost of ploughing per square. meter }}$
$=\frac{5775}{1.50}$
=3850m2
Given: A field is in the form of a circle.
Let the radius of the field = r
∴ Area of the field = πr2
⇒ 3850 = πr2
$\Rightarrow 3850=\frac{22}{7} \mathrm{r}^{2}$
$\Rightarrow \mathrm{r}^{2}=\frac{3850 \times 7}{22}$
⇒ r2 = 1225⇒ r = √1225
⇒ r = √(5×5×7×7)
⇒ r = 35m
Now,
Circumference = 2πr
$=2 \times \frac{22}{7} \times 35$
= 2 × 22 × 5
= 220 m
Cost of fencing 1m of the field = Rs 8.50
Cost of fencing 220 of the field = Rs 8.50 × 220
= Rs 1870
Q22 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 22
A chord of a circle of radius 10cm subtends a right angle at the centre. Use π=3.14 and find:
(i) area of the minor sector
(ii)area of the major sector
Sol :(i) area of the minor sector
(ii)area of the major sector
Given: Radius of the circle = OA = OB = 10 cm
and θ = 90°
(i) Area of the minor sector
"Area of minor sector "$=\frac{\theta}{360} \times \pi r^{2}$
$=\frac{90}{360} \times 3.14 \times\left(10^{2}\right)$
$=\frac{1}{4} \times 3.14 \times 100$
= 3.14 × 25= 78.5cm2
(ii) Area of major sector
Area of major sector = Area of circle - Area of minor sector = πr2 – 78.5
= 3.14 × (10)2 – 78.5
= 314 – 78.5
= 235.5 cm2
Q23 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 23
In a circle of radius 21 cm, and arc subtends an angle of 60° at the centre. Find
(i) length of arc
(ii) area of the sector formed by the arc
(iii) area of the segment formed by the corresponding chord of the arc.
Sol :(i) length of arc
(ii) area of the sector formed by the arc
(iii) area of the segment formed by the corresponding chord of the arc.
Given: Radius of the circle = OA =OB = 21cm
and θ = 60°
(i) Length of the arc
Length of the $\operatorname{arc} \mathrm{APB}=\frac{\theta}{360} \times 2 \mathrm{nr}$
$=\frac{60}{360} \times 2 \times \frac{22}{7} \times 21$
$=\frac{1}{6} \times 2 \times 22 \times 3$
= 22cm(ii) Area of the sector formed by this arc
Area of minor sector OAPB $=\frac{\theta}{360} \times \pi r^{2}$
$=\frac{60}{360} \times \frac{22}{7} \times 21 \times 21$
$=\frac{1}{6} \times 22 \times 3 \times 21$
= 11 × 21= 231 cm2
(iii) area of the segment formed by the corresponding chord of the arc
In ΔOAB,
∠OAB = ∠OBA (As OA = OB)
∠OAB + ∠AOB + ∠OBA = 180°
2∠OAB + 60° = 180°
∠OAB = 60°
∴ ΔOAB is an equilateral triangle.
