Question 1: A bucket is raised from a well by a means of a rope which is wound around a wheel of diameter 77 \ cm . Given that the bucket ascents in 1 minute 28 seconds with a uniform speed of 1.1 \frac{m}{s} , calculate the number of complete revolution the wheel will make in raising the bucket.    [1997]

Answer:

Radius of the wheel r = \frac{77}{2} = 38.5 \ cm

Length of the rope = 1.1 \frac{m}{s} \times 88 = 96.8 m

Circumference of the wheel = 2 \pi r = 2 \times \frac{22}{7} \times 38.5 = 242 cm = 2.42 \ m

Therefore No of revolutions = \frac{96.8}{2.42} = 40

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Question 2: The wheel of a cart is making 5 revolutions per second. If the diameter of the wheel is 84 \ cm , find the speed in \frac{km}{hr} . Give your answer, correct to the nearest km.    [1998]

Answer:

No of revolutions = 5 / sec

Radius of the wheel = \frac{84}{2} = 42 cm

Speed = 2 \pi r \times 5 \frac{cm}{s}

= 2 \times \frac{22}{7} \times 42 \times 5 = 1320 \frac{cm}{s}

= 1320 \times \frac{3600}{100000} \frac{km}{hr} = 47.52 \frac{km}{hr} = 48 \frac{km}{hr}

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211Question 3: In the given figure. AB is the diameter of the circle with center O and OA = 7 \ cm . Find the area of the shaded region.    [2006]

Answer:

OA = 7 \ cm, AB = 14 \ cm and OB = 7 \ cm

Area of smaller circle = \pi (\frac{7}{2})^2 = \frac{22}{7} \times  (\frac{7}{2})^2 = 38.5 \ cm^2

Area of the triangle = \frac{1}{2} \times base \times height = \frac{1}{2} \times 14 \times 7 = 49 \ cm^2

Area of semi-circle = \frac{1}{2} \pi (7)^2 = 77 \ cm^2

Therefore the area of the shaded area = 38.5 + 77 - 49 = 66.5 \ cm^2

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212Question 4: AC and BD are two mutually perpendicular diameters of a circle ABCD . Given the area of the shaded region is 308 Calculate the (i) the length of AC and (ii) the circumference of the circle [Take \pi = \frac{22}{7}    [2009]

Answer:

Given Area of the shaded region : \frac{1}{2} \times \pi r^2 = 308

\Rightarrow r = 2 \times \frac{7}{22} \times 308 = 14 \ cm

(i) Therefore AC = 28 \ cm

(ii) Circumference of the circle =2 \pi (14) = 28 \ cm

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2010-3Question 5: A doorway is decorated as shown in the figure. There are four semi-circles. BC , the diameter of the larger Semi-circles. BC the diameter of the larger semi-circle is of length 84 \ cm . Center of the three equal semi-circles lie on BC . ABC is an isosceles triangle with AB = AC . If BO = OC , find the area of the shaded, region.   [2010]

Answer: 

Let AB = AC = x \ cm

As angle in semi-circle is 90^o

\therefore \ \angle A = 90^o

In right angled \triangle ABC , by Pythagoras theorem, we get

AB^2+AC^2 = BC^2

x^2+x^2 = 84^2

2x^2 = 84 \times 84

x^2 = 84 \times 42

Also, Area of \triangle ABC = \frac{1}{2} \times AB \times AC

= \frac{1}{2} \times 84 \times 42 = 1764 \ cm^2

Diameter of semicircle (2r) = 84 \ cm

Radius(r) = \frac{1}{2} \times 84  = 42 \ cm

Area of semi-circle = \frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 42 \times 42 = 2772 \ cm^2

Diameter of each (three equal) semi-circles = \frac{1}{3} \times 84 = 28 \ cm

Radius of the 3 equal semi-circles = \frac{1}{2} \times 28 = 14 \ cm

Therefore Area of three equal semi circles = \frac{1}{3} \pi r^2 = \frac{1}{3} \times \frac{22}{7} \times 14 \times 14 = 924 \ cm^2

