Rd Sharma 2022 Solutions for Class 10 Maths Chapter 13 Areas Related To Circles are provided here with simple step-by-step explanations. These solutions for Areas Related To Circles are extremely popular among Class 10 students for Maths Areas Related To Circles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2022 Book of Class 10 Maths Chapter 13 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2022 Solutions. All Rd Sharma 2022 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.
Page No 575:
Question 1:
Find the circumference and area of a circle of radius 4.2 cm.
Answer:
We know that the circumference C and area A of a circle of radius r are given by and respectively.
Here,
So substituting the value of r in above formulas,
Circumference of the circle
Area of the circle
Page No 575:
Question 2:
Find the circumference of a circle whose area is 301.84 cm2.
Answer:
Let r cm be the radius of the circle. Then
Area of a circle is
We know that the Circumference of circle of radius r is
So substituting the value of r in above formula
Page No 575:
Question 3:
Find the area of a circle whose circumference is 44 cm.
Answer:
Let r be the radius of the circle. Then
Circumference of the circle
We know that the area of a circle of radius r is
Substituting the value of r in above formula
Page No 575:
Question 4:
A road which is 7 m wide surrounds a circular park whose circumference is 352 m . Find the area of the road .
Answer:
Width of the road = 7 m
circumference of the circular park = 352 m
Area of the road = Area of the circular park including the path − area of the circular park
Page No 575:
Question 5:
The circumference of two circles are in the ratio 2 : 3. Find the ratio of their areas.
Answer:
Let the radius of two circles beandrespectively. Then their circumferences are and respectively and their areas are and respectively.
It is given that,
Now we will calculate the ratio of their areas,
Substituting the value of ,
Hence the ratio of their Areas is.
Page No 575:
Question 6:
The sum of the radii of two circles is 140 cm and the difference of their circumferences is 88 cm. Find the diameters of the circles.
Answer:
Let the radius of two circles beandrespectively. Then their circumferences are and respectively and their areas are and respectively.
It is given that the sum of the radii of two circles is 140 cm and difference of their circumferences is 88 cm. So,
……(A)
……(B)
Now, solving (A) and (B)
Thus diameters of circles are,
Page No 575:
Question 7:
The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having its area equal to the sum of the areas of the two circles.
Answer:
Let the radius of circles be , andrespectively. Then their areas are, and respectively.
It is given that,
We have, and
Substituting the values of ,
Hence, the radius of circle is.
Page No 575:
Question 8:
The radii of two circles are 19 cm and 9 cm respectively. Find the radius and area of the circles which has it circumference equal to the sum of the circumferences of the two circles.
Answer:
Let the radius of circles be, andrespectively. Then their circumferences are, and respectively.
It is given that,
We have, and
Substituting the values of,
Hence the radius of the circle is.
We know that the area A of circle is
Substituting the value of r
Hence the area of the circle is.
Page No 575:
Question 9:
The outer circumference of a circular race-track is 528 m . The track is everywhere 14 m wide. Calculate the cost of levelling the track at the rate of 50 paise per square metre .
Answer:
Let the radius of the inner circle and the race track be R m.
Outer circumference of the race track = 528 m
Total radius of the outer circle = 84 − 14 = 70 m
Area of the circular track = Area of the outer circle − area of inner circle
Cost of levelling the track =
Page No 575:
Question 10:
A circular park is surroundeed by a rod 21 m wide. If the radius of the park is 105 m, find the area of the road .
Answer:
Radius of the park, r = 105 m
Radius of the park with the road, R = 105 + 21 = 126 m
Area of the road = Area of the park with the road − Area of the park
Page No 575:
Question 11:
A circle of radius 7.5 cm is inscribed in a square. Find the area outside the circle and inside the square.
Answer:
Given that, a circle of radius 7.5 cm is inscribed in a square.
Area of circle =
Since, the side of square = diameter of circle = 15 cm
Area of square =
Thus, the area outside the circle and inside the square is .
Hence, the required area is 48.375 cm2.
Page No 576:
Question 12:
The area of a circle inscribed in an equilateral triangle is 154 cm2. Find the perimeter of the triangle. [Use π = 22/7 and = 1.73]
Answer:
It is given that the area A of circle inscribed in an equilateral triangle is 154 cm2.
We know that the area A of circle inscribed in an equilateral triangle is
Now, we will find the value of r.
Substituting the value of area,
Let the height of triangle be h. Then
If a is the side of triangle, then
Substituting the value of h,
Hence perimeter of triangle is .
Page No 576:
Question 13:
A car travels 1 kilometre distance in which each wheel makes 450 complete revolutions. Find the radius of the its wheels.
Answer:
Let the radius of wheel be r. Thus, circumference C of the wheel
Since car travels 1 km distance in which wheel makes 450 complete revolutions. Then
We know that,
Hence the radius of wheel is.
Page No 576:
Question 14:
A square of diagonal 8 cm is inscribed in a circle. Find the area of the region lying outside the circle and inside the square .
Answer:
DISCLAIMER: There is some error in the given question.
We have solved the question by taking a square inscribed in a circle. Then finding the area inside the circle and outside the square.
Diagonal of the square = 8 cm.
Let the side of the square be a cm.
In triangle BCD,
Radius of the circle will be R = 4 cm
Now area between the circle and the square will be
Page No 576:
Question 15:
A path of 4 m width runs round a semi-circular grassy plot whose circumference is . Find:
(i) the area of the path
(ii) the cost of gravelling the path at the rate of ₹ 1.50 per square metre
(iii) the cost of turfing the plot at the rate of 45 paise per m2.
Answer:
We have given AB = 4m and circumference of semicircle with radius OA as.
We are asked to find the area between the two semi-circles.
For that we will first find OA.
Now we will substitute
Now we will find OB.
Now we will find the area between two semi-circles as given below,
Therefore, area of the path is 339.43 m2.
Now we will find the cost of gravelling the path.
Therefore, cost of gravelling the path is Rs 509.14.
Now we will find the cost of turfing the plot. For that we will find the area of the plot.
Therefore, cost of the turfing the plot is Rs 1912.11.
Disclaimer: Due to some error in the question, we get the different answers.
Page No 576:
Question 16:
A path of width 3.5 m runs around a semi-circular grassy plot whose perimeter is 72 m . Find the area of the path .
Answer:
Let the radius of the semicircular plot be r.
Perimeter of the semi-circular grassy plot =
Given that the width of the plot = 3.5 m
Thus, the outer radius = 3.5 + 14 = 17.5 m
Area of the path =
Page No 576:
Question 17:
Find the area enclosed between two concentric circles of radii 3.5 cm and 7 cm. A third concentric circle is drawn outside the 7 cm circle , such that the area enclosed between it and the 7 cm circle is same as that between the two inner circles . Find the radius of the third circle correct to one decimal place.
Answer:
The area enclosed between the two circles of radii 3.5 cm and 7 cm
Let the radius of the outermost circle be r cm.
