RS Aggarwal Solutions Class 10 Ex 18C

Q.1: A sector is cut from a circle of radius 21 cm. The angle of the sector is \(150^{\circ}\). Find the length of the arc and the area of the sector.

Sol:

Given:

Radius = 2 cm

Angle of sector = \(150^{\circ}\)

Now,

Length of arc = \(= \frac{2 \pi r \theta }{360}\)

\(= \frac{2 \times \frac{22}{7} \times 21 \times 150 }{360}\)

= 55 cm

Area of the sector = \(= \frac{\pi r^{2} \theta }{360}\)

\(= \frac{\frac{22}{7} \times 21\times 21 \times 150 }{360}\) = 577.5 \(cm^{2}\)

Q.2: The area of the sector of a circle of radius 10.5 cm is 69.3 \(cm^{2}\). Find the central angle of the sector.

Sol:

Given

Area of the sector = 63 \(cm^{2}\)

Radius = 10.5 cm

Now,

Area of the sector = \(= \frac{\pi r^{2} \theta }{360}\)

\( \Rightarrow 69.3 = \frac{\frac{22}{7} \times  10.5 \times 10.5 \times \theta }{360}\)

\(\Rightarrow \theta =\frac{69.3 \times 7 \times 360}{22 \times 10.5 \times 10.5}\)

\(\Rightarrow \theta = 72^{\circ}\)

Therefore, Central angle of the sector = \(72^{\circ}\)

Q.3 The length of an arc of a circle, subtending an angle of \(54^{\circ}\) at the centre is 165 cm. Calculate the radius, circumference and area of the circle.

Sol:

Length of the arc = 16.5 cm

\(\theta = 54^{\circ}\)

Radius = ?

Circumference = ?

We know:

Length of the arc = \(= \frac{2 \pi r \theta }{360}\)

\(\Rightarrow 16.5 = \frac{2 \times \frac{22}{7} \times r \times 54}{360}\)

\(\Rightarrow r = \frac{16.5 \times 360 \times 7}{44 \times 54}\)

\(\Rightarrow r = 17.5\) cm

\(c = 2 \pi r\)

\(c = 2 \times \frac{22}{7} \times 17.5\) = 110 cm

Circumference = 110 cm

Now,

Area of the circle = \(= \pi r^{2}\)

= \(\frac{22}{7} \times 17.5 \times 17.5\) = \(962.5 \; cm^{2}\)

Q.4: The radius of a circle with center O is 7 cm. Two radii OA and OB are drawn at right angles to each other. Find the areas of minor and major segments.

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Area of minor segment = Area of sector AOBC – Area of right triangle AOB

\(= \frac{90}{360} \pi (OA)^{2} – \frac{1}{2} \times OA \times OB\)

\(= \frac{1}{4} \times \frac{22}{7} \times (7)^{2} – \frac{1}{2} \times 7 \times 7\)

\(= 38.5 – 24.5\)

= 14 \(cm^{2}\)

Area of major segment APB = Area of circle – Area of minor segment

\(\pi (OA)^{2} – 14\)

\(= \frac{22}{7} \times (7)^{2} – 14\)

\(= 154 – 14\)

\(= 140 \; cm^{2}\)

Hence, the area of major segment is \(= 140 \; cm^{2}\)

Q.5: Find the length of the arcs cut off from a circle of radius 12 cm by a chord 12 cm long. Also, find the area of the minor segment.   (\(\pi = 3.14 \; and \; \sqrt{3} = 1.73\))

Sol:

https://lh4.googleusercontent.com/Cs9eBA9vZyeWPFEXUQ6U5ib3HdMxngAbOtNRKt-2DVm_yvbsA8fZK95z-DibXu4pn0_G4WmCs1WNqYalxblp0qhgiHFABvYOSQBvsdS4zuF3eYGzN_2gCBW2uiNlsI6O-sQSKK6U

Let AB be the chord. Joining A and B to O, we get an equilateral triangle OAB.