Area of shaded region = Area of semi-circles + Area of three equal circles – Area of \triangle ABC

= 2772 + 924 - 1764 = 3696 - 1764 = 1932 \ cm^2

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213Question 6: The shaded part of the given figure shows the shape of the top of a table in a restaurant which is a segment of a circle with center O , \angle BOD = 90^o and BO = OD = 60 \ cm . Find (i) the area of the top of the table (ii) the perimeter of the table (\pi = 3.14 ).    [2002]

Answer:

Area of the table = \frac{3}{4} \times \pi (60)^2 

= \frac{3}{4} \times 3.14 \times (60)^2 = 8478 \ cm^2 

Perimeter of the table = \frac{3}{4} \times 2 \pi (60) +OD + OB 

=  \frac{3}{4} \times 2 \times 3.14 \times (60) +60 + 60 

= 282.6+ 120 = 402.6 \ cm 

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214Question 7: Calculate the area of the shaded portion. The quadrants shown in the figure are each of radius 7 \ cm    [2000]

Answer:

Area of the shaded region = (2r) \times (2r) - 4 \times (\frac{1}{4}  \times \pi r^2) 

= 14 \times 14 - 4 \times (\frac{1}{4}  \times \pi \times 7^2) 

= 196 - 154 = 42 \ cm^2 

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icse 14 2Question 8:  In the figure given below ABCD  is a rectangle AB=14 \ cm, \ BC=7\ cm . From the rectangle a quarter circle BFEC  and a semicircle DGE  are removed Calculate the area of the remaining piece of the rectangle (Take \pi = \frac{22}{7} ) [2014]

Answer: 

Area of rectangle ABCD= 14 \times  7 = 98cm

Area of quarter circle BFEC=\frac{1}{4}\pi (7^{2})=\frac{49}{4}\pi

Area of semicircle DGE= \frac{1}{2}\pi (\frac{7}{2})^2=\frac{1}{2}\times \frac{49}{4}\pi 

Area of remaining piece of rectangle = 98-[\frac{49}{4}\pi +\frac{1}{2}\times \frac{49}{4}\pi ]

= 98-\frac{49}{4}\pi \lbrack 1+\frac{1}{2}\rbrack 

= 98-\frac{49}{4}\times \frac{22}{7}\times \frac{3}{2}=98-\frac{231}{4}

= 98-57.75 = 40.25 cm^2

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Question 9: The given figure shows a running track surrounding a grass enclosure PQRSTU . The enclosure consists of a rectangle PQST with a semicircular region at each end. PQ = 200 \ m and PT = 70 \ m . (i) calculate the area of the grass enclosure in m^2 (ii) given that the track is of constant width 7 \ m , calculate the outer perimeter ABCDEF of the track.    [1999]215

Answer:

PQ = 200 \ m, PT = 70 \ m

Radius of the small semi-circle = 35 \ m

Radius of the large semi-circle = 35 + 7 = 42 \ m

(i) Area of grass = \pi (35)^2 + 70 \times 200 = 3850+ 14000 = 17850 \ m^2

(ii) Outer perimeter = 2\pi (42) + 200 + 200 = 264 + 400 = 664 \ m

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Question 10: A rectangular playground has two semi circles added to the outside with its smaller sides as diameters. If the sides of the rectangle are 120 \ m and 21 \ m , find the area of the playground (\pi = \frac{22}{7} ).    [2000]

Answer:

Area of the playground = \pi (10.5)^2 + 21 \times 200 = 346.5 + 2520 = 2866.5 \ m^2

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Question 11:   16-10In the figure along side OAB is a quadrant of a circle, The radius OA=3.5 \ cm \ and \  OD=2 \ cm , Calculate the area of the shaded portion. (Take \pi = \frac{22}{7} ) [2013]

Answer: 