Area betweent the circles with radius r and 7 cm=Area between the circles with radius 7 cm and 3.5 cm
Page No 576:
Question 18:
An archery target has three regions formed by three concentric circles as shown in figure15.8. If the diameters of the concentric circles are in the ratios 1 : 2 : 3 , then find the ratio of the areas of three regions .
Answer:
Let the three regions be A , B and C.
The diameters are in the ratio 1 : 2 : 3.
Let the diameters be 1x, 2x and 3x
Then the radius will be
Area of region A =
Area of region B =
Area of region C =
Thus, ratio of the areas of regions A, B and C will be
::
⇒1:3:5
Page No 585:
Question 1:
Find, in terms of π, the length of the arc that subtends an angle of 30° at the centre of a circle of radius 4 cm.
Answer:
The arc length l of a sector of an angle θ in a circle of radius r is given by
It is given that and. Substituting the value of r and θ in above equation,
Page No 585:
Question 2:
Find the angle subtended at the centre of a circle of radius 5 cm by an arc of length (5π/3) cm.
Answer:
We know that the arc length l of a sector of an angle θ in a circle of radius r is
It is given that and length . Substituting these value in above equation,
Hence, the angle subtended at the centre of circle is.
Page No 585:
Question 3:
An arc of length 20π cm subtends an angle of 144° at the centre of a circle. Find the radius of the circle.
Answer:
We know that the arc length l of a sector of an angle θ in a circle of radius r is
It is given and angle.
Now we substitute the value of l and θ in above formula to find the value of radius r of circle.
Page No 585:
Question 4:
An arc of length 15 cm subtends an angle of 45° at the centre of a circle. Find in terms of π, the radius of the circle.
Answer:
We know that the arc length l of a sector of an angle θ in a circle of radius r is
It is given that and angle.
Now we substitute the value of l and θ in above formula to find the value of radius r of circle.
Page No 585:
Question 5:
A sector of a circle of radius 4 cm contains an angle of 30°. Find the area of the sector.
Answer:
We know that the area A of a sector of an angle θ in the circle of radius r is given by
It is given that and angle.
Now we substitute the value of r and θ in above formula,
Page No 585:
Question 6:
The area of sector is one-twelfth that of the complete circle. Find the angle of the sector .
Answer:
We know, .....(i)
Let the area of the circle be Ar.
Thus area of the sector = .....(ii)
From (i) and (ii) we have
Page No 585:
Question 7:
In a circle of radius 35 cm, an arc subtends an angle of 72° at the centre. Find the length of the arc and area of the sector.
Answer:
We know that the arc length l and area A of a sector of an angle θ in the circle of radius r is given by and respectively.
It is given that, and.
We will calculate the arc length using the value of r and θ,
Now, we will find the value of area A of the sector
Page No 585:
Question 8:
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Answer:
Angle make by the minute hand in 1 minute = 6∘
Angle make by the minute hand in 5 minute = 5 ⨯ 6∘ = 30∘
Area of the sector having central angle is given by
Hence, the area swept by minute hand in 5 minutes is 51.33 cm2
Page No 585:
Question 9:
A sector is cut-off from a circle of radius 21 cm. The angle of the sector is 120°. Find the length of its arc and the area.
Answer:
We know that the arc length l and area A of a sector of circle at an angle θ of radius r is given by and angle.
Let OAB is the given sector.
It is given that and angle.
Now using the value of r and θ, we will find the value of l and A,
Arc length,
Area of sector,
Page No 585:
Question 10:
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find (i) the length of the arc (ii) area of the sector formed by the arc. (Use π = 22/7)
Answer:
Here, we have θ = 60° and r = 21 cm
(i) The length of the arc is given by
(ii) Area of the sector formed by the arc is given by
Page No 585:
Question 11:
The length of minute hand of a clock is 5 cm . Find the area swept by the minute hand during the time period 6:05 am and 6:40 am.
Answer:
Minute hand of the clock describes a circle of radius equal to its length i.e 5 cm.
The minute hand rotates through 6o in one minute.
So, the minute hand will sweep 210o in one minute.
Area swept by the minute hand in one minute is the area of a sector of angle 6o in a circle of radius 5 cm.
Area swept by the minute hand in 35 minutes is the area of a sector of angle 210o in a circle of radius 5 cm
Page No 585:
Question 12:
The perimeter of a sector of a circle of radius 5.7 m is 27.2 m. Find the area of the sector.
Answer:
We know that the area A of a sector of circle of radius r and arc length l is given by
Let OAB is the given sector. Then,
arc AB = 15.8 m
So, l = 15.8 m
Now substituting the value of r and l in above formula,
Page No 585:
Question 13:
AB is a chord of a circle with centre O and radius 4 cm. AB is of length 4 cm. Find the area of the sector of the circle formed by chord AB.
Answer:
We have to find the area of the sector AOB formed by the chord AB.
We have and. So,
Let. Then,
In, we have
Hence,
Now, using the value of and r we will find the area of sector AOB,
Page No 585:
Question 14:
In a circle of radius 6 cm, a chord of length 10 cm makes an angle of 110° at the centre of the circle. Find:
(i) the circumference of the circle
(ii) the area of the circle
(iii) the length of the arc AB,
(iv) the area of the sector OAB.
Answer:
It is given that the radius of circle, and angle at the centre of circle.
(i) We know that the Circumference C of circle of radius r is,
(ii) We know that the Area A of circle of radius r is,
(iii) We know that the arc length l of a sector of an angle θ in a circle of radius r is
(iv) We know that the area A of a sector of an angle θ in the circle of radius r is given by
Page No 586:
Question 15:
In the following figure, shows a sector of a circle, centre O, containing an angle 0°. Prove that:
(i) Perimeter of the shaded region is
(ii) Area of the shaded region is
Answer:
It is given that the radius of circle is r and the angle.
In,
It is given that.
(i) We know that the arc length l of a sector of an angle θ in a circle of radius r is
Now we substitute the value of OC, OA and l to find the perimeter of sector AOC,
Hence,
(ii) We know that area A of the sector at an angle θ in the circle of radius r is
.
Thus
Hence,
Page No 586:
Question 16:
Figure 15.18 shows a sector of a circle of radius r cm containing an angle . The area of the sector is A cm2 and perimeter of the sector is 50 cm. Prove that
(i)
(ii) A = 25r − r2
Answer:
It is given that the radius of circle is r cm and angle.
(i) We know that the arc length l of a sector of an angle θ in a circle of radius r is
Now we substitute the value of OB, OA and l to find the perimeter of sector AOB,
(ii) We know that area A of the sector at an angle θ in the circle of radius r is
. Thus
Substituting the value of θ,
Page No 590:
Question 1:
A chord PQ of length 12 cm subtends an angle of 120° at the centre of a circle. Find the area of the minor segment cut off by the chord PQ.
