Extra Questions for Class 10 Maths Areas Related to Circles with Answers
Extra Questions for Class 10 Maths Chapter 12 Areas Related to Circles. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.
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Areas Related to Circles Class 10 Extra Questions Very Short Answer Type
Question 1.
Find the ratio of the area of a circle and equilateral triangle whose diameter and side are respectively equal.
Answer:
Question 2.
Find the area of a ring whose internal and external radii are 15 cm and 20 cm respectively.
Answer:
r = 15 cm, R = 20 cm
Area of ring = π(R2 – r2)
= \(\frac{22}{7}\)[(20)2 – (15)2]
= \(\frac{22}{7}\) × 175 = 550 cm2
Question 3.
A wire is looped in the form of a circle of r = 28 cm. If it is rebent into a square form, then find the side of the square.
Answer:
Circumference of circle = Perimeter of square
2πr = 4a
2 × \(\frac{22}{7}\) × 28 = 4a
⇒ a = 44 cm
Question 4.
Find the diameter of a circle whose area is I equal to the stun of the areas of the two circles of radii 24 cm and 7 cm.
Answer:
According to the question,
πR2 = πr12 + πr22
R2 = r12 + r22
R2 = (24)2 + (7)2
R2 = 576 + 49
R = \(\sqrt{625}\) = 25 cm
∴ Diameter of circle = 2R = 2 × 25 = 50 cm
Question 5.
Find the area of square that can be inscribed in a circle of r = 7 cm.
Answer:
d = 2r = 7 × 2 = 14 cm
AC = 14 cm
Let side of square = a
In ΔABC, AC2 = AB2 + BC2
(14)2 = a2 + a2
a2 = 196
a2 = 98
a = (\(\sqrt{2}\))2 = 98 cm2
Question 6.
Find the area of circle that can be inscribed in a square of side 24 cm.
Answer:
Side = 24 cm
Diameter of circle = Side = 24 cm
Radius of circle = \(\frac{24}{2}\)
= 12 cm
Area of circle = πr2
= π × (12)2 = 144 π cm2
Question 7.
The diameter of a circular field is 70 m. Find the number of rounds that a person is required to go around the field so as to cover a distance of 1540 m.
Answer:
Diameter of circular field (d) = 70 m
Radius of circular field (r) = 35 m
Perimeter of circular field = 2 πr
= 2 × \(\frac{22}{7}\) × 35 = 220 m
Total distance = 1540 m
No. of rounds = \(\frac{1540}{220}\) = 7 rounds
Question 8.
A path of width 7 m runs around a circular park whose radius is 18 m. Find the area of path.
Answer:
Raidus of park (r) = 18 m
Width of path = 7 m
Radius of outer circle (R) = 18 + 7 = 25 m
Area of path = πR2 – πr2
= π(R2 – r2)
= \(\frac{22}{7}\)[(25)2 – (18)2]
= \(\frac{22}{7}\)[625 – 324]
= \(\frac{22}{7}\) × 301 = 946 m2
Areas Related to Circles Class 10 Extra Questions Short Answer Type-1
Question 1.
Find the area of circle whose circumference is 22 cm. [CBSE 2019 SQP (Basic)]
Answer:
Circumference = 22
2πr = 22
\(\frac{2 \times 2 \times 2 r}{7}\) = 22
r = \(\frac{7}{2}\)
Area of circle = πr2
= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}=\frac{77}{2}\) cm2
Question 2.
Prove that tangents drawn at the ends of a diameter of a circle are parallel to each other. [CBSE Delhi 2017]
Answer:
Given: A circle cord with AOB as diameter. PQ and RS are tangents at A and B respectively.
To prove: PQ ∥ RS
Proof: ∠PAO = 90° [∵ radius ⊥ tangent]
∠OBS = 90°
⇒ ∠PAO = ZOBS
But these are alternate interior angles
∴ PQ ∥ RS
Hence Proved
Question 3.
A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.