Thus we have:

\(\angle O = \angle A = \angle B = 60^{\circ}\)

Let the arc ACB:

\(= 2 \pi \times 12 \times \frac{60}{360}\) = \(4\pi\) = 12.56 cm

Length of the arc ADB:

Circumference of the circle – length of the arc ACB

= \(2\pi \times 12 – 4\pi\)\(= 20 \pi\) = 62.80 cm

Now, Area of the minor segment:

Area of the sector – Area of the triangle

\(= \left [ \pi \times (12)^{2} \times \frac{60}{360} – \frac{\sqrt{3}}{4} \times (12)^{2}\right ]\) = 13.08 \( cm^{2}\)

Q.6: A chord 10 cm long is drawn in a circle whose radius is \(5 \sqrt{2} \; cm\). Find the areas of both the segments.  [Take \(\pi = 3.14\) ]

Sol:

Let O be the centre of the circle and AB be the chord.

Consider \(\bigtriangleup AOB\)

OA = OB = \(5\sqrt{2} \; cm\)

\(OA^{2} + OB^{2} = 50+50 =100\)

Now,

\(\sqrt{100}= 10 \; cm = AB\)

Thus \(\bigtriangleup AOB\) is a right isosceles triangle.

Thus, we have :

Area of \(\bigtriangleup OAB = \frac{1}{2} \times 5\sqrt{2} \times 5\sqrt{2} = 25 cm^{2} \)

Area of the minor segment = Area of the sector – Area of the triangle

\(= \left ( \frac{90}{360} \times \pi \times (5\sqrt{2})^{2} \right ) – 25\) = \(14.25 \; cm^{2}\)

Area of the major segment = Area of the circle – Area of the minor segment

= \(\pi \times (5\sqrt{2})^{2} – 14.25\) = \(14.75 \; cm^{2}\)

Q.7: Find the area of both the segments of a circle of radius 42 cm with central angle \(120^{\circ}\).   (\(\sin 120^{\circ} = \frac{\sqrt{3}}{2} \; and \; \sqrt{3} = 1.73\) )

Sol:

Area of the minor sector = \(\frac{120}{360} \times \pi \times 42 \times 42\)

= \(\frac{1}{3} \times \pi \times 42 \times 42\) = 1848 \(cm^{2}\)

Area of the triangle = \(\frac{1}{2} R^{2}\sin \theta\)

Here, R is the measure of the equal sides  of the isosceles triangle and \(\theta\) is the angle enclosed by the equal sides.

Thus, we have

\(\frac{1}{2} \times 42 \times 42 \times \sin (120^{\circ})\)

= 762.93 \(cm^{2}\)

Area of the minor segment = Area of the sector – Area of the triangle

= 1848 – 762.93 = 1085.07 \(cm^{2}\)

Area of the major segment = Area of the circle – Area of the minor segment

= \((\pi \times 42 \times 42) – 1085.07\)

= 5544 – 108.07 = 4458.93 \(cm^{2}\)

Q.8: A chord of a circle of radius 30 cm makes an angle of \(60^{\circ}\) at the centre of the circle. Find the areas of the minor and major segments.   [Take \(\pi = 3.14 \; and \; \sqrt{3} = 1.732\) ]

Sol:

https://lh4.googleusercontent.com/GC65lCsUfIzsK9ZpAD4j_utbBZ4vsS6On5T9RUs4ONbs0UhtdJ4KI3nVxqgRyxuxtCae5P3e5i4qWe21MSQqWZaIcq3gvT4Kqqb76TRtxNjXniKiVl9FyHk_U-UooDKwiof2oTho

Let the chord be AB. The ends of the chord are connected to the centre of the circle O to give the triangle OAB.

OAB is an isosceles triangle. The angle at the centre is \(60^{\circ}\)

Area of the triangle = \(\frac{1}{2} \times (30)^{2} \sin 60^{\circ} = 450 \times \frac{\sqrt{3}}{2} = 389.25 \; cm^{2}\)

Area of the sector OACBO = \(\frac{60}{360} \times \pi \times 30 \times 30 = 150 \pi = 471 \; cm^{2}\)

Area of the minor segment = Area of the segment – Area of the triangle = 471 – 389.25 = 81.75 \(cm^{2}\)

Area of the major segment = Area of the circle – Area of the minor segment

= \((\pi \times 30 \times 30) – 81.29\) = 2744.71 \(cm^{2}\)

Q.9: In a circle of radius 10.5 cm, the major arc is one-fifth of the major arc. Find the area of the sector corresponding to the major arc.

