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**Very short answer type questions : -**

**Q.1 In the fig., O is the centre of a circle. The area of sector OAPB is of the area of the circle. Find x. **

**[Delhi 2008]**

Area of sector OAPB =

Area of circle =

Given

Area of sector OAPB = area of circle

**Q.2 Find the perimeter of the given figure, where arc AED is a semicircle and ABCD is a rectangle. **

**[AI 2008]**

**Sol.**

AB + BC + CD = 20 + 14 + 20 = 54 cm

Diameter of semicircle = 14 cm

Radius of semicircle i.e. r = 7 cm

Permimeter of Arc AED = =

Hence, perimeter of given figure = cm

**Q.3 A chord of a circle of radius 14 cm subtends a right angle at the centre. What is the area of the minor sector ? **

**[Delhi 2008 C]**

**Sol.**

Area of sector = [Where ]

Hence, area of minor sector is .

**Q.4 What is the perimeter of a sector of angle 45° of a circle with radius 7 cm ? **

**[AI 2008 C]**

**Sol.**

Let O be the centre of the circle and be 45ºPerimeter of a sector OAB =

cm

Hence , perimeter of circle = 19.5 cm

**Q.5 If the diameter of a semicircular protractor is 14 cm, then find its perimeter. **

**[AI 2009]**

**Sol.**

Diameter of semicircle = 14 cm

radius(r) = 7 cmPerimeter of semicircle =

= 22 + 14

= 36 cm

Hence , perimeter of semicircular protractor is 36 cm.

**Short answer type questions - I**

**Q.1 In figure APB and CQD are semicircles of diameter 7 cm each, while ARC and BSD are semicircles of diameter 14cm each. Find the perimeter of the shaded region. [use ] **

**[Delhi 2011]**

Perimeter of semicircle APB =

cm

Perimeter of semicircle ARC =

cm

Perimeter of semicircle CQD =

cm

Perimeter of semicircle BSD =

Perimeter of shaded region = Perimeter of semicircle (APB + ARC + CQD + BSD)

= 66 cm

**Q.2 Find the perimeter of the shaded region in figure, if ABCD is a square of side 14 cm and APB and CPD are semicircles. [use ] **

**[AI 2011]**

Given : Side of square = 14 cm = diameter of semicircle(APB and CPD)

AB = BC = CD = DA = 14 cm

radius of semicircle(r)

Perimeter of semicircle DPC =

cm

Perimeter of semicircle APB =

cm

AD + BC = 14 + 14 = 28 cm

Perimeter of shaded region = AD + BC + Perimeter of semicircle (APB + DPC)

= 28 + 44 = 72 cm

Hence,the perimeter of the shaded region is 72 cm.

**Q.3 In given figure, a semicircle is drawn with O as centre and AB as diameter. Semicircles are drawn with AO and OB as diameters. If AB = 28 m, find the perimeter of the shaded region. [use ] **

**[AI 2011]**

Let radius of large semicircle be cm and radius of smaller semicircle be cm and cm.

Therefore = 14 cm, and = 7 cm

Perimeter of shaded region = [Perimeter of semicircle]

= 88 cm

Hence,the perimeter of the shaded region is 88 cm.

**Q.4 In given figure, ABC is a triangle right-angled at B, with AB = 14 cm and BC = 24 cm. With the vertices A, B and C as centres, arcs are drawn each of radius 7 cm. Find the area of the shaded region. [use ] **

**[AI 2011]**

Radius of each sector (r) = 7 cm

Let ,and be the angles of sectors (angle of triangle)

(Angle sum-property)

Area of shaded region = Area of – Area of 3 sectors

[Area of sector=]

[AB = 14 cm, BC = 24 cm(given)]

= 168 – 77

**Q.5 In the given figure, the shape of the top of a table is that a sector of a circle with centre O and . If AO = OB = 42 cm, then find the perimeter of the top of the table. [use ] **

**[Delhi 2012]**

Given : AO = OB = 42 cm (radius of circle)

Perimeter of the top of table = Length of major arc + 2r

[]

= 198 + 84 = 282 cm

**Q.6 Two circular pieces of equal radii and maximum area, touching each other are cut out from a rectangular card board of dimensions 14 cm × 7 cm. Find the area of the remaining cardboard. [use ] **

**[Delhi 2013]**

Let length of cardboard be l and height be h and radius of semicircle be r cm.

l = 14 cm , h = 7 cm and r = 7/2 cm

Area of remaining cardboard = Area of rectangle – Area of two semicircle

= 98 – 38.5 = 59.5

**Q.7 In the given figure, the area of the shaded region between two concentric circles is 286 . If the difference of the radii of the two circles is 7 cm, find the sum of their radii. [use ] **

**[Foreign 2013]**

Given : Area of shaded region = 286

Area of shaded region = Area of large circle – Area of small circle

Let radius of large circle be R

and radius of small circle be r

R-r = 7 cm [given]

Therefore area of shaded region =

286

286

cm

**Short answer type questions - II**

**Q.1 In the fig., find the perimeter of shaded region where ADC, AEB and BFC are semicircles on diameter AC, AB and BC respectively.**

**Or**

** Find the area of the shaded region in the fig., where ABCD is a square of side 1.4 cm. **

**[Delhi 2008]**

* Let * be the radius of semicircle ADC,AEB and BFC .

cm

cm

cm

Perimeter of shaded region = Perimeter of semicircles (ADC+AEB+BFC)

[Perimeter of semicircle = ]

= 13.2 cm

AB = BC = CD = DA = 1.4 cm(sides of square)

radius of circle (r) = 7/2 cm

Area of shaded region = Area of square – Area of 4 circles

= 196 – 4 × 38.5

= 196 – 154 = 42 .