Answer:
Here radius (r) = 25 cm
θ = 115°
As number of wipers is 2
So area cleaned = 2 × (Area of sector)
= 2 × \(\frac{\pi r^2 \theta}{360^{\circ}}\)
= 2 × \(\frac{22 \times 25 \times 25 \times 115}{7 \times 360}\)
= \(\frac{158125}{126}\) cm2
= 1255 cm2
Question 4.
The area of an equilateral triangle is 100√3 cm2. Taking each vertex as centre of a circle which is described with 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).
Answer:
Since area of equilateral triangle = \(\frac{\sqrt{3}}{4}\)a2
⇒ 100√3 = \(\frac{\sqrt{3}}{4}\)BC2
⇒ BC2 = 400
⇒ BC = 20 cm
Now radius of each circle = \(\frac{\mathrm{BC}}{2}=\frac{20}{2}\) = 10 cm
Reqd. Area = Area of ΔABC – 3 × area of sector with central angle 60°
= 100√3 – 3 × \(\frac{\pi r^2}{360^{\circ}}\)θ
= 100(1.732) – \(\frac{3 \times 3.14 \times 10 \times 10 \times 60}{360}\)
= 173.2 – 157
= 16.2 cm2
Question 5.
Find the area of the largest triangle that can be inscribed in a semicircle of radius r cm.
Answer:
Base of largest triangle = AB = diameter of semicircle 2r
For the largest area, its third vertex should be at C, so that its height = CO = r
∴ Area = \(\frac{1}{2}\) (AB) (CO) = \(\frac{1}{2}\)(2r)r = r2cm2
Question 6.
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 described by minute hand in 5 minutes = \(\frac{360}{60}\) × 5 = 30°
Now radius of sector generated (r) = 14 cm
Required area = Area of sector with central angle 30°
= \(\frac{\pi r^2 \theta}{360^{\circ}}=\frac{22 \times 14 \times 14 \times 30}{7}=\frac{154}{3}\)cm2
= 51.33 cm2.
Areas Related to Circles Class 10 Extra Questions Short Answer Type-2
Question 1.
Find the area of the shaded region in fig., if ABCD is a rectangle with sides 8 cm and 6 cm and O is the centre of circle. (Take π = 3.14) [CBSE 2019]
Answer:
ΔABC is a right-angled triangle.
∴ By Pythagoras Theorem
AC2 = AB2 + BC2
⇒ AC = \(\sqrt{64+36}\) = 10 cm.
∴ Radius of the circle (r) = \(\frac{\mathrm{AC}}{2}\) = 5 cm
Area of shaded region
= Area of circle – ar(ABCD) = πr2 – l × b
= 3.14 × (5)2 – 6 × 8
= 3.14 × 25 – 6 × 8
= 78.5 – 48 = 30.5 cm2
Question 2.
In figure, ABC is a right-angled triangle at A. Semi-circles are drawn on AB, AC and BC as diameters. Find the area of the shaded region. [CBSE 2019 (c)]
Answer:
Area of semicircle with diameter AB,
ar(i) = \(\frac{\pi}{2}\left(\frac{6}{2}\right)^2=\frac{9 \pi}{2}\) ……..(i)
Area of semicircle with diameter AC,
ar(ii) = \(\frac{\pi}{2}\left(\frac{8}{2}\right)^2\) = 8Ttcm2 ……….(ii)
Area of semicircle with diameter BC,
ar(iii) = \(\frac{\pi}{2}\left(\frac{10}{2}\right)^2=\frac{25 \pi}{2}\)cm ……..(iii)
[∵ BC = \(\sqrt{\mathrm{AB}^2+\mathrm{AC}^2}=\sqrt{6^2+8^2}\) = 10 cm]
Area of ΔABC, ar (iv) = \(\frac{1}{2}\) × AB × AC
= \(\frac{1}{2}\) × 6 × 8 = 24cm2
∴ Area of shaded region = (ar(i) + ar(ii) + ar(m)} – ar(in)
Question 3.