Area of $\Delta \mathrm{OAB}=\frac{\sqrt{3}}{4} \times(\text { side })^{2}$
$=\frac{\sqrt{3}}{4} \times(21)^{2}$
$=\frac{441 \sqrt{3}}{4} \mathrm{cm}^{2}$
Area of segment APB = Area of sector OAPB − Area of ΔOAB
$=\left(231-\frac{441 \sqrt{3}}{4}\right) \mathrm{cm}^{2}$
Q24 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 24
A chord of a circle of radius 12cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle [Use π=3.14,√3=1.73]
Sol :
Given: Radius of the circle = OA = OB = 12cm
and θ = 120°
To find: Area of the corresponding segment of the circle
i.e. Area of segment APB = Area of sector OAPB – Area of ΔAOB
So, firstly we find the Area of sector OAPB
Area of minor sector $=\frac{\theta}{360} \times \pi r^{2}$
$=\frac{120}{360} \times 3.14 \times(12)^{2}$
$=\frac{1}{3} \times 3.14 \times 12 \times 12$
= 150.72cm2Now, we have to find the area of ΔAOB
We draw OM ⊥ AB
∴∠OMB = ∠OMA = 90°
In ΔOMA and ΔOMB
∠OMA = ∠OMB [both 90°]
OA = OB [both radius]
OM = OM [common]
∴ OMA ≅ ΔOMB [by RHS congruency]
⇒ ∠AOM = ∠BOM [CPCT]
$\therefore \angle \mathrm{AOM}=\angle \mathrm{BOM}=\frac{1}{2} \angle \mathrm{BOA}$
$=\frac{1}{2} \times 120=60^{\circ}$
∴In right triangle OMA, we have
$\sin 60^{\circ}=\frac{\mathrm{AM}}{\mathrm{OA}}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{\mathrm{AM}}{12}$
⇒ AM = 6√3 cm⇒ 2AM =12√3 cm
⇒ AB =12√3 cm
and
$\cos 60^{\circ}=\frac{\mathrm{OM}}{\mathrm{OA}}$
$\Rightarrow \frac{1}{2}=\frac{\mathrm{OM}}{12}$
⇒ OM = 6cm$\therefore$ Area of $\Delta \mathrm{AOB}=\frac{1}{2} \times \mathrm{AB} \times \mathrm{OM}$
$=\frac{1}{2} \times 12 \sqrt{3} \times 6$
= 36√3
= 36 × 1.73
= 62.28 cm2
Area of segment APB = Area of sector OAPB – Area of ΔAOB
= (150.72 – 62.28)
= 88.44 cm2
Q25 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 25
A brooch is made with silver wire in the form of a circle with diameter 35mm. 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
Sol :
Here, brooch is made up of silver wire in the form of a circle.
The diameter of the brooch = 35mm
"⇒ The radius of the brooch, r" $=\frac{35}{2} \mathrm{mm}$
Since the wire is used in making 5 diameters and circle
So,
The total length of the silver wire required = length of wire in circle + wire used in 5 diameters
= 2πr + 5 × 2r
= 2r(π + 5)
$=2 \times \frac{35}{2}\left(\frac{22}{7}+5\right)$
$=35\left(\frac{22+35}{7}\right)$
= 5×57= 285 mm
Now, Area of each sector of the brooch
$=\frac{1}{10} \times \Pi r^{2}$
$=\frac{1}{10} \times \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2}$
$=\frac{385}{4}$
∴Area of each sector=$=\frac{385}{4} \mathrm{mm}^{2}$Q26 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 26
A horse is tied to a pole at one corner of a square grass field of side 15 m using 10 m long rope. Find
(i) the area of that part of the field in which the horse can graze.
(ii) the decrease in the grazing area if the rope was 5 m long instead of 10m. [Use π=3.14]
Sol :(i) the area of that part of the field in which the horse can graze.