Sol:

Let the length of the major arc be x cm.

Radius of the circle = 10.5 cm

Therefore, Length of the minor arc = \(\frac{x}{5} \; cm\)

Circumference = \(x + \frac{x}{5} = \frac{6x}{5} \; cm\)

Using the given data, we get :

\(\frac{6x}{5} = 2 \times \frac{22}{7} \times \frac{21}{2}\)

\(\Rightarrow \frac{6x}{5} = 66\)

\(\Rightarrow x = 55\)

Therefore, Area of the sector corresponding to the major arc = \(\frac{1}{2} \times 55 \times \frac{21}{2} = 288.75 \; cm^{2}\)

Q.10: The short and long hands of a clock are 4 cm and 6 cm respectively. Find the sum of distances travelled by their tips in 2 days.     [Take \(\pi = 3.14\)]

Sol:

In 2 days, the short hand will complete 4 rounds.

Length of the short hand = 4 cm

Distance covered by the short hand = \(= 4 \times 2\pi \times 4 = 32 \pi\) cm

In the same 2 days, the long hand will complete 48 rounds.

Length of the long hand = 6 cm

Distance covered by the long hand = \(= 48 \times 2\pi \times 4 = 576 \pi\) cm

Therefore, Total distance covered by both the hands = Distance covered by the short hand + Distance covered by the long hand

= \(32 \pi +576 \pi\) = \(608 \pi\) = \(608 \times 3.14\) = \(1909.12\) cm

Q.11: Find the area of a quadrant of a circle whose circumference is 88 cm.

Sol:

Let the radius of the circle be r.

Now,

Circumference = 88 cm

\(\Rightarrow 2 \pi r = 88\)

\(\Rightarrow r = 14\) cm

Now,

Area of quadrant \(= \frac{1}{4} \pi r^{2} = \frac{1}{4} \times \frac{22}{7} \times (14)^{2} = 154 \; cm^{2}\)

Hence, the area of the quadrant is \( 15 cm^{2}\).

Q.12: A rope by which a cow is tethered is increased from 16 cm to 23 cm. How many additional grounds does it have now to graze?

Sol:

https://lh6.googleusercontent.com/LTLLH62ssrA6UaZP8JPn5p9zMp34wy6_KiszCw4qoZBvMyoceZqGcBH_2wKIyQNAzoorlOVG_A_SgnPzzLuGD5EF4xwaaU_8HGSOniYuyjet5fhhQuxbH4sYtJ3oWtZYRMJCzKbz

\(r_{1} = 16 \; cm\)

\(r_{2} = 23 \; cm\)

Amount of additional ground available = Area of the bigger circle – Area of the smaller circle

\(= \pi (r_{1}^{2} – r_{2}^{2})\)

\(= \pi (23^{2} – 16^{2})\)\(= \pi (23 + 16) (23 – 16)\)\(= 858 \; m^{2} \)

Q.13: A horse is placed for grazing inside a rectangular field 70 m by 52 m. It is tethered to one corner by a rope 21 m long. On how much area can it graze? How much area is left unglazed?

Sol:

Radius of the quadrant of the circle = 21 m

The shaded portion shows the part of the field the horse can graze.

https://lh4.googleusercontent.com/HM5c1mwqwHMHIn89Q-K9XWrUOfxYdsw8TRpvq5EOmnGp307feY9cK1RHxonoHSqjW7J2YU2FS3NGdNARzLHd7XKHjGNAXn06d52RzDIxeWlrVQgKi1flD-KBvZVXXGhPkGgU1KNJ

Area of the grazed field = Area of the quadrant OPQ

\(= \frac{1}{4}\) of the circle having radius OP.

\(= \frac{1}{4} \pi r^{2}\)

\(= \frac{1}{4} \times \frac{22}{7} \times 21 \times 21\)

= 346.5 \(m^{2}\)

Total area of the field = \(70 \times 52 = 3640 \; m^{2}\)

Area left ungrazed  = Area of the field – Area of the grazed field

= 3640 – 346.5 = 3293.5 \(m^{2}\)

Q.14: A horse is tethered to one corner of a field which is in the shape of an equilateral triangle of side 12 m. If the length of the rope is 7 m, find the area of the field which the horse cannot graze. Take \(\sqrt{3} = 1.732\). Write the answer correct to 2 decimal places.