In the given 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 shaded region. [CBSE 2017]
Answer:
Area shaded Region = {Area of quadrant OACB) – {Area of AODB}
Question 4.
In fig. PSR, RTQ and PAQ are three semicircles of diameters 10 cm, 3 cm and 7 cm respectively. Find the perimeter of the shaded region. [Use π = 3.14]
Answer:
PQ = 7 cm, QR = 3cm, PR = 3 + 7 = 10 cm
d1 = 7 cm, d2 = 3 cm, D = 10 cm, r1 = 3.5 cm, r2 = 1.5 cm, R = 5 cm
Perimeter of shaded region = Perimeter of semicircle PSR+ Perimeter of semicircle PAQ + Perimeter of semicircle QTR.
= πR + πr1 + πr2
= π[R + r1 + r2]
= 3.14 [5 + 3.5 + 1.5]
= 3.14 × 10 = 31.4 cm
Question 4.
In fig. PSR, RTQ and PAQ are three semicircles of diameters 10 cm, 3 cm and 7 cm respectively. Find the perimeter of the shaded region. [Use π = 3.14]
Answer:
Side of equilateral ΔPQR = 8 cm
∴ Area of equilateral ΔPQR = \(\frac{\sqrt{3}}{4}\) × (Side)2
= \(\frac{\sqrt{3}}{4}\) × 8 × 8 = 1.732 × 16
= 27.712 cm2
∴ Area of each sector of (0 = 60 )
and radius = 4 cm
= \(\frac{60}{360}\) × πr2 = \(\frac{1}{6}\) × 3.14 × 4 × 4
= \(\frac{1}{6}\) × 3.14 × 16cm2
∴ Area of the shaded region
= Area of ΔPQR – Area of 3 Sectors
= 27.712 – 25.12
= 2.592 cm2
Question 7.
In figure, ABCD is a square of side 14 cm. Semi-circles are drawn with each side of square as diameter. Find the area of the shaded region, (use n = y )[CBSE Delhi 2016]
Answer:
Let the four shaded regions be I, II, III and IV and the centres of the semicircles be P, Q, R, and S, as shown in the figure. It is given that the side of the square is 14 cm.
Now,
Area of region I + Area of region III = Area of the square – Areas of the semicircles with centres S and Q.
= 14 × 14 – 2 × \(\frac{1}{2}\) × π × 72
(∵ Radius of the semicircle = 7 cm)
= 196 – 49 × \(\frac{22}{7}\) = 196 – 154 = 42 cm2
Similarly, ar (II) + ar (IV) = Area of the square – Area of the semicircles with centres P and R.
= 14 × 14 – 2 × \(\frac{1}{2}\) × π × 72
(∵ Radius of the semicircle = 7 cm)
= 196 – 49 × \(\frac{22}{7}\)
= 196 – 154 = 42 cm2
Thus,
Area of the shaded region = ar (I) + ar (II) + ar (III) + ar (IV)
= 42 cm2 + 42 cm2 = 84 cm2
Question 8.
In figure, O is the centre of a circle such that diameter AB = 13 cm and AC = 12 cm. BC is joined. Find the area of the shaded region. (Take π = 3.14) [CBSE Outside Delhi 2016]
Answer:
∠BCA = 90° [angle in semicircle]
∴ In right angled ΔABC
AB2 = BC2 + AC2
⇒ 132 = BC2 + 122
⇒ BC = 5
Area of shaded region
= ar (semicircle) – ar (ΔABC)
= \(\frac{\pi r^2}{2}-\frac{1}{2}\) × BC × AC
= \(\frac{3.14}{2} \times \frac{13}{2} \times \frac{13}{2}-\frac{1}{2}\) × 5 × 12
= 66.3325 – 30
= 36.33 cm2.
Question 9.