(ii) the decrease in the grazing area if the rope was 5 m long instead of 10m. [Use π=3.14]
(i)
Let ABCD be square field
and length of rope, r = 10m
We need to find the area of the field which horse can graze, i.e. the area of sector QBP
As we know that in a square all angles are of 90°
Hence, ∠QBP = 90°
Area of sector $\mathrm{QBP}=\frac{\theta}{360} \times \pi \mathrm{r}^{2}$
Hence, the area of the field which horse can graze = 78.5 m2
(ii) the decrease in the grazing area if the rope was 5 m long instead of 10m
Length of rope is decreased to 5m
Area grazed by a horse now = Area of sector HBG
Area of sector $\mathrm{HBG}=\frac{\theta}{360} \times \pi r^{2}$
$=\frac{90}{360} \times 3.14 \times 10 \times 10$
$=\frac{1}{4} \times 314$
= 78.5 m2Hence, the area of the field which horse can graze = 78.5 m2
(ii) the decrease in the grazing area if the rope was 5 m long instead of 10m
Length of rope is decreased to 5m
Area grazed by a horse now = Area of sector HBG
Area of sector $\mathrm{HBG}=\frac{\theta}{360} \times \pi r^{2}$
$=\frac{90}{360} \times 3.14 \times 5 \times 5$
$=\frac{1}{4} \times 3.14 \times 25$
= 19.625 m2
So, the decrease in the grazing area
= Area of sector QBP – Area of sector HBG
= 78.5 – 19.625
= 58.875 m2
Given: Radius of circle = 32cm
Area of design = Area of circle – Area of ΔABC
Firstly, we find the area of a circle
Area of circle = πr2
= Area of sector QBP – Area of sector HBG
= 78.5 – 19.625
= 58.875 m2
Q27 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 27
In a circular table cover of radius 32cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the given figure. Find the area of the design (shaded region)
Sol :Given: Radius of circle = 32cm
Area of design = Area of circle – Area of ΔABC
Firstly, we find the area of a circle
Area of circle = πr2
$=\frac{22}{7} \times(32)^{2}$
$=\frac{22}{7} \times 32 \times 32$
$=\frac{22528}{7} \mathrm{cm}^{2}$ …(a)
Now, we will find the area of equilateral ΔABC
Construction:
Draw OD ⊥ BC
In ΔBOD and ΔCOD
OB = OC (radii)
OD = OD (common)
∠ODB = ∠ODC (90°)
∴ ΔBOD ≅ ΔCOD [by RHS congruency]
⇒ BD = DC [by CPCT]
or BC = 2BD …(i)
and, $\angle \mathrm{BOD}=\angle \mathrm{COD}=\frac{1}{2} \angle \mathrm{BOC}=\frac{120}{2}=60^{\circ}$
Now, In ΔBOD, we have
From (i), BC = 2BD ⇒ BC = 32√3 cm
Now, Area of equilateral ΔABC
Construction:
Draw OD ⊥ BC
In ΔBOD and ΔCOD
OB = OC (radii)
OD = OD (common)
∠ODB = ∠ODC (90°)
∴ ΔBOD ≅ ΔCOD [by RHS congruency]
⇒ BD = DC [by CPCT]
or BC = 2BD …(i)
and, $\angle \mathrm{BOD}=\angle \mathrm{COD}=\frac{1}{2} \angle \mathrm{BOC}=\frac{120}{2}=60^{\circ}$
Now, In ΔBOD, we have
$\sin 60^{\circ}=\frac{\mathrm{BD}}{\mathrm{OB}}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{\mathrm{BD}}{32}$
⇒ BD = 16√3 cmFrom (i), BC = 2BD ⇒ BC = 32√3 cm
Now, Area of equilateral ΔABC
$=\frac{\sqrt{3}}{4}(\text { side })^{2}$
$=\frac{\sqrt{3}}{4} \times(32 \sqrt{3})^{2}$
= 768√3 cm2 …(b)
Therefore, Area of design = Area of circle – Area of ΔABC
$=\frac{22528}{7}-768 \sqrt{3} \mathrm{cm}^{2}$ [from (a) and (b)]
Given: PQ = 24cm and PR = 7cm
Since QR is a diameter, it forms a semicircle
We know that angle in a semicircle is a right angle.