Answer:
We know that the area of minor segment of angle θ in a circle of radius r is,
It is given that the chord PQ divides the circle in two segments.
We have and. So,
Since,
In, we have
Thus the radius of circle is .
Now using the value of radius r and angle θ we will find the area of minor segment
Page No 590:
Question 2:
A chord AB of a circle, of radius 14 cm makes an angle of 60° at the centre of the circle. Find the area of the minor segment of the circle. (Use π = 22/7)
Answer:
We know that the area of minor segment of angle θ in a circle of radius r is,
It is given that,
Substituting these values in above formula
Page No 590:
Question 3:
A chord of a circle of radius 20 cm sub tends an angle of 900 at the centre . Find the area of the corresponding major segment of the circle
( Use )
Answer:
We know area of minor segment of the circle is
Area of the major segment = Area of the circle − area of the minor segment
DISCLAIMER: The answer given in the book is incorrect. If we take the radius 10 instead of 20, then the answer will match.
Page No 590:
Question 4:
A chord 10 cm long is drawn in a circle whose radius is 5 cm. Find area of both the segments.
Answer:
We know that the area of minor segment of angle θ in a circle of radius r is,
It is given that the chord AB divides the circle in two segments.
We have and. So,
Let. Then,
In, we have
Hence,
Now using the value of r and θ, we will find the area of minor segment
Page No 590:
Question 5:
The radius of a circle with centre O is 5 cm in the given figure. Two radii OA and OB are drawn at right angles to each other. Fin the areas of the segments made by the chord AB (Take π = 3.14).
Answer:
We know area of minor segment of the circle is
Area of the major segment = Area of the sector − area of minor segment
Page No 590:
Question 6:
In the given figure, AB is the diameter of a circle, centre O, C is a point on the circumference such that ∠COB = . The area of the minor segment cut off by AC is equal to twice the area of the sector BOC. Prove that
Answer:
We know that the area of minor segment of angle θ in a circle of radius r is,
.
It is given that,
So,
Area, A of minor segment cutoff by AC at angle
Now, since and
We know that the area of sector of a circle of radius r at an angle θ is
So, the area of sector BOC,
It is given that,
Page No 590:
Question 7:
A chord of a circle subtends an angle of at the centre of the circle. The area of the minor segment cut off by the chord is one eighth of the area of the circle. Prove that
Answer:
We know that the area of circle and area of minor segment of angle θ in a circle of radius r is given by, andrespectively.
It is given that,
Page No 604:
Question 1:
A plot is in the form of a rectangle ABCD having semi-circle on BC as shown in the following figure. If AB = 60 m and BC = 28 m, find the area of the plot.
Answer:
It is given that a plot is in form of rectangle ABCD having a semicircle on BC.
Since BC is the diameter of semicircle. Then, radius of semicircle is
=308 m2
= 1680 m2
Now,
Area of plot = Area of rectangle + Area of semicircle
= 1680 + 308 m2
= 1988 m2
Page No 604:
Question 2:
In the following figure, PQRS is a square of side 4 cm. Find the area of the shaded square.
Answer:
Area of the shaded region is equal to the area of the square minus area of four sectors with the same radius minus area of the circle.
We know that area of the four sectors with radius is equal to area of one circle.
Therefore, area of the shaded region is
Page No 604:
Question 3:
Four cows are tethered at four corners of a square plot of side 50 m, so that they just cannot reach one another. What area will be left ungrazed? (Fig. 15.66)
Answer:
It is given that four cows are tethered at four corner of square ABCD. We have to find the area of plot that will left ungrazed.
Let the side of square is a.
a = 25 + 25 m = 50 m
Area of square = a2
Page No 604:
Question 4:
A cow is tied with a rope of length 14 m at the corner of a rectangular field of dimensions , find the area of the field in which the cow can graze .
Answer:
Shaded portion shows the area that is grazed by the cow. It is in the form of a quadrant of a circle with radius 14 m.
Page No 604:
Question 5:
A calf is tied with a rope of length 6 m at the corner of a square grassy lawn of side 20 m . If the length of the rope is increased by 5.5 m , find the increase in area of the grassy lawn in which the calf can graze .
Answer:
The area grazed by the calf is in the form of a quadrant of a circle with radius 6 m.
When the rope length is increased then total rope length = 6 + 5.5 m = 11.5 m
Hence, increase in area of grassy lawn that is grazed = 103.91 − 28.28 = 75.63 m2
Page No 605:
Question 6:
In Fig. 13.113, ABCD is a square with side cm and inscribed in a circle. Find the area of the shaded region. (Use π = 3.14)
Answer:
ABCD is a square.
Side of the square = AB =
We know
Length of the diagonal of square = × Side of the square
∴ BD = Diameter of the circle = Length of the diagonal of square = = 4 cm
⇒ Radius of the circle = = 2 cm
Now,
Area of shaded region
= Area of the circle − Area of the square ABCD
Thus, the area of the shaded region is 4.56 cm2.
Page No 605:
Question 7:
In Fig. 13.112, a square OABC is inscribed in a quadrant OPBQ. If OA = 15 cm, find the area of shaded region (use π = 3.14).
Answer:
OABC is a square.
Side of the square = OA = 15 cm
We know
Length of the diagonal of square = × Side of the square
∴ OB = Radius of the quadrant of the circle = Length of the diagonal of square =
Now,
Area of shaded region
= Area of quadrant OPQO − Area of the square OABC
Thus, the area of the shaded region is 128.25 cm2.
Page No 605:
Question 8:
In the following figure, from a rectangular region ABCD with AB = 20 cm, a right triangle AED with AE = 9 cm and DE = 12 cm, is cut off. On the other end, taking BC as diameter, a semicircle is added on outside' the region. Find the area of the shaded region. [Use π =] [CBSE 2014]
Answer:
AD2 = AE2 + DE2
= (9)2 + (12)2
= 81 + 144
= 225
∴ AD2 = 225
⇒ AD = 15 cm
We know that the opposite sides of a rectangle are equal
AD = BC = 15 cm
Area of the shaded region = Area of rectangle − Area of triangle AED + Area of semicircle
Hence, the area of shaded region is 334.39 cm2
Page No 605:
Question 9:
From each of the two opposite corners of a square of side 8 cm, a quadrant of a circle of radius 1.4 cm is cut. Another circle of radius 4.2 cm is also cut from the centre as shown in the following figure. Find the area of the remaining (Shaded) portion of the square. (Use π = 22/7)
Answer:
It is given that a circle of radius 4.2 cm and two quadrants of radius 1.4 cm are cut from a square of side 8 cm.
Let the side of square be a. Then,
Since the radius of circle is 4.2 cm. So,
Now area of quadrant of circle of radius 1.4 cm is,
Page No 605:
Question 10:
In the following figure, ABCD is a rectangle with AB = 14 cm and BC = 7 cm. Taking DC, BC and AD as diameters, three semi-circles are drawn as shown in the figure. Find the area of the shaded region.