In figure are shown two arcs PAQ and PBQ. Arc PAQ is a part of circle with centre O and radius OP while arc PBQ is a semi-circle drawn on PQ as diameter with centre M. If OP = PQ = 10 cm show that area of shaded region is 25(√3 – \(\frac{\pi}{6}\))cm2 [CBSE 2016]
Answer:
Given OP = OQ = 10 cm
Tangents drawn from an external point to a circle are equal in length.
OP = OQ = 10 cm
Therefore, AABC is an equilateral triangle.
⇒ ∠POQ = 60°
Now, Area of part II = Area of the sector – Area of the equilateral triangle POQ
Area of the semicircle on diameter PQ = ar(II) + ar(III)
Question 10.
In figure, find the area of the shaded region, enclosed between two concentric circles of radii 7 cm and 14 cm where ∠AOC = 40°. (use π = \(\frac{22}{7}\)) [CBSE Outside Delhi 2016]
Answer:
Area of shaded region = Area of Ring – area (BDAC)
Question 11.
Following figure depicts a park where two opposite sides are parallel and left and right ends are semi-circular in shape. It has a 7 m wide track for walking. Two friends Seema and Meena went to the park. Meena said that area of the track is 4066 m2. Is she right? Explain. [CBSE SQP 2019 (Basic)]
Answer:
Let us mark various points of two rectangles, outer and inner rectangles as P, Q R, S and A, B, C, D.
AD = PS – (DS + AP)
= 70 – (7 + 7) = 56 m
Outer radius of ring = \(\frac{70}{2}\) = 35 m
BC = AD = 56 m
∴ Inner radius of each of two semicircular rings
Now area of track = ar (outer rectangle PQRS) – ar (inner rectangle ABCD) + area of semicircular ring PSDA + area of semicircular ring QRCB
= (120 × 70) – (120 × 56) + \(\frac{\pi}{2}\)[(35)2 – (28)2]
= 8400 – 6720 + \(\frac{\pi}{2}\) [(35 – 28)(35 + 28)]
= 1680 + \(\frac{22}{7}\) [7 × 63]
= 1680 + 1386
= 3066 m2
Yes; Meena is wrong.
Areas Related to Circles Class 10 Extra Questions Long Answer Type 1
Question 1.
A chord PQ of a circle of radius 10 cm subtends an angle of 60° at the centre of circle. Find the area of major and minor segments of the circle. [CBSE Delhi 2017]
Answer:
Area of minor segment = \(\frac{\pi r^2 \theta^{\circ}}{360^{\circ}}\) ar ∆OPQ
= \(\frac{\pi r^2 \times \theta^{\circ}}{360^{\circ}}-\frac{\sqrt{3} a^2}{4}\)
Area of major segment = area of circle – area of minor segment
Question 2.
In the given fig., AB and PQ are perpendicular diameters of the circle whose centre is O and radius OA = 7 cm. Find the area of the shaded region [Use π = \(\frac{22}{7}\)]
Answer:
Diameter of the smaller circle = 7 cm 7
Area of shaded region
= {Area of semicircle – Area of ∆APB] + [Area of circle]
= {77 – 49} + 38.5
= 28 + 38.5 = 66.5 cm2.
Question 3.
Find the area of the shaded region in the fig. where ABCD is a square of side 10 cm and semicircles are drawn with each side of the square as diameter. [Take π = 3.14]
Answer:
Mark the four unshaded regions as I, II, III and IV respectively as shown in figure.
Let radius of each semicircle be r.
∴ r = \(\frac{10}{2}\) = 5 cm
Area of each semicircle = \(\frac{1}{2}\) πr2
= \(\frac{1}{2}\) × 3.14 × 5 × 5 = 39.25 cm2
Area of square = a2 = 102 = 100 cm2
Area of I + Area of II = Area of square ABCD – Area of two semicircles
= 100 – 2 × 39.25
= 100 – 78.5 = 21.5 cm2
Similarly, Area of III + Area of IV = 21.5 cm2
∴ Area of the shaded portion = Area ABCD – Area of regions (I + II + III + IV)
= 100 – 2 × 21.5
= 100 – 43 = 57 cm2.