Hence, ∠RPQ = 90°
Hence, ΔRPQ is a right triangle
In ΔRPQ, by Pythagoras theorem
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
(QR)2 = (PQ)2 + (PR)2
⇒ (QR)2 = (24)2 + (7)2
⇒ (QR)2 = 576 + 49
⇒ (QR)2 = 625
⇒ (QR)2 = (25)2
⇒ QR = 25cm
∴ Diameter, QR = 25cm
$\Rightarrow$ Radius $=\frac{25}{2}$
So,
Area of semicircle $=\frac{1}{2} \pi r^{2}$
$=\frac{1}{2} \times \frac{22}{7} \times \frac{25}{2} \times \frac{25}{2}$
$=\frac{11 \times 25 \times 25}{28}$
$=\frac{6875}{28} \mathrm{cm}^{2}$
Now, Area of ΔPQR
Area of $\Delta \mathrm{PQR}=\frac{1}{2} \times \mathrm{b} \times \mathrm{h}$
$=\frac{1}{2} \times \mathrm{PQ} \times \mathrm{PR}$
$=\frac{1}{2} \times 24 \times 7$
= 12 × 7
= 84cm2
Area of shaded region = Area of semicircle – Area of ΔPQR
$=\frac{6875}{28}-84$
$=\frac{6875-2352}{28}$
$=\frac{4523}{28} \mathrm{cm}^{2}$
Area of the shaded region
= Area of sector AOB – Area of sector COD
Area of sector AOB
Here, radius = 21cm and θ = 30°
Area of minor sector $=\frac{\theta}{360} \times \pi r^{2}$
$=\frac{30}{360} \times \frac{22}{7} \times 21 \times 21$
$=\frac{1}{12} \times 22 \times 3 \times 21$
$=\frac{231}{2} \mathrm{cm}^{2}$
Area of sector COD
Here, radius = 7cm and θ = 30°
Area of minor sector $=\frac{\theta}{360} \times \pi r^{2}$
$=\frac{30}{360} \times \frac{22}{7} \times 7 \times 7$
$=\frac{1}{12} \times 22 \times 21$
$=\frac{77}{6} \mathrm{cm}^{2}$
Now, shaded region = Area of sector AOB – Area of sector COD
$=\frac{231}{2}-\frac{77}{6}$
$=\frac{693-77}{6}$
$=\frac{616}{6}$
$=\frac{308}{3} \mathrm{cm}^{2}$
Hence, area of shaded region is $\frac{308}{3}$ cm2
Area of shaded region = Area of square ABCD– Area of semicircle APD– Area of semicircle BPC
Area of square ABCD
Given: Side of square = 14cm
Area of square = Side × Side
= 14 × 14
= 196 cm2
Area of semicircle APD
Diameter = AD = 14cm
So,
Radius $=\frac{14}{2}=7 \mathrm{cm}$
Area of semicircle $\mathrm{APD}=\frac{1}{2} \pi \mathrm{r}^{2}$
$=\frac{1}{2} \times \frac{22}{7} \times 7 \times 7$
= 11 × 7
= 77cm2
Similarly, Area of semicircle BPC = 77cm2
Area of shaded region = Area of square ABCD– Area of semicircle APD– Area of semicircle BPC
= 196 – 77 -77
= 42cm2
Hence, area of shaded region is 42cm2
Area of shaded region = Area of quadrant OBPQ – Area of square OABC
Area of square OABC
Given: Side of square = 20cm
Area of square = Side × Side
= 20 × 20
= 400 cm2
Area of quadrant
We need to find the radius
Joining OB
Also, all angles of a square are 90°
∴∠BAO = 90°
Hence, ΔOBA is a right triangle
In ΔOBA, by Pythagoras Theorem
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
(OB)2 = (AB)2 + (OA)2
⇒ (OB)2 = (20)2 + (20)2
⇒ (OB)2 = 400 + 400
⇒ (OB)2 = 800
⇒ OB = √(10×10×2×2×2)
⇒ OB = 20√2cm
Area of quadrant $=\frac{1}{4} \pi r^{2}$
$=\frac{1}{4} \times 3.14 \times(20 \sqrt{2})^{2}$
$=3.14 \times 5 \times 20 \times 2$
= 628 cm2
Area of shaded region = Area of quadrant OBPQ– Area of square OABC
= 628 – 400
= 228cm2
ABCD is a square lawn of side 58m. AED and BFC are two circular ends.
Now, diagonal of the lawn = √(58)2 + (58)2 = 58√2m
It is given that diagonal of square = Diameter of circle
∴The radius of a circle having a centre at the point of intersection of diagonal
$=\frac{58 \sqrt{2}}{2}=29 \sqrt{2} \mathrm{m}$
$=\frac{22528}{7}-768 \sqrt{3} \mathrm{cm}^{2}$ [from (a) and (b)]
Q28 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 28
Find the area of the shaded region in the figure, if PQ=24 cm, PR = 7cm and O is the centre of the circle.