Answer:
Area of the shaded region can be calculated as shown below,
Area of the shaded region = Area of rectangle − area of the semi-circle with diameter DC triangle + 2 area of two semicircles with diameters AD and BC
Substituting we get,
Therefore, area of the shaded region is.
Page No 605:
Question 11:
In the following figure, ABCD is a rectangle, having AB = 20 cm and BC = 14 cm. Two sectors of 180° have been cut off. Calculate:
(i) the area of the shaded region.
(ii) the length of the boundary of the shaded region.
Answer:
(i) We have given two semi-circles and a rectangle.
Area of the shaded region = Area of the rectangle − Area of the two semicircles ……..(1)
Substituting we get,
Therefore, area of shaded region is.
(ii) Now we will find length of the boundary of the shaded region.
Therefore, length of the boundary of the shaded region is.
Page No 605:
Question 12:
In the following 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.
Answer:
It is given that OACB is a quadrant of circle with centre at O and radius 3.5 cm.
(i) We know that the area of quadrant of circle of radius r is,
Substituting the value of radius,
Hence, the area of OACB is.
(ii) It is given that radius of quadrant of small circle is 2 cm.
Let the area of quadrant of small circle be.
It is clear from the above figure that area of shaded region is the difference of larger quadrant and the smaller one. Hence,
Page No 605:
Question 13:
In the following figure a square OABC is inscribed in a quadrant OPBQ of a circle. If OA = 21 cm, find the area of the shaded region.
Answer:
Construction: Join OB
In right triangle AOB
OB2 = OA2 + AB2
= 212 + 212
= 441 + 441
= 882
∴ OB2 = 882
Area of the shaded region = Area of quadrant OPBQ − Area of Square OABC
Hence, the area of the shaded region is 252 cm2.
Page No 605:
Question 14:
(i) In the given figure, OABC is a square of side 7 cm. If OAPC is a quadrant of a circle with centre O, then find the area of the shaded region. (Use π = 22/7)
(ii) Find the area of the shaded region in the given figure, if AC = 24 cm, BC = 10 cm and O is the centre of the circle. (Use π = 3.14)
Answer:
(i) Area of shaded region = Area of square OABC − Area of quadrant OAPC
Hence, the area of the shaded region is 10.5 cm2
(ii) It is given a triangle ABC is cut from a circle.
In,
, Since any angle inscribed in semicircle is always right angle.
By applying Pythagoras theorem,
We know that the area A of circle of radius r is
Substituting the value of radius r,
Page No 606:
Question 15:
ABCDEF is a regular hexagon with centre O (in the following figure). If the area of triangle OAB is 9 cm2, find the area of : (i) the hexagon and (ii) the circle in which the haxagon is incribed.
Answer:
We know that a regular hexagon is made up of 6 equilateral triangles.
We have given area of the one of the triangles.
We know that if a regular hexagon is inscribed in the circle, then the radius of the circle is same as the side of the regular hexagon.
We also know that a regular hexagon is made up of 6 equilateral triangles and we have area of one of the equilateral triangle.
Substituting the value of the given equilateral triangle we get,
Now we will find the area of the circle.
Substituting the values we get,
Now we will substitute we get,
Therefore, area of the hexagon and area of the circle are and respectively.
Page No 606:
Question 16:
A child makes a poster on a chart paper drawing a square ABCD of side 14 cm. She draws four circles with centre A, B, C and D in which she suggests different ways to save energy. The circles are drawn in such a way that each circle touches externally two of the three remaining circles (in the following figure). In the shaded region she write a message 'Save Energy'. Find the perimeter and area of the shaded region.
(Use π = 22/7)
Answer:
Perimeter of the shaded portion = 4 ⨯ Length of the arc having central angle 90∘
Area of shaded portion = Area of square ABCD − 4 ⨯ Area of the arc having central angle 90∘
Page No 606:
Question 17:
In the following figure, AB and CD are two diameters of a circle perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.
Answer:
It is given that AB and CD are two diameters of a circle perpendicular to each other and OD is the diameter of small circle.
It is given that,
So, radius r of small circle is
We know that the area A of circle of radius r is.
Substituting the value of r in above formula,
Now, let the area of large circle be.
Using the value radius OA,
Hence,
Page No 606:
Question 18:
In the the following figure, PSR, RTQ and PAQ are three semicircles of diameter 10 cm, 3 cm and 7 cm respectively. Find the perimeter of shaded region. [CBSE 2014]
Answer:
Perimeter of shaded region = Length of the arc PAQ + Length of the arc PSR + Length of the arc RTQ
Hence, the perimeter of shaded region is 31.4 cm.
Page No 606:
Question 19:
In the following figure, two circles with centres A and B touch each other at the point C. If AC = 8 cm and AB = 3 cm, find the area of the shaded region.
Answer:
Area of the shaded region can be calculated as shown below,
Area of the shaded region = Area of circle with radius AC − area of circle with radius BC
We have given radius of the outer circle that is 8 cm but we don’t know the radius of the inner circle.
We can calculate the radius of the inner circle as shown below,
Substituting we get,
Therefore, area of the shaded region is.
Page No 606:
Question 20:
In the following figure, O is the centre of a circular arc and AOB is a straight line. Find the perimeter and the area of the shaded region correct to one decimal place. (Take π = 3.142)
Answer:
(i) Let us find the perimeter of the shaded region.
Therefore, perimeter of the shaded region is .
Now we will find the area of the shaded region can be calculated as shown below,
Area of the shaded region = Area of the semi-circle − area of the right angled triangle
First, we will find the length of AB as shown below,
Substituting we get,
Therefore, area of the shaded region is.
Page No 606:
Question 21:
In the following figure, shows a kite in which BCD is the shape of a quadrant of a circle of radius 42 cm. ABCD is a square and Δ CEF is an isosceles right angled triangle whose equal sides are 6 cm long. Find the area of the shaded region.
Answer:
We will find the area of the shaded region as shown below,
Area of the shaded region = area of quadrant + area of isosceles triangle ……..(1)
Substituting we get,
Therefore, area of the shaded region is.
Page No 606:
Question 22:
In the following figure, ABCD is a trapezium of area 24.5 cm2 , If AD || BC, ∠DAB = 90°, AD = 10 cm, BC = 4 cm and ABE is quadrant of a circle, then find the area of the shaded region. [CBSE 2014]
Answer:
Area of shaded region = Area of trapezium ABCD − Area of quadrant ABE
Hence, the area of shaded region is 14.875 cm2
Page No 606:
Question 23:
From a thin metallic piece, in the shape of a trapezium ABCD, in which AB || CD and ∠BCD = 90°, a quarter circle BEFC is removed (in the following figure). Given AB = BC = 3.5 cm and DE = 2 cm, calculate the area of the remaining piece of the metal sheet.