Question 4.
It is proposed to add to a square lawn with each side 50 m, 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]
Answer:
Let ABCD be a square lawn whose side is 50 m. Diagonals of this lawn intersect at O. Then, by geometry
OA = OB = OC = OD
and ∠AOD = ∠BOC = 90°
With O as centre and radius equal to \(\frac{1}{2}\) AC = \(\frac{1}{2}\)
BD, \(\widehat{\mathrm{AED}}\) and \(\widehat{\mathrm{BFC}}\) have been drawn.
Now diagonal
AC = √2 AB = √2 (50) = 50 √2m
∴ Radius of each arc = \(\frac{1}{2}\) (diagonal) = 25 √2
Area of whole lawn = (Area of the sq. ABCD) + 2
(Area of segment with r = 25√2 and θ = 90°)
Question 5.
Find the area of the shaded portion shown in figure. The four corners are quadrants and at the centre there is a circle. [Take π = \(\frac{22}{7}\)]
Answer:
Area of square ABCD = (4)2 = 16 cm2
Sum of areas of 4 quadrants at each corner of square = Area of the circle at the centre of square ABCD
= πr2 = n(1)2 = π cm2
Thus area of shaded portion = Area of square ABCD – {Area of four quadrants + Area of circle at the centre of square}
= (4)2 – (π + π)
= 16 – 2π = 16 – 2 × \(\frac{22}{7}\)
= 16 – 6.28 = 9.72 cm2
Question 6.
Three semicircles each of diameter 3 cm, a circle of diameter 4.5 cm and a semicircle of radius 4.5 cm are drawn in the given figure. Find the area of the shaded region. [A.I. CBSE-2017]
Answer:
Let us mark various regions of given figure by P, Q, R, A, B, C & D as shown below:
Question 7.
In fig., a circle is inscribed in an equilateral triangle ABC of side 12 cm. Find the radius of inscribed circle and the area of the shaded region. [User π = 3.14]
Answer:
Area of shaded region = Area of ∆ABC – Area of circle
= 62.28 – πr2 = 62.28 – 3.14 × 2√3 × 2√3
= 62.28 – 37.68 = 24.6 cm2.
Question 8.
In the adjacent figure, ABCD is a trapezium with AB || DC and ∠BCD = 60°. If BFEC is a sector of a circle with centre C and AB = BC = 7 cm and DE = 4 cm, find the area of the shaded 22 region. (Use π = \(\frac{22}{7}\) and √3 = 1.732]
Answer:
For trapezium ABCD
AB = 7 cm, DC = DE + EC = 4 + 7 = 11 cm
Now Area of trapezium = \(\frac{1}{2}\) (AB + DC) × BL
= \(\frac{1}{2}\) (7 + 11) × 6.062 = 9 × 6.062 = 54.558 cm2
For sector θ = 60°
Area of sector ECBFE = \(\frac{60}{360}\) × π(7)2
= \(\frac{1}{6} \times \frac{22}{7}\) × 49 = \(\frac{11}{3}\) × = \(\frac{77}{3}\) = 25.667 cm2
Area of shaded region = 54.558 – 25.667
= 28.891 cm2.
Areas Related to Circles Class 10 Extra Questions HOTS
Question 1.
A wire in the shape of a regular hexagon encloses an area 726 √3 cm2. If the same wire is bent to form a circle. Find the area of the circle.
Answer:
Let the regular hexagon be named as PQRSTU and PQ = QR = RS = ST = TU = UP = a cm
Area of regular hexagon PQRSTU = 6 (area of equilateral ∆POQ)
Question 2.
In a circle with centre at O and radius 8 cm, AB is a chord of length 8√3 cm. Find the area of the sector AOB.
Answer:
Here chord AB = 8√3 cm. radius OA = OB = 8 cm.
Draw OM ⊥ AB as perpendicular from centre of circle bisects the chord.