Sol :Given: PQ = 24cm and PR = 7cm
Since QR is a diameter, it forms a semicircle
We know that angle in a semicircle is a right angle.
Hence, ∠RPQ = 90°
Hence, ΔRPQ is a right triangle
In ΔRPQ, by Pythagoras theorem
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
(QR)2 = (PQ)2 + (PR)2
⇒ (QR)2 = (24)2 + (7)2
⇒ (QR)2 = 576 + 49
⇒ (QR)2 = 625
⇒ (QR)2 = (25)2
⇒ QR = 25cm
∴ Diameter, QR = 25cm
$\Rightarrow$ Radius $=\frac{25}{2}$
So,
Area of semicircle $=\frac{1}{2} \pi r^{2}$
$=\frac{1}{2} \times \frac{22}{7} \times \frac{25}{2} \times \frac{25}{2}$
$=\frac{11 \times 25 \times 25}{28}$
$=\frac{6875}{28} \mathrm{cm}^{2}$
Now, Area of ΔPQR
Area of $\Delta \mathrm{PQR}=\frac{1}{2} \times \mathrm{b} \times \mathrm{h}$
$=\frac{1}{2} \times \mathrm{PQ} \times \mathrm{PR}$
$=\frac{1}{2} \times 24 \times 7$
= 12 × 7
= 84cm2
Area of shaded region = Area of semicircle – Area of ΔPQR
$=\frac{6875}{28}-84$
$=\frac{6875-2352}{28}$
$=\frac{4523}{28} \mathrm{cm}^{2}$
Q29 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 29
AB and CD are arcs of two concentric circles of radii 21 cm and 7 cm respectively and centre O. If ∠AOB =30°, find the area of the shaded region.
Sol :Area of the shaded region
= Area of sector AOB – Area of sector COD
Area of sector AOB
Here, radius = 21cm and θ = 30°
Area of minor sector $=\frac{\theta}{360} \times \pi r^{2}$
$=\frac{30}{360} \times \frac{22}{7} \times 21 \times 21$
$=\frac{1}{12} \times 22 \times 3 \times 21$
$=\frac{231}{2} \mathrm{cm}^{2}$
Area of sector COD
Here, radius = 7cm and θ = 30°
Area of minor sector $=\frac{\theta}{360} \times \pi r^{2}$
$=\frac{30}{360} \times \frac{22}{7} \times 7 \times 7$
$=\frac{1}{12} \times 22 \times 21$
$=\frac{77}{6} \mathrm{cm}^{2}$
Now, shaded region = Area of sector AOB – Area of sector COD
$=\frac{231}{2}-\frac{77}{6}$
$=\frac{693-77}{6}$
$=\frac{616}{6}$
$=\frac{308}{3} \mathrm{cm}^{2}$
Hence, area of shaded region is $\frac{308}{3}$ cm2
Q30 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 30
In the given figure ABCD is a square whose each side is 14cm. APD and BPC are semicircles. Find the area of the shaded region.
Sol :Area of shaded region = Area of square ABCD– Area of semicircle APD– Area of semicircle BPC
Area of square ABCD
Given: Side of square = 14cm
Area of square = Side × Side
= 14 × 14
= 196 cm2
Area of semicircle APD
Diameter = AD = 14cm
So,
Radius $=\frac{14}{2}=7 \mathrm{cm}$
Area of semicircle $\mathrm{APD}=\frac{1}{2} \pi \mathrm{r}^{2}$
$=\frac{1}{2} \times \frac{22}{7} \times 7 \times 7$
= 11 × 7
= 77cm2
Similarly, Area of semicircle BPC = 77cm2
Area of shaded region = Area of square ABCD– Area of semicircle APD– Area of semicircle BPC
= 196 – 77 -77
= 42cm2
Hence, area of shaded region is 42cm2
Q31 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 31
In the given figure a square OABC is inscribed in a quadrant OPBQ. If OA =20cm, find the area of the shaded region.