**Q.2 In fig., ABC is a right-angled triangle, right-angled at A. Semicircles are drawn on AB, AC and BC as diameters. Find the area of the shaded region. **

**[AI 2008]**

Area of shaded region = Area of semicircle or side AB + Area of semicircle side AC – Area of semicircle on side BC + Area of .............. (1)

By using Pythagoras theorem in

[AB = 3 units ,AC = 4 units]

cm

Area of semicircle on side AB =

units .............. (2)

Area of semicircle on side AC =

units ............. (3)

Area of semicircle on side BC =

units ............... (4)

Area of

units ............. (5)

Put values from equation (1), (2), (3), (4) and (5), we get

Area of shaded region = = 6 sq. units

**Q.3 In the fig., ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region. **

**[Foreign 2008, Delhi 2008 C]**

Area of shaded = Area of semicircle of side BC – [Area of quadrant ABC – Area of ] .......... (1)

In

[AB = AC = 14 cm (given)]

cm

Area of semicircle on side BC =

[]

= 154 cm .............(2)

Area of =

............. (3)

Area of quadrant ABC =

................(4)

Put values from equation (2), (3) and (4) in equation (1)

Area of shaded region = 154 – [154 – 98]

= 154 – 154 + 98

= 98

**Q.4 Find the area of the shaded region in fig. If PR = 24 cm, PQ = 7 cm and O is the centre of the circle. **

**[Delhi 2008C, 2009]**

Given : PR = 24 cm , PQ = 7 cm

[Angle in the semicircle]

In

cm

Area of shaded region = Area of circle – [Area of semicircle of side QR] – [Area of ] .......(1)

Area of =

............ (2)

Area of circle =

...... (3)

Area of semi circle on side QR =

.............. (4)

Put values from equation (2), (3) and (4) in equation (1) -

Area of shaded region =

= 161.53

**Q.5 Find the area of the segment of a circle of radius 14 cm, if the length of the corresponding arc APB is 22 cm. [use ]**

** Or**

** A square OABC is inscribed in a quadrant OPBQ of a circle as shown in fig. If OB = 14 cm, find the area of the shaded region. [use ] **

**[AI 2008 C]**

1st method : -

Area of segment of a circle =

= 154 – 98

= 56

2nd method : -

Area of sector =

Area of

Area of segment = 154 – 98 = 56

** OR**

Area of quadrant

= 154

Let side of a square be x cm

Now, in

cm

Area of square = = 98

Area of shaded region = Area of quadrant OQBP - Area of square

= (154 - 98)

= 56

**Q.6 The area of an equilateral triangle in . Taking each angular point as centre, circles are drawn with radius equal to half the length of the side of the triangle. Find the area of triangle not included in the circles. [Take ] [AI 2009]**

* Sol.* Let side of an equilateral be a

Area of an equilateral

cm

Radius of circle cm

Area of sector =

Area of 3 sectors =

Area of not included in the circle = Area of – Area of 3 secotrs

**Q.7 In fig., the shape of the top of a table in restaurant is that of a sector of a circle with centre O and , if BO = OD = 60 cm find : **

**Sol. **

Given : Radius(r) =OB = OD = 60 cm** ** (i) Area of top of the table = Area of sector

= 8485

(ii) Perimeter of top of the table =

=282.85 + 120

= 402.6 cm

**Q.8 In figure the boundary of shaded region consists of four semicircular arcs, two smallest being equal. If diameter of the largest is 14 cm and that of the smallest is 3.5 cm, calculate the area of the shaded region. [use ]**

**Or**

** Find the area of the shaded region in figure, if AC = 24 cm, BC = 10 cm and O is the centre of the circle. **

**[Delhi 2010]**

Radius of largest semicircle = 7 cm

Radius of smallest semicircle = cm

Radius of middle semicircle = cm

Area of shaded region

= Area of largest semicircle – Area of 2 small semicircles + Area of middle semicircle ............ (1)

Area of largest semicircle =

.............. (2)

Area of smallest semicircle =

............ (3)

Area of middle semicircle =

.......... (4)

Put values from equation (2), (3) and (4) in equation (1), we get

Area of shaded region = 77 – 9.625 + 19.25

= 86.625

= 86.63

**OR**

Given : AC = 24 cm , BC = 10 cm

In

cm

Area of shaded region = Area of semicircle or side AB – Area of

[Where cm]

= 265.33 – 120

= 145.33

**Q.9 Find the area of the shaded region in figure, where a circular arc of radius 7 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm, as centre. [Foreign 2010]**