∴ AM = BM = \(\frac{1}{2}\)(8√3) = 4√3 cm
Let ∠AOB = 2θ
⇒ ∠AOM = ∠BOM = θ
Question 3.
A circle is inscribed in a given square and another circle is circumscribed about the square. What is the ratio of the areas of inscribed circle to the circumscribed circle?
Answer:
Both inscribed and circumscribed circles are concentric whose centre lies at point of intersection of diagonals of the square. Let ‘a’ be the side of the square.
∴ Required ratio = π(OE)2 : π(OA)2
Question 4.
Find the area enclosed between two concentric circles of radii 3.5 cm, 7 cm. A third concentric circle is drawn outside the 7 cm circle so that the area enclosed between it and the 7 cm circle is same as that between two inner circles. Find the radius of the third circle.
Answer:
Area between first two circles
= π × 72 – π × 3.52
= 49π – 12.25π ……. (i)
Area between next two circles
= πr2 – π × 72
= πr2 – 49π …(it)
(i) and (ii) are equal,
49π – 12.25π = πr2 – 49π
πr2 = 49π + 49π – 12.25π
∴ r2 = 98 – 12.25 = 85.75
r2 = \(\frac{8575}{100}=\frac{343}{4}\)
r = \(\frac{\sqrt{343}}{2}\) cm
(i) ⇒ Area between first two circles
= 49π – 12.25 π = \(\frac{22}{7}\) (49 – 12.25)
= \(\frac{22}{7}\) × 36.75 = 22 × 5.25 = 115.5 cm2.
Question 5.
Two circles touch externally. The sum of their areas is 130n sq. cm and the distance between their centres is 14 cm. Find the radii of the circle.
Answer:
Let r be R be the radii of two circles respectively. Distance between two centres = 14 cm (Given)
r + R = 14 cm
r = (14 – R) ……… (1)
According to question,
(Area of circle having radius r) + (Area of circle having radius R) = 130π cm2
πr2 + πR2 = 130 π
r2 + R2 = 130
(14 – R)2 + R2 = 130
196 + R2 – 28R + R2 = 130
2R2 – 28R + 66 = 0
R2 – 11R – 3R + 33 = 0
R2 – 11R – 3R + 33 = 0
R (R – 11) – 3 (R – 11) = 0
(R – 11) (R – 3) = 0
R = 3, 11
When R = 3, r = 14 – 3 = 11 cm.
When r = 11 then r = 14 -11 = 3 cm
Hence, radii of circles are 3 cm, 11 cm or 11 cm, 3 cm.
Multiple Choice Questions
Choose the correct option for each of the following:
Question 1.
The circumference of a circle is 100 cm. The side of a square inscribed in the circle is:
(a) 50√2
(b) \(\frac{50 \sqrt{2}}{\pi}\) cm
(c) \(\frac{50 \sqrt{2}}{\pi}\) cm
(d) \(\frac{100 \sqrt{2}}{\pi}\) cm
Answer:
(c) \(\frac{50 \sqrt{2}}{\pi}\) cm
Question 2.
The diameter of a circle whose area is equal to the sum of areas of the two circles of radii 40 cm and 9 cm is:
(a) 41 cm
(b) 49 cm
(c) 82 cm
(d) 62 cm
Answer:
(c) 82 cm
Question 3.
If the circumference of a circle of radius V and the perimeter of a square of side ‘a’ are equal, then the ratio of area of the circle to that of the square is:
(a) 4: π
(b) π: 4
(c) π2:16
(d) π2:4
Answer:
(a) 4: π
Question 4.
If the area and circumference of a circle are numerically equal, then the diameter of the circle is:
(a) 3 units
(b) 5 units
(c) 4 units
(d) 2 units
Answer:
(c) 4 units
Question 5.
The area of circle that can be inscribed in a square of side 6 cm is:
(a) 36 π cm2
(b) 18 π cm2
(c) 12 π cm2
(d) 9 π cm2
Answer:
(d) 9 π cm2
Question 6.