Sol :Area of shaded region = Area of quadrant OBPQ – Area of square OABC
Area of square OABC
Given: Side of square = 20cm
Area of square = Side × Side
= 20 × 20
= 400 cm2
Area of quadrant
We need to find the radius
Joining OB
Also, all angles of a square are 90°
∴∠BAO = 90°
Hence, ΔOBA is a right triangle
In ΔOBA, by Pythagoras Theorem
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
(OB)2 = (AB)2 + (OA)2
⇒ (OB)2 = (20)2 + (20)2
⇒ (OB)2 = 400 + 400
⇒ (OB)2 = 800
⇒ OB = √(10×10×2×2×2)
⇒ OB = 20√2cm
Area of quadrant $=\frac{1}{4} \pi r^{2}$
$=\frac{1}{4} \times 3.14 \times(20 \sqrt{2})^{2}$
$=3.14 \times 5 \times 20 \times 2$
= 628 cm2
Area of shaded region = Area of quadrant OBPQ– Area of square OABC
= 628 – 400
= 228cm2
Q32 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 32
It is proposed to add to a square lawn with the side 58m, two circular ends (the centre of each circle being the point of intersection of the diagonals of the square.) Find the area of the whole lawn
[Take π=3.14]
Sol :ABCD is a square lawn of side 58m. AED and BFC are two circular ends.
Now, diagonal of the lawn = √(58)2 + (58)2 = 58√2m
It is given that diagonal of square = Diameter of circle
∴The radius of a circle having a centre at the point of intersection of diagonal
$=\frac{58 \sqrt{2}}{2}=29 \sqrt{2} \mathrm{m}$
It is given that square ABCD is inscribed by the circle with centre O.
∴Area of 4 segments = Area of circle – Area of square
= πr2 – (side)2
$=\frac{22}{7} \times 29 \sqrt{2} \times 29 \sqrt{2}-58 \times 58$
$=\frac{22}{7} \times 29 \times 29 \times 2-29 \times 29 \times 4$
$=29 \times 29 \times 4\left(\frac{11}{7}-1\right)$
$=29 \times 29 \times 4 \times\left(\frac{4}{7}\right)$
$\Rightarrow$ Area of two segments $=\frac{1}{2} \times 29 \times 29 \times 4 \times \left(\frac{4}{7}\right)$m2
$=\frac{29 \times 29 \times 8}{7}$
= 961.14m2
Area of whole lawn = Area of circle – Area of two segments
$=\frac{22}{7} \times 29 \sqrt{2} \times 29 \sqrt{2}-\frac{29 \times 29 \times 8}{7}$
=5286.28 – 961.14
= 4325.14 m2
Q33 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 33
Find the difference between the area of a regular hexagonal plot each of whose side is 72 m and the area of the circular swimming tank inscribed in it. $\left(\right.$ Take $\left.\pi=\frac{22}{7}\right)$
Sol :
The side of hexagonal plot = 72m
Area of equilateral triangle $\mathrm{OAB}=\frac{\sqrt{3}}{4}(\text { side })^{2}$
$=\frac{\sqrt{3}}{4} \times 72 \times 72$
= 1296√3cm2
∴ Area of hexagonal plot = 6 × Area of triangle OAB
= 6 × 1296√3
= 7776(1.732)
=13468.032m2
In ΔOCA, by Pythagoras theorem
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
(OA)2 = (OC)2 + (AC)2
⇒ (72)2 = (OC)2 + (36)2
⇒ (OC)2 = 5184 – 1296
⇒ (OC)2 = 3888
⇒ r2 = 3888
∴Area of inscribed circular swimming tank = πr2
$=\frac{22}{7} \times 3888$
= 12219.429m2
∴ Required difference = 13468.032 – 12219.429
= 1248.603m2
Hence, the difference between the area of a regular hexagonal plot and the area of the circular swimming tank inscribed in it is 1248.60.3m2
Q34 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 34
In the figure, ABC is a quadrant of a circle of radius 14cm, and a semicircle is drawn with BC as diameter. Find the area of the shaded region.