Sol:

Side of the equilateral triangle = 12 m

Area of the equilateral triangle = \(\frac{\sqrt{3}}{4} \times (Side)^{2}\)

\(\frac{\sqrt{3}}{4} \times 12 \times 12\)

= 62.28 \(m^{2}\)

Length of the rope = 7 m

Area of the field that horse can graze is the area of the sector of radius 7 m. Als, the angle subtended at the centre is \(60^{\circ}\)

\(\frac{\theta }{360} \times \frac{22}{7} \times (7)^{2}\) = 25.67 \(m^{2}\)

Area of the field that horse cannot graze = Area of the equilateral triangle – Area of the field the horse can graze = 62.28 – 25.67 = 36.61 \(m^{2}\)

Q.15: Four cows are tethered at the four corners of a square field of side 50 m such that each can graze the maximum unshared area. What area will be left unglazed?    [Take \(\pi = 3.14\) ]

Sol:

https://lh4.googleusercontent.com/-6do9xuCwPwaZrpNMg-79HISfI6EBKXJWIdn0uivPK1VpJqQDBoMGRFPkovJ6alNX7XygpPw2cF9fTAf_SyUT6OPw3uQYgX0bx7BTLIJqVthpJDMgsBiVtSkQ03IOHyPhERXub3_

Each cow can graze a region that cannot be accessed by the cows.

Therefore, Radius of the region grazed by each cow \(= \frac{50}{2} = 25\)

Area that each cow grazes = \(\frac{1}{4} \pi r^{2}\)

\(\frac{1}{4} \times 3.14 \times 25 \times 25\)

\(= 490.625 cm^{2}\)

Total area grazed = \(4\times 490.625 = 1963.49 \; m^{2}\)

Area of the square = \((Side)^{2}\)

\(=(50)^{2}\) = 2500 \(cm^{2}\)

Now,

Area left ungrazed = Area of the square – Grazed area = 2500 – 1963.49 = 563.51 \(m^{2}\)

Q.16: In the given figure, OPQR is a rhombus, three of whose vertices lie on a circle with centre O. If the area of the rhombus is \(32\sqrt{3} \; cm^{2}\), find the radius of the circle.

                              https://lh3.googleusercontent.com/7MkcJEH4AkK30WUKejFPJFFU39qqAWi4gKLQWJMEAkFerQQi_81irX5kgpIoiBoZ148xRxAPbMjsYUW9R7kqqD_X2NLsKLR6ASJ2rsc9rATSE7oj1h-S0zF1O3BvlsQsktffakI5

Sol:

In a rhombus, all sides are congruent to each other.

Thus, we have:

OP = PQ = QR = RO

Now, consider \(\bigtriangleup QOP\)

OQ = OP (Both the radii)

Therefore, \(\bigtriangleup QOP\) is equilateral.

Similarly, \(\bigtriangleup QOR \) is equilateral and \(\bigtriangleup QOP \cong \bigtriangleup QOR\)

Ar. (QROP) = Ar. (\(\bigtriangleup QOP\)) + Ar. (\(\bigtriangleup QOR\)) = 2 Ar. (\(\bigtriangleup QOP\))

Ar. \((\bigtriangleup QOP) = \frac{1}{2} \times 32 \sqrt{3} = 16 \sqrt{3}\)

Or,

\(16\sqrt{3} = \frac{\sqrt{3}}{4} s^{2}\)   ( where s is the side of the rhombus )

Or,

\(s^{2} = 16 \times 4 = 64\)

\(s = 8\) cm

\(Therefore, OQ = 8\) cm

Hence, the radius of the circle is 8 cm.


Practise This Question

A mathematician trying to cross a street happened to witness a bank robbery. When the police questioned him, he stated that the number plate of the van in which the thieves escaped had its last four digits as follows:
The first digit is 5. The last digit is the square of the second digit. The third digit is twice the second digit. Also he noticed the sum of digits to be “9”. What is the quadratic equation that the police need to frame to find the 2nd digit, given that ‘b’ is the second digit?