**Sol.**

Radius of circle = 7 cm

Area of shaded region = Area of circle – Area of sector + Area of an equilateral ........... (1)

Area of circle =

............ (2)

Area of sector =

........... (3)

Area of an equilateral where a is side of an equilateral ABC [Where a = 12 cm]

............ (4)

Put values from equation (2), (3) and (4) in equation (1), we get

Area of shaded region =

**Q.10 Find the area of the major segment APB, in figure of a circle of radius 35 cm and . [use ] **

**[Delhi 2011]**

Given : Radius of circle (r) = 35 cm

Area of circle =

Area of minor segement of circle =

[Since ]

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

**Q.11 In the given figure, O is the centre of the circle with AC = 24 cm, AB = 7 cm and . Find the area of the shaded region. [use ] **

**[AI 2012]**

[Given]

[Angle is semicircle]

In

[By Pythagoras Theorem]

cm

Area of shaded region = Area of circle – Area of – Area of sector COD ................ (1)

Area of circle =

[ Since r= cm]

.............. (2)

Area of

............... (3)

Area of sector

................ (4)

Put values from equation (2), (3) and (4) in equation (1)

Area of shaded region

**Q.12 In the given figure, AB and CD are two diameters of a circle with centre O, which are perpendicular to each other. OB is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region. [use ] **

**[Delhi 2013]**

Area of shaded region = Area of circle – Area of – Area of semicircle on side CD + Area of smaller circle ............... (1)

Area of circle =

............ (2)

Area of smaller circle =

.............. (3)

Area of = Area of + Area of

.................. (4)

Area of semicircle on side CD =

............ (5)

Put values from equation (2), (3), (4), (5) in equation (1), we get

Area of shaded region

**Q.13 In the given figure, from each corner of a square of side 4 cm, a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut. Find the area of the shaded region. [use ] **

**[Foreign 2013]**

Area of one quadrant

Therefore area of 4 quadrants =

Area of circle =

Area of square

= 4 ×4

= 16

Area of shaded region = Area of square – Area of circle – Area of 4 quadrants

= (16 – 3.14 – 3.14 )

= 16 - 6.28

**Long answer type questions : -**

**Q.1 ABC is a right triangle, right angled at A. Find the area of shaded region if AB = 6 cm, BC = 10 cm and O is the centre of the incircle of . [Take ] **

**[Delhi 2009]**

Given : AB = 6 cm , BC = 10 cm

Area of shaded region = Area of right angle ABC – Area of circle ............. (1)

Area of right angle ABC

............. (2)

Therefore in

cm

and [Because a tangent to the circle is perpendicular to the radius through the point of contact]

Therefore OQAR is a square

Let OR = x

Because OQAR is a square

Therefore OR = OQ = AQ = AB = x cm

BP = BQ [Because length of tangents drawn from a external point to a circle are equal]

and CR = CP [Because length of tangents drawn from a external point to a circle are equal]

Therefore BP = BQ = 6 – x

and CP = CR = 8 – x

and BC = CP + BP

10 = 8 – x + 6 – x

10 = 14 – 2x

2x = 4

x = 2 cm

Therefore radius of circle = 2 cm

Area of circle =

............ (3)

Put values from equation (2) and (3) in equation (1), we get

Area of shaded region = 24 – 12.56 = 11.44

**Q.2 In the given figure, three circles each of radius 3.5 cm are drawn in such a way that each of them touches the other two. Find the area of shaded region enclosed between these three circles. [use ] **

**[AI 2011]**

As radius of each circle is 3.5 cm

Therefore AC = 3.5 + 3.5 = 7 cm

BC = 3.5 + 3.5 = 7 cm

AB = 3.5 + 3.5 = 7 cm

ABC is an equilateral with side 7 cm.

Area of an equilateral

Area of one sector =

[ ia an angle of equilateral ]

Therefore area of 3 sectors =

Area of shaded region = Area of an equilateral – Area of 3 sectors

**Q.3 In given figure, an equilateral triangle has been inscribed in a circle of radius 6 cm. Find the area of the shaded region. [Use ] **

**[AI 2011]**

As ABC is an equilateral

Therefore [Because angle subitendend at centre is double the angle at alternate segment]

Now, in

OB = OC [Radii of circle]

[Because angle opposite to equal sides are equal]

Now in

Now, in

cos30°=

[ OB is radius of circle]

cm

As

Therefore BM = CM [Because perpendicular from the centre to the chord bisects the chord]

cm

Area of an equilateral =

Area of circle =

Area of shaded region = Area of circle – Area of an equilateral

**Q.4 From a thin metallic piece, in the shape of a trapezium ABCD in which AB || CD and , a quarter circle BFEC is removed (See figure). Given AB = BC = 3.5cm and DE = 2 cm, calculate the area of the remaining (shaded) part of the metal sheet. [use ] **

**[Foreign 2011]**

AB = BC = 3.5 cm [Given]

AE = BC = 3.5 cm

Area of shaded region = Area of trapezium – Area of quarter circle