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:

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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

State true or false.
The splitting of white light into its component colors is called dispersion.