Sol :Area of shaded region = Area of segment with chord BC
Now, AC = BA = 14cm
∴ BC = √(14)2 + (14)2 = 14√2 cm
$\therefore$ Area of semicircle on $\mathrm{BC}=\frac{1}{2} \pi \mathrm{r}^{2}$
$=\frac{1}{2} \times \frac{22}{7} \times 14 \sqrt{2} \times 14 \sqrt{2}$
= 154cm2
Area of $\triangle \mathrm{ABC}=\frac{1}{2} \times \mathrm{b} \times \mathrm{h}$
$=\frac{1}{2} \times 14 \times 14$
= 98
So, Area of shaded region = 154 – 98
= 56cm2
Q35 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 35
The area of an equilateral triangle is 100√3cm2. Taking each vertex as centre, a circle is described with a radius equal to half the length of the side of the triangle, as shown in the figure. Find the area of that part of the triangle which is not included in the circles
[Take π=3.14 and √3=1.732]
Area of the shaded region
= Area of an equilateral triangle – Area of 3 sectors
Given: Area of equilateral ΔABC = 100√3cm2
$\Rightarrow \frac{\sqrt{3}}{4} \mathrm{a}^{2}=100 \sqrt{3}$
⇒ a2 = 400
⇒ a = 20cm
It is given that radius is equal to half the length of the side
i.e. $\mathrm{r}=\frac{\mathrm{a}}{2}=\frac{20}{2}=10 \mathrm{cm}$
Now,
Area of 3 sectors $=3 \times \frac{\theta}{360} \times \pi r^{2}$
Area of 3 sectors $=3 \times \frac{60}{360} \times \frac{22}{7} \times 10 \times 10$
$=\frac{11}{7} \times 100$
= 157.14cm2
Hence, the area of the shaded region
= Area of ΔABC – Area of 3 sectors
= 100√3 – 157.14
= 100×1.732 – 157.14
=173.2 – 157.14
= 16.06 cm2
Q36 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 36
In an equilateral triangle of side 12cm, a circle is inscribed touching its sides. Find the area of the portion of the triangle not included in the circle.
[Take √3=1.73 and π=3.14]
Sol :
Area of shaded region = Area of ΔABC – Area of circle
Given side of triangle = 12cm
$\therefore$ Area of equilateral triangle $=\frac{\sqrt{3}}{4}(\text { side })^{2}$
$=\frac{\sqrt{3}}{4} \times 12 \times 12$
= 36√3cm2
Now, we have to find the area of a circle. For that we need a radius.
Draw AD ⊥ BC
So, In BDO
$\tan 30^{\circ}=\frac{\mathrm{OD}}{\mathrm{BD}}$
$=\frac{\sqrt{3}}{4} \times 12 \times 12$
= 36√3cm2
Now, we have to find the area of a circle. For that we need a radius.
Draw AD ⊥ BC
So, In BDO
$\tan 30^{\circ}=\frac{\mathrm{OD}}{\mathrm{BD}}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{r}}{6}$ $\left[\because \mathrm{AD} \perp \mathrm{BC} \Rightarrow \mathrm{BD}=\frac{1}{2} \mathrm{BC}=6 \mathrm{cm}\right]$
⇒ r = 2√3cm
Now, Area of circle = πr2
= 3.14 × (2√3)2
= 37.68cm2
Area of shaded region = Area of ΔABC – Area of circle
= 36√3 – 37.68
= 36(1.73) – 37.68
= 24.6cm2
Hence, the area of the portion of the triangle not included in the circle is 24.6cm2
Given: Radius of circle = 16cm
Area of shaded region = Area of circle – Area of ΔABC
Firstly, we find the area of a circle
⇒ r = 2√3cm
Now, Area of circle = πr2
= 3.14 × (2√3)2
= 37.68cm2
Area of shaded region = Area of ΔABC – Area of circle
= 36√3 – 37.68
= 36(1.73) – 37.68
= 24.6cm2
Hence, the area of the portion of the triangle not included in the circle is 24.6cm2
Q37 | Ex-13 |Class 10 | Area related to circle | KC Sinha Mathematics | Chapter 13| myhelper
Question 37
In a circular table-cover of radius 16cm, 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 in the figure).
