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]

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]

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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Question 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]

$OA = 7 \ cm, AB = 14 \ cm$ 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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Question 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]

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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Question 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]

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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Question 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]

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

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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Question 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]

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]

$PQ = 200 \ m, PT = 70 \ m$

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]

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

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Question 11:   In 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]

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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Question 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]

$\triangle ABC$ 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]

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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Question 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]

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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Question 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]

$= (Area \ of \ ACDE + \ Area \ of \ semi \ circle \ EFD) - \\ (Area \ of \ Quadrant \ ABE + Area \ of \ Quadrant \ BCD)$
$= (14 \times 7 + \frac{1}{2} \pi \times 7^2) - (\frac{1}{4} \times \pi \times 7^2 + \frac{1}{4} \times \pi \times 7^2 )$
$= 98 \ cm^2$
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