Answer:
We have given a trapezium. We are asked to find the area of the shaded region.
We can find the area of the remaining part that is area of the shaded region as shown below.
……..(1)
Now we find the value of CD.
………(Since, CE is radius of the sector, therefore, CE = 3.5)
Substituting the values of CD and in equation (1),
Therefore, area of the remaining part is.
Page No 607:
Question 24:
In the following figure, ABC is an equilateral triangle of side 8 cm. A, B and C are the centres of circular arcs of radius 4 cm. Find the area of the shaded region correct upto 2 decimal places. (Take π =3.142 and = 1.732).
Answer:
Area of the shaded region can be calculated as shown below,
Area of the shaded region = Area of equilateral triangle − 3 area of circular arc
Substituting and we get,
Therefore, area of the shaded region is.
Page No 607:
Question 25:
The inside perimeter of a running track (shown in the following figure) is 400 m. The length of each of the straight portion is 90 m and the ends are semi-circles. If the track is everywhere 14 m wide. find the area of the track. Also find the length of the outer running track.
Answer:
It is given that, and
We know that the circumference C of semicircle of radius be r is
The inside perimeter of running track is the sum of twice the length of straight portion and circumferences of semicircles. So,
inside perimeter of running track = 400 m
Thus, radius of inner semicircle is 35 m.
Now,
radius of outer semi circle r' = 35 + 14 = 49 m
Area of running track =
Hence, the area of running track = 6216 m2
Now, length L of outer running track is
L = 2 × l + 2r'
Hence, the length L of outer running track is 488 m.
Page No 607:
Question 26:
In the following figure, the square ABCD is divided into five equal parts, all having same area. The central part is circular and the lines AE, GC, BF and HD lie along the diagonals AC and BD of the square. If AB = 22 cm, find:
(i) the circumference of the central part.
(ii) the perimeter of the part ABEF.
Answer:
We have a square ABCD.
We have,
(i)We have to find the perimeter of the triangle. We have a relation as,
So,
So perimeter of the circular region,
(ii)We have to find the perimeter of ABEF. Let O be the centre of the circular region.
Use Pythagoras theorem to get,
Similarly,
Now length of arc EF,
So, perimeter of ABFE,
Page No 607:
Question 27:
In the following figure find the area of the shaded region. (Use π = 3.14)
Answer:
Area of shaded region = Area of square − Area of 4 semicircle having diameter 4 cm − Area of square having side 4 cm
Hence, the area of shaded region is 154.88 cm2
Page No 607:
Question 28:
A circle is inscribed in an equilateral triangle ABC is side 12 cm, touching its sides (the following figure). Find the radius of the inscribed circle and the area of the shaded part.
Answer:
We have to find the area of the shaded portion. We havewhich is an equilateral triangle and.
We have O as the incentre and OP, OQ and OR are equal.
So,
Thus,
So area of the shaded region,
Page No 607:
Question 29:
In the following figure, an equilateral triangle ABC of side 6 cm has been inscribed in a circle. Find the area of the shaded region. (Take π = 3.14).
Answer:
We have to find the area of the shaded portion. We havewhich is an equilateral triangle and. Let r be the radius of the circle.
We have O as the circumcentre.
So,
Thus,
So area of the shaded region,
Page No 607:
Question 30:
Find the area of a shaded region in the the following figure,where a circular arc of radius 7 cm has been drawn with vertex A of an equilateral triangle ABC of side 14 cm as centre. (Use π = 22/7 and = 1.73)
Answer:
In equilateral traingle all the angles are of 60°
∴ ∠BAC = 60°
Area of the shaded region = (Area of triangle ABC − Area of sector having central angle 60°) + Area of sector having central angle (360° − 60°)
Hence, the area of shaded region is 187.45 cm2
Page No 607:
Question 31:
In the following figure, ABCD is a square of side 2a, Find the ratio between
(i) the circumferences
(ii) the areas of the in circle and the circum-circle of the square.
Answer:
We have a square ABCD having. From the given diagram we can observe that,
Radius of incircle
Radius of circumcircle
(i) We have to find the ratio of the circumferences of the two circles. So the required ratio is,
(ii) We have to find the ratio of the areas of the two circles. So the required ratio is,
Page No 607:
Question 32:
In the following figure, there are three semicircles, A, B and C having diameter 3 cm each, and another semicircle E having a circle D with diameter 4.5 cm are shown. Calculate:
(i) the area of the shaded region
(ii) the cost of painting the shaded region at the rate of 25 paise per cm2 , to the nearest rupee.
Answer:
(i) Area of the shaded region can be calculated as shown below,
Area of the shaded region = Area of the semi-circle with diameter of 9 cm − area of 2 semi-circles with radius 3cm − area of the circle with centre D + area of semi-circle with radius 3 cm
Substituting we get,
Therefore, area of the shaded region is.
Now we will find the cost of painting the shaded region at the rate of 25 paise per cm2.
paise
Therefore, cost of painting the shaded region to the nearest rupee is.
Page No 608:
Question 33:
In the following figure, ABC is a right-angled triangle, ∠B = 90°, AB = 28 cm and BC = 21 cm. With AC as diameter a semicircle is drawn and with BC as radius a quarter circle is drawn. Find the area of the shaded region correct to two decimal places.
Answer:
We have given two semi-circles and one circle.
Area of the shaded region = area of semicircle with diameter AC + area of right angled triangle ABC − area of sector
First we will find the hypotenuse of right angled triangle ABC.
Substituting we get,
Therefore, area of shaded region is.
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Question 34:
In the following figure, the boundary of the shaded region consists of four semi-circular arcs, the smallest two being equal. If the diameter of the largest is 14 cm and of the smallest is 3.5 cm, find
(i) the length of the boundary.
(ii) the area of the shaded region.
Answer:
(i) We will first find the length of the boundary.
Length of the boundary perimeter of semi-circle with diameter AB + boundary of semi-circle with diameter 7 cm
Therefore, length of the boundary is.
Now we will find the area of the shaded region as shown below,
Area of the shaded region = Area of the semi-circle with AB as a diameter − area of the semi-circle with radius AE − area of the semi-circle with radius BC + area of the semi-circle with diameter 7 cm.
Therefore, area of the shaded region is.
Page No 608:
Question 35:
In the following figure, AB = 36 cm and M is mid-point of AB. Semi-circles are drawn on AB, AM and MB as diameters. A circle with centre C touches all the three circles. Find the area of the shaded region.
Answer:
We have given two semi-circles and one circle.
Area of the shaded region = area of semicircle with diameter AB − area of two semicircles with diameters AM and MB - area of circle ……..(1)
Let us calculate the area of the semi-circle with AB as a diameter.
Now we will find the area of the semi-circle with AM as a diameter.
Area of the semi-circle with MB as a diameter is same as the area of the semi-circle with diameter with AM as a diameter.