Sol :Given: Radius of circle = 16cm
Area of shaded region = Area of circle – Area of ΔABC
Firstly, we find the area of a circle
Area of circle = πr2
$=\frac{22}{7} \times(16)^{2}$
$=\frac{22}{7} \times 16 \times 16$
$=\frac{5632}{7} \mathrm{cm}^{2}$ …(a)
Now, we will find the area of equilateral ΔABC
Construction:
Draw OD ⊥ BC
In ΔBOD and ΔCOD
OB = OC (radii)
OD = OD (common)
∠ODB = ∠ODC (90°)
∴ ΔBOD ≅ ΔCOD [by RHS congruency]
⇒ BD = DC [by CPCT]
or BC = 2BD …(i)
$=\frac{22}{7} \times(16)^{2}$
$=\frac{22}{7} \times 16 \times 16$
$=\frac{5632}{7} \mathrm{cm}^{2}$ …(a)
Now, we will find the area of equilateral ΔABC
Construction:
Draw OD ⊥ BC
In ΔBOD and ΔCOD
OB = OC (radii)
OD = OD (common)
∠ODB = ∠ODC (90°)
∴ ΔBOD ≅ ΔCOD [by RHS congruency]
⇒ BD = DC [by CPCT]
or BC = 2BD …(i)
and, $\angle \mathrm{BOD}=\angle \mathrm{COD}=\frac{1}{2} \angle \mathrm{BOC}=\frac{120}{2}=60^{\circ}$
Now, In ΔBOD, we have
$\sin 60^{\circ}=\frac{\mathrm{BD}}{\mathrm{OB}}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{\mathrm{BD}}{16}$
⇒ BD = 8√3 cm
From (i), BC = 2BD ⇒ BC = 16√3 cm
Now, Area of equilateral ΔABC
$=\frac{\sqrt{3}}{4}(\text { side })^{2}$
$=\frac{\sqrt{3}}{4} \times(16 \sqrt{3})^{2}$
= 192√3 cm2 …(b)
Therefore, Area of design = Area of circle – Area of ΔABC
$=\frac{5632}{7}-192 \sqrt{3} \mathrm{cm}^{2}$ [from (a) and (b)]
= 804.57 – 332.544
= 472.03cm2
Now, In ΔBOD, we have
$\sin 60^{\circ}=\frac{\mathrm{BD}}{\mathrm{OB}}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{\mathrm{BD}}{16}$
⇒ BD = 8√3 cm
From (i), BC = 2BD ⇒ BC = 16√3 cm
Now, Area of equilateral ΔABC
$=\frac{\sqrt{3}}{4}(\text { side })^{2}$
$=\frac{\sqrt{3}}{4} \times(16 \sqrt{3})^{2}$
= 192√3 cm2 …(b)
Therefore, Area of design = Area of circle – Area of ΔABC
$=\frac{5632}{7}-192 \sqrt{3} \mathrm{cm}^{2}$ [from (a) and (b)]
= 804.57 – 332.544
= 472.03cm2
| S.no | Chapters | Links |
|---|---|---|
| 1 | Real numbers | Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 |
| 2 | Polynomials | Exercise 2.1 Exercise 2.2 Exercise 2.3 |
| 3 | Pairs of Linear Equations in Two Variables | Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 |
| 4 | Trigonometric Ratios and Identities | Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4 |
| 5 | Triangles | Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5 |
| 6 | Statistics | Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4 |
| 7 | Quadratic Equations | Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5 |
| 8 | Arithmetic Progressions (AP) | Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4 |
| 9 | Some Applications of Trigonometry: Height and Distances | Exercise 9.1 |
| 10 | Coordinates Geometry | Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4 |
| 11 | Circles | Exercise 11.1 Exercise 11.2 |
| 12 | Constructions | Exercise 12.1 |
| 13 | Area related to Circles | Exercise 13.1 |
| 14 | Surface Area and Volumes | Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4 |
| 15 | Probability | Exercise 15.1 |
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