Radius of quadrant OACB, r=3.5 \ cm

Area of quadrant OACB=\frac{1}{4} \pi r^2 = \frac{1}{4}. \frac{22}{7}. (3.5)^2 = 9.625 \ cm^2

Here, \angle AOD=90^o

base =3.5 \ cm and height =2 \ cm

Then area of \triangle AOD = \frac{1}{2} \times base \times height = \frac{1}{2} \times 3.5 \times 2 = 3.5 \ cm^2

Area of shaded portion =Area of quadrant – Area of triangle =9.625-3.5 =6.125 \ cm^2

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icse 2012 9.jpgQuestion 12: ABC is an is isosceles right-angled triangle with \angle ABC = 90^o . A semi-circle is drawn with AC as the diameter. If AB=BC=7 \ cm find the area of the shaded region. ( \pi = \frac{22}{7})     [2012]

Answer: 

\triangle ABC is a right angled triangle. Therefore

AC^2 = AB^2+BC^2 = (7)^2 + (7)^2 = 98

\Rightarrow AC = 7 \sqrt{2}

Area of semi circle = \frac{1}{2} \times \frac{22}{7} \times (\frac{7 \sqrt{2}}{2})^2 = 38.5 cm^2

Area of \triangle ABC = \frac{1}{2} \times 7 \times 7 = 24.5 cm^2

Area of the shaded region = Area of the semi circle  – Area of \triangle ABC = 38.5 - 24.5 = 14 cm^2

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Question 13:   (i) From a rectangular card board ABCD , two circles and one semi-circle of the largest sizes are cut out as shown below. Calculate the ratio of the area of the remaining card board and the area of the card board cut.

(ii) If the figure of part (i), given above, find the area of the shaded portion within the rectangle, if radius of each circle is 3 \ cm and \pi = 3.14   [2008]

216

Answer:

Let the radius of the circle = r

Therefore BC = 2r and AB = 5r

Area of rectangle = 5r \times 2r = 10r^2

Area of the cardboard cut out = 2.5 \times \pi r^2 = 7.85 r^2

Area of remaining cardboard = 10r^2 - 7.85r^2 = 2.15r^2

(i) Ratio = \frac{Area \ of \ remaining \ cardboard}{Area \ of \ the \ cardboard \ cut out} = \frac{2.15r^2}{7.85r^2} = 0.2727

(ii) Area of the shaded region = 10(3)^2 - 7.85(3)^2 = 90-70.65 = 19.35 \ cm^2

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217Question 14:  In an equilateral \triangle ABC of side 14 \ cm , side BC is the diameter of a semi-circle as shown in the given figure. Find the area of the shaded region. [Take \pi = \frac{22}{7}   and \sqrt{3} = 1.732    [2007]

Answer:

Area of the semi-circle = \frac{1}{2} \times \pi (7)^2 = \frac{1}{2} \times \frac{22}{7} \times (7)^2 = 77 \ cm^2

Height of the triangle = \sqrt{14^2 -7^2} = \sqrt{147} = 7 \sqrt{3} = 7 \times 1.732 = 12.124 \ cm

Area of triangle = \frac{1}{2} \times base \times height =  \frac{1}{2} \times 14 \times 12.124 = 84.868 \ cm^2

Therefore the area of the shaded region = 77 + 84.868 = 161.868 \ cm^2 

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2011-2Question 15: Calculate the area of the shaded region, if the diameter of the semi circle is equal to 14 \ cm .   (Take \pi = \frac{22}{7}   [2011]

Answer:

Area of shaded portion = Total area – area of the two quadrants

= (Area \ of \ ACDE + \ Area \ of \ semi \ circle \ EFD) - \\ (Area \ of  \ Quadrant \ ABE + Area \ of \ Quadrant \ BCD)

  = (14 \times 7 + \frac{1}{2} \pi \times 7^2) - (\frac{1}{4} \times \pi \times 7^2 + \frac{1}{4} \times \pi \times 7^2 ) 

 = 98 \ cm^2

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