Now we will find the area of the circle with centre C.
We know that radius of the circle is one sixth of AB.
Now we will substitute all these values in equation (1).
Therefore, area of shaded region is.
Page No 608:
Question 36:
In the following figure, ABC is a right angled triangle in which ∠A = 90°, AB = 21 cm and AC = 28 cm. Semi-circles are described on AB, BC and AC as diameters. Find the area of the shaded region.
Answer:
We have given three semi-circles and one right angled triangle.
……..(1)
Let us calculate the area of the semi-circle with AB as a diameter.
Now we will find the area of the semi-circle with AC as a diameter.
Now we will find the length of BC.
In right angled triangle ABC, we will use Pythagoras theorem,
Now we will calculate the area of the right angled triangle ABC.
Now we will find the area of the semi-circle with BC as a diameter.
Now we will substitute all these values in equation (1).
Therefore, area of shaded region is.
Page No 608:
Question 37:
In the following figure, shows the cross-section of railway tunnel. The radius OA of the circular part is 2 m. If ∠AOB = 90°, calculate:
(i) the height of the tunnel
(ii) the perimeter of the cross-section
(iii) the area of the cross-section.
figure
Answer:
We have a cross section of a railway tunnel. is a right angled isosceles triangle, right angled at O. let OM be perpendicular to AB.
(i) We have to find the height of the tunnel. We have,
Use Pythagoras theorem into get,
Let the height of the tunnel be h. So,
Thus,
Therefore,
(ii)
Perimeter of cross-section is,
(iii)
Page No 608:
Question 38:
In the given figure, ABCD is a trapezium with AB || DC, AB = 18 cm DC = 32 cm and the distance between AB and DC is 14 cm. Circles of equal radii 7 cm with centres A, B, C and D have been drawn. Then find the area of the shaded region.
(Use [CBSE 2014]
Answer:
Hence, the area of shaded region is 196 cm2
Page No 608:
Question 39:
Sides of a triangular field are 15 m , 16 m , 17 m . With three corners of the field a cow , a buffalo and a horse are tied separately with ropes of length 7 m each to graze in the field . Find the area of the field which cannot be grazed by three animals.
Answer:
The area grazed by the cow, buffalo and the horse are in the form of sectors of the circle with radius 7 m.
Let the angle formed in the three sectors be
Now the area of these sectors will be
Area of the triangle ABC wil be
Area of the field not grazed by the animals = Area of the triangle ABC − Area of the three sectors
Page No 608:
Question 40:
In the given figure, the side of square is 28 cm and radius of each circle is half of the length of the side of the square where O and O' are centres of the circles. Find the area of shaded region.
Answer:
We have,
Side of square = 28 cm and radius of each circle = cm
Area of the shaded region
= Area of the square + Area of the two circles − Area of the two quadrants
Therefore, the area of the shaded region is 1708 cm2.
Page No 608:
Question 41:
In a hospital used water is collected in a cylindrical tank of diameter 2 m and height 5 m. After recycling, this water is used to irrigate a park of hospital whose length is 25 m and breadth is 20 m. If tank is filled completely then what will be the height of standing water used for irrigating the park.
Answer:
Diameter of cylinder (d) = 2 m
Radius of cylinder (r) = 1 m
Height of cylinder (H) = 5 m
Volume of cylinderical tank, Vc = =
Length of the park (l) = 25 m
Breadth of park (b) = 20 m
height of standing water in the park = h
Volume of water in the park = lbh =
Now water from the tank is used to irrigate the park. So,
Volume of cylinderical tank = Volume of water in the park
Page No 610:
Question 1:
What is the ratio of the areas of a circle and an equilateral triangle whose diameter and a side are respectively equal?
Answer:
We are given that diameter and side of an equilateral triangle are equal.
Let d and a are the diameter and side of circle and equilateral triangle respectively.
Therefore d = a
We know that area of the circle
Area of the equilateral triangle
Now we will find the ratio of the areas of circle and equilateral triangle.
We know that radius is half of the diameter of the circle.
Now we will substitute in the above equation,
Therefore, ratio of the areas of circle and equilateral triangle is.
Page No 610:
Question 2:
If the circumference of two circles are in the ratio 2 : 3, what is the ratio of their areas?
Answer:
We are given ratio of circumferences of two circles. If and are circumferences of two circles such that
…… (1)
Simplifying equation (1) we get,
Let and are the areas of the respective circles and we are asked to find their ratio.
…… (2)
We know that substituting this value in equation (2) we get,
Therefore, ratio of their areas is.
Page No 610:
Question 3:
Write the area of the sector of a circle whose radius is r and length of the arc is l.
Answer:
We know that area of the sector of the circle of radius r
Length of the arc
But we have given that length of the arc
…… (1)
Area of the sector
Now we will adjust 2 in the following way,
Area of the sector
Area of the sector
From equation (1) we will substitute
Area of the sector
Area of the sector
Therefore, area of the sector =.
Page No 610:
Question 4:
What is the length (in terms of π) of the arc that subtends an angle of 36° at the centre of a circle of radius 5 cm?
Answer:
We have
We have to find the length of the arc.
Substituting the values we get,
……….(1)
Now we will simplify the equation (1) as below,
Therefore, length of the arc is
Page No 610:
Question 5:
What is the angle subtended at the centre of a circle of radius 6 cm by an arc of length 3 π cm?
Answer:
We have
We will find the angle subtended at the centre of a circle.
Substituting the values we get,
…… (1)
Now we will simplify the equation (1) as below,
Therefore, angle subtended at the centre of the circle is.
Page No 610:
Question 6:
What is the area of a sector of a circle of radius 5 cm formed by an arc of length 3.5 cm?
Answer:
We have
Now we will find the area of the sector.
Substituting the values we get,
…… (1)
Now we will simplify the equation (1) as below,
Therefore, area of the sector is.
Page No 610:
Question 7:
In a circle of radius 10 cm, an arc subtends an angle of 108° at the centre. what is the area of the sector in terms of π?
Answer:
We have given the radius of the circle and angle subtended at the centre of the circle.
Now we will find the area of the sector.
Substituting the values we get,
…… (1)
Now we will simplify the equation (1) as below,
Therefore, area of the sector is.
Page No 610:
Question 8:
If a square is inscribed in a circle, what is the ratio of the areas of the circle and the square?
Answer:
We have the following situation
Let BD be the diameter and diagonal of the circle and the square respectively.
We know that area of the circle
Area of the square
As we know that diagonal of the square is the diameter of the square.
…… (1)
Substituting in equation (1) we get,
Now we will find the ratio of the areas of circle and square.
Now we will simplify the above equation as below,
Therefore, ratio of areas of circle and square is.
Page No 610:
Question 9:
Write the formula for the area of a sector of angle (in degrees) of a circle of radius r.
Answer:
Let r be the radius of the circle and angle θ subtended at the centre of the circle.