If the circumference of a circle and the perimeter of a square are equal, then:
(a) Area of the circle = Area of the square
(b) Area of the circle > Area of the square
(c) Area of the circle < Area of the square (d) Nothing definite can be said about the relation between the areas of the circle and square Answer: (b) Area of the circle > Area of the square
Question 7.
The area of the square that can be inscribed in a circle of radius 8 cm is :
(a) 256 cm2
(b) 128 cm2
(c) 64√2 cm2
(d) 64 cm2
Answer:
(b) 128 cm2
Question 8.
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:
(a) 56 cm
(b) 42 cm
(c) 28 cm
(d) 16 cm
Answer:
(c) 28 cm
Question 9.
Two parallel lines touch the circle at points A and B respectively. If area of the circle is 25 p cm2. Then AB is equal to:
(a) 5 cm
(b) 8 cm
(c) 10 cm
(d) 25 cm
Answer:
(c) 10 cm
Question 10.
The area of a circle whose circumference is 44 cm is:
(a) 152 cm2
(b) 153 cm2
(c) 154 cm2
(d) 150 cm2
Answer:
(c) 154 cm2
Question 11.
It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameter 16 m and 12 m in a locality. The radius of the new park would be:
(a) 10 m
(b) 15 m
(c) 20 m
(d) 24 m
Answer:
(a) 10 m
Question 12.
If the sum of the areas of two circles with radii r1 and r2 is equal to the area of a circle of radius r, then:
(a) r1 + r2 = r
(b) r1 + r2 = r2
(c) r1 + r2 < r
(d) r12 + r22 < r2
Answer:
(b) r1 + r2 = r2
Question 13.
If the sum of the circumferences of two circles with radii and r2 is equal to the circumference of a circle of radius r, then:
(a) r1 + r2 = r
(b) r1 + r2 > r
(c) r1 + r2 < r
(d) Nothing definite can be said about the relation among r1, r2 and r.
Answer:
(a) r1 + r2 = r
Question 14.
Area of the largest triangle that can be inscribed in a semicircle of radius r units is :
(a) r2 sq.units
(b) r2 sq.units
(c) √2 r2 sq.units
(d) r2 sq.units
Answer:
(a) r2 sq.units
Question 15.
The area of sector of central angle ‘x°’ of a circle with radius ‘4r’ is:
(a) \(\frac{4 \pi x}{360}\)
(b) \(\frac{2 \pi x r^2}{45}\)
(c) \(\frac{\pi r^2 x}{360}\)
(d) \(\frac{2 \pi r x}{360}\)
Answer:
(b) \(\frac{2 \pi x r^2}{45}\)
Fill in the Blanks
Question 1.
Area of a circle having circumference 39.6 cm is ___________
Answer:
124.74 cm2
Question 2.
If the area of a circle is 154 cm2, then its perimeter is ___________.
Answer:
44 cm
Question 3.
Area of a quadrant of a circle having circumference 44 cm is ___________.
Answer:
38.5 cm2
Question 4.
The distance travelled by a road roller of radius r and length l in 20 rotations is ___________.
Answer:
40 π r
Question 5.
The areas of two circular fields are in the ratio 16: 49. If the radius of the bigger circle is 14 cm, then radius of the smaller circle is ___________.
Answer:
8 cm
Question 6.
Ratio of the areas of a circle and a rectangle, such that diameter of circle and diagonal of rectangle are equal is ___________.
Answer:
275 : 168
Question 7.
If the perimeter and area of a circle are numerically equal, then the radius of the circle is ___________.
Answer:
2 units
Question 8.
Area of the largest triangle that can be inscribed in a semicircle of radius r is ___________.
Answer:
r2 sq.units
Question 9.
If the perimeter of a circle is equal to that of a square, then the ratio of their areas is ___________.
Answer:
14: 11
Question 10.
If the length of an arc subtending an angle of 72° at the centre is 44 cm, then the area of the circle is ___________
Answer:
a(1 + \(\frac{\pi}{2}\)) units