Area of the sector of the circle
Therefore, area of the sector is.
Page No 610:
Question 10:
Write the formula for the area of a segment in a circle of radius r given that the sector angle is (in degrees).
Answer:
In this figure, centre of the circle is O, radius OA = r and
We are going to find the area of the segment AXB.
…… (1)
We know that
We also know that
Substituting these values in equation (1) we get,
Therefore, area of the segment is.
Page No 610:
Question 11:
If the adjoining figure is a sector of a circle of radius 10.5 cm, what is the perimeter of the sector? (Take )
Answer:
Given figure is a quadrant of a circle. We have given radius of sector that is 10.5 cm. Arc AB subtended an angle of 60° at the centre of the circle.
Perimeter of the sector
Substituting the values we get,
Perimeter of the sector …… (1)
Now we will simplify equation (1) as shown below,
Now we will substitute.
Therefore, perimeter of the given sector is.
Page No 611:
Question 12:
If the diameter of a semi-circular protractor is 14 cm, then find its perimeter.
Answer:
Let AB be the diameter of the semi-circular protractor.
We know that perimeter of the semicircle ……(1)
We have given the diameter of the protractor.
Therefore, radius of the protractor
So, radius of the protractor
Substituting the value of r in equation (1) we get,
Substituting we get,
Therefore, perimeter of the semi-circular protractor is.
Page No 611:
Question 13:
A piece of wire 22 cm long is bent into the form of an arc of a circle subtending angle of 60° at its centre. Find the radius of the circle
Answer:
Length of the wire = 22 cm
Angle subtended at the centre = = 60º
Length of the arc =
Page No 611:
Question 14:
Find the area of the largest triangle that can be inscribed in a semi-circle of radius r units. [CBSE 2015]
Answer:
Radius of the semi circle r units
When a largest triangle inscribed in a semicircle, then base = r + r = 2r units
Thus, Area of triangle is given by
Page No 611:
Question 15:
An arc subtends an angle of 90° at the centre of the circle of the radius 14 cm. Write the area of minor sector thus formed in terms of π.
Answer:
We have given an angle subtended by an arc at the centre of the circle and radius of the circle.
Now we will find the area of the minor sector.
Area of the minor sector =
Substituting the values we get,
Area of the minor sector = …… (1)
Now we will simplify the equation (1) as below,
Area of the minor sector =
Area of the minor sector =
Area of the minor sector = π × 7 × 7
Area of the minor sector = 49π
Therefore, area of the minor sector is.
Page No 611:
Question 16:
Find the area of sector of circle of radius 21 cm and central angle 1200.
Answer:
Area of a sector of a circle =
Page No 611:
Question 17:
What is the area of a square inscribed in a circle of diameter p cm ?
Answer:
The diameter of the circle = Diagonal of the square = p
We know that diagonal of a square =
Thus, the area of the square =
Page No 611:
Question 18:
Is it true to say that area of segment of a circle is less than the area of its corresponding sector ? Why ?
Answer:
A circle has two segments, a major segment and a minor segment.
The area of the minor segment is less than the area of the corresponding sector.
But same is not true for the major segment.
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Question 19:
If the numerical value of the area of a circle is equal to the numerical value of its circumference , find its radius.
Answer:
Given that area of the circle = circumference
Since radius cannot be equal to 0 so, r = 2
Page No 611:
Question 20:
How many revolutions a circular wheel of radius r metres makes in covering a distance of s metres?
Answer:
In one revolution, the wheel covers a distance of
So, to cover a distance of s meters, the wheel has to make revolutions.
Page No 611:
Question 21:
Find the ratio of the area of the circle circumscribing a square to the area of the circle inscribed in the square .
Answer:
Let the side of the square inscribed in a square be a units.
Diameter of the circle outside the square = Diagonal of the square =
Radius =
So, the area of the circle circumscribing the square = .....(i)
Now, the radius of the circle inscribed in a square =
Hence, area of the circle inscribed in a square = .....(ii)
From (i) and (ii)
Hence, the required ratio is 2 : 1.
Page No 611:
Question 1:
If the area of a circle is 154 cm2, then its perimeter is __________.
Answer:
Let the radius of the circle be r cm.
Area of the circle = 154 cm2
∴ Perimeter of the circle
= Circumference of the circle
If the area of a circle is 154 cm2, then its perimeter is ___44 cm___.
Page No 611:
Question 2:
If the perimeter of a circle is equal to that of a square, then the ratio of their areas is ___________.
Answer:
Ans
Page No 611:
Question 3:
The area of the circle that can be inscribed in a square of side 6 cm is __________.
Answer:
If a circle is inscribed in a square, then the diameter of the circle is equal to the side of the square.
Let the radius of the circle be r cm.
Now,
Diameter of the circle = Side of the square
⇒ 2r = 6 cm
⇒ r = 3 cm
∴ Area of the inscribed circle
The area of the circle that can be inscribed in a square of side 6 cm is .
Page No 611:
Question 4:
The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm is __________.
Answer:
Let the radius of the circle be r cm.
Radius of first circle, r1 = 24 cm
Radius of second circle, r2 = 7 cm
Now,
Area of the circle = Area of first circle + Area of second circle
∴ Diameter of circle = 2r = 2 × 25 = 50 cm
The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm is __50 cm__.
Page No 611:
Question 5:
The area of the square that can be inscribed in a circle of radius 8 cm is ___________.
Answer:
If a square is inscribed in a circle, then the length of diagonal of square is equal to the diameter of the circle.
Let the side of the square be a cm.
∴ Length of diagonal of square = × Side of square =
Radius of the circle, r = 8 cm
∴ Diameter of the circle = 2r = 2 × 8 = 16 cm
Now,
Length of diagonal of square = Diameter of the circle
∴ Area of the square
The area of the square that can be inscribed in a circle of radius 8 cm is .
Page No 611:
Question 6:
It is proposed to build a single park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be ___________.
Answer:
Let the radius of the new circular park be r m.
Suppose the radii of two circular parks be r1 m and r2 m.
Diameter of the first circular park = 16 m
⇒ 2r1 = 16
⇒ r1 = 8 m
Diameter of the second circular park = 12 m
⇒ 2r2 = 12
⇒ r2 = 6 m
Now,
Area of the new circular park = Area of first circular park + Area of second circular park
= 10 m
Thus, the radius of the new park is 10 m.
It is proposed to build a single park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be __10 m__.
Page No 611:
Question 7:
The radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters 36 cm and 20 cm, is __________.
Answer:
Let the radius of the circle be r cm.
Suppose the radii of two given circles be r1 cm and r2 cm.
Diameter of the first circle = 36 cm
⇒ 2r1 = 36 .....(1)
Diameter of the second circle = 20 cm
⇒ 2r2 = 20 .....(2)
Now,
Circumference of the circle = Circumference of the first circle + Circumference of the second circle
Thus, the radius of the circle is 28 cm.
The radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters 36 cm and 20 cm, is ___28 cm___.
Page No 611:
Question 8:
If the difference between outer and inner radii of a circular ring is 14 cm, then the difference between outer and inner circumferences, is __________.
Answer:
Let r1 and r2 be the radii of the inner and outer circular rings, respectively.
Given: r2 − r1 = 14 cm .....(1)
∴ Difference between their outer and inner circumferences
If the difference between outer and inner radii of a circular ring is 14 cm, then the difference between outer and inner circumferences, is ___88 cm___.
Page No 611:
Question 9:
The area of a sector whose perimeter is four times its radius r, is ___________.
Answer:
Let l be the length of the arc of the corresponding sector.
Radius of the sector = r units
We know
Perimeter of the sector = l + 2r
∴ l + 2r = 4r (Given)
⇒ l = 4r − 2r = 2r .....(1)
∴ Area of the sector [Using (1)]
The area of a sector whose perimeter is four times its radius r, is ___r2 units___.
Page No 612:
Question 10:
Two circles touch each other externally. The sum of their areas is 490 π cm2. Their centres are separated by 28 cm. The difference of their perimeters is ___________.
Answer:
Let R and r be the radii of the two circles. Suppose R > r.
We know that if two circles touch each other externally, then the distance between their centres is equal to the sum of the radii of the two circles.
∴ R + r = 28 cm .....(1)
Also,
Sum of the areas of the two circles = cm2
Now,
∴ Difference between their perimeters
Two circles touch each other externally. The sum of their areas is 490 π cm2. Their centres are separated by 28 cm. The difference of their perimeters is _____88 cm______.
Page No 612:
Question 11:
If l is the arc length of a sector of a circle of radius r and A is the area of the sector, then A : l = __________.
Answer:
Arc length of the sector = l units
Radius of the sector = r units
We know
Area of the sector, A
⇒ A : l = r : 2
If l is the arc length of a sector of a circle of radius r and A is the area of the sector, then A : l = ___r : 2___.
Page No 612:
Question 12:
If the circumferences of two circles are in the ratio c1 : c2, then the ratio of their areas is __________.
Answer:
Let R and r be the radii of the two circles.
Now,
Circumference of first circle : Circumference of second circle = c1 : c2 (Given)
Squaring on both sides, we get
∴ Area of first circle : Area of second circle =
If the circumferences of two circles are in the ratio c1 : c2, then the ratio of their areas is .
Page No 612:
Question 13:
The area of a sector of angle θ° of a circle of radius r is _________.
Answer:
Radius of the sector = r
Angle of the sector = θ
∴ Area of the sector =
The area of a sector of angle θ° of a circle of radius r is .
Page No 612:
Question 14:
The length of the arc of a sector of angle θ° of a circle of radius r is ___________.
Answer:
Radius of the sector = r
Angle of the sector = θ
∴ Length of arc of the sector =
The length of the arc of a sector of angle θ° of a circle of radius r is .
Page No 612:
Question 15:
The area of the minor segment of angle θ° of a circle of radius r is ___________.
Answer:
Radius of the sector = r
Angle of the minor segment = θ
∴ Area of the minor segment =
The area of the minor segment of angle θ° of a circle of radius r is .
Page No 612:
Question 16:
The difference in the areas of the regular hexagon circumscribing a circle of radius 10 cm and the regular hexagon inscribed in the circle is ___________.
Answer:
In the figure, O is the centre of circle of radius 10 cm. ABCDEF is a regular hexagon circumscribing the circle and PQRSTU is a regular hexagon inscribed in the circle.
Here, ∆OPQ is an equilateral triangle.
Length of each side of equilateral ∆OPQ = Radius of the circle = 10 cm
Now,
Area of the regular hexagon PQRSTU
= 6 × Area of equilateral ∆OPQ
Also,
∆OAB is an equilateral triangle.
Length of an altitude of equilateral ∆OAB = Radius of the circle = 10 cm
We know
Length of an altitude of an equilateral triangle =
Area of the regular hexagon ABCDEF
= 6 × Area of equilateral ∆OAB
∴ Required difference in areas
= Area of the regular hexagon ABCDEF − Area of the regular hexagon PQRSTU
The difference in the areas of the regular hexagon circumscribing a circle of radius 10 cm and the regular hexagon inscribed in the circle is .
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Question 17:
In Fig. 13.121, ABCD is a square of side 10 cm and a circle is inscribed in it. The area of the shaded part is __________.
Answer:
Let O be the centre of the circle. Suppose the circle touches the sides BC and CD of the square at E and F, respectively.
OE = OF (Radius of the circle)
Now, ∠OEC = ∠OFC = 90º (Radius is perpendicular to the tangent at the point of contact)
Also, OE = OF = CF = CE
∴ OECF is a square.
We know that if a circle is inscribed in a square, then the diameter of the circle is equal to the side of the square.
⇒ Length of each side of the square OECF = 5 cm
Now,
Area of the shaded portion
= (Area of the square OECF − Area of the quadrant OEFO)
In Fig. 13.121, ABCD is a square of side 10 cm and a circle is inscribed in it. The area of the shaded part is .
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Question 18:
In Fig. 13.122, two circles of radii 7 cm each are shown. ABCD is a rectangle and AD and BC are the radii. The area of the shaded region is __________.
Answer:
In the given figure,
BC = CE = DE = 7 cm (Radius of each circle)
Also, the two circles touch each other externally.
We know that if two circles touch each other externally, then the distance between their centres is equal sum of their radii.
CD = CE + ED = 7 + 7 = 14 cm
Now,
Area of the shaded region
= Area of the rectangle ABCD − (Area of quadrant BCEB + Area of quadrant ADEA)
In Fig. 13.122, two circles of radii 7 cm each are shown. ABCD is a rectangle and AD and BC are the radii. The area of the shaded region is ___21 cm2___.
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Question 19:
The area of the largest circle that can be drawn inside a rectangle of length a cm and breadth b cm (a > b) is __________.
Answer:
The largest circle that can be drawn inside a rectangle of length a cm and breadth b cm (a > b) will have its diameter equal to the breadth of the rectangle.
Let the radius of the largest circle that can be drawn inside a rectangle of length a cm and breadth b cm (a > b) be r cm.
∴ Diameter of the circle = b
∴ Area of the required circle
The area of the largest circle that can be drawn inside a rectangle of length a cm and breadth b cm (a > b) is .
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Question 20:
The number of revolutions made by a circle of radius r to cover a distance s is ___________.
Answer:
In one revolution, the distance covered by a circle of radius r is equal to the circumference of the circle.
Circumference of the circle of radius r =
Number of revolutions made by a circle of radius r to cover a distance = 1
∴ Number of revolutions made by a circle of radius r to cover a distance s =
The number of revolutions made by a circle of radius r to cover a distance s is .
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