Class 10: Circumference and Area – ICSE Board Problems

$\displaystyle \textbf{Question 1: }\text{A bucket is raised from a well by means of a rope}$
$\displaystyle \text{which is wound around a wheel of diameter }77\text{ cm}.$
$\displaystyle \text{Given that the bucket ascends in }1\text{ minute }28\text{ seconds}$
$\displaystyle \text{with a uniform speed of }1.1\frac{\text{m}}{\text{s}},\text{ calculate the number of complete}$
$\displaystyle \text{revolutions the wheel will make in raising the bucket.}\hfill \text{[ICSE 1997]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of the wheel }r=\frac{77}{2}=38.5\text{ cm}$
$\displaystyle \text{Time taken}=1\text{ minute }28\text{ seconds}=88\text{ s}$
$\displaystyle \text{Length of the rope}=1.1\times88=96.8\text{ m}$
$\displaystyle \text{Circumference of the wheel}=2\pi r$
$\displaystyle =2\times\frac{22}{7}\times38.5=242\text{ cm}=2.42\text{ m}$
$\displaystyle \text{Number of revolutions}=\frac{96.8}{2.42}=40$
$\displaystyle \therefore \text{The wheel makes }40\text{ complete revolutions.}$
$\\$

$\displaystyle \textbf{Question 2: }\text{The wheel of a cart is making }5\text{ revolutions per second.}$
$\displaystyle \text{If the diameter of the wheel is }84\text{ cm, find the speed in }$
$\displaystyle \frac{\text{km}}{\text{hr}}.\text{ Give your answer correct to the nearest km.}\hfill \text{[ICSE 1998]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Number of revolutions}=5\text{ per second}$
$\displaystyle \text{Radius of the wheel }r=\frac{84}{2}=42\text{ cm}$
$\displaystyle \text{Speed}=2\pi r\times5\frac{\text{cm}}{\text{s}}$
$\displaystyle =2\times\frac{22}{7}\times42\times5=1320\frac{\text{cm}}{\text{s}}$
$\displaystyle =1320\times\frac{3600}{100000}\frac{\text{km}}{\text{hr}}$
$\displaystyle =47.52\frac{\text{km}}{\text{hr}}\approx48\frac{\text{km}}{\text{hr}}$
$\displaystyle \therefore \text{The speed of the cart is }48\frac{\text{km}}{\text{hr}}\text{ (nearest km).}$
$\\$

$\displaystyle \textbf{Question 3: }\text{In the given figure, }AB\text{ is the diameter of the circle}$
$\displaystyle \text{with center }O\text{ and }OA=7\text{ cm. Find the area of the shaded region.}$
$\displaystyle \hfill \text{[ICSE 2006]}$ $\displaystyle \text{Answer:}$
$\displaystyle OA=7\text{ cm},\quad AB=14\text{ cm},\quad OB=7\text{ cm}$
$\displaystyle \text{Radius of the smaller circle}=\frac{7}{2}\text{ cm}$
$\displaystyle \text{Area of the smaller circle}=\pi\left(\frac{7}{2}\right)^2$
$\displaystyle =\frac{22}{7}\times\left(\frac{7}{2}\right)^2=38.5\text{ cm}^2$
$\displaystyle \text{Area of }\triangle CBD=\frac{1}{2}\times14\times7=49\text{ cm}^2$
$\displaystyle \text{Area of the right semicircle}=\frac{1}{2}\pi(7)^2$
$\displaystyle =\frac{1}{2}\times\frac{22}{7}\times49=77\text{ cm}^2$
$\displaystyle \text{Area of the shaded region}$
$\displaystyle =\text{Area of smaller circle}+\text{Area of right semicircle}-\text{Area of }\triangle CBD$
$\displaystyle =38.5+77-49=66.5\text{ cm}^2$
$\displaystyle \therefore \text{The area of the shaded region is }66.5\text{ cm}^2.$
$\\$

$\displaystyle \textbf{Question 4: }AC\text{ and }BD\text{ are two mutually perpendicular}$
$\displaystyle \text{diameters of a circle }ABCD.\text{ Given the area of the shaded region is }308\text{ cm}^2,$
$\displaystyle \text{calculate (i) the length of }AC\text{ and (ii) the circumference of the circle.}$
$\displaystyle \text{Take }\pi=\frac{22}{7}.\hfill \text{[ICSE 2009]}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{The shaded region is half of the circle.}$
$\displaystyle \therefore \frac{1}{2}\pi r^2=308$
$\displaystyle \frac{1}{2}\times\frac{22}{7}\times r^2=308$
$\displaystyle r^2=308\times2\times\frac{7}{22}$
$\displaystyle r^2=196$
$\displaystyle r=14\text{ cm}$
$\displaystyle \text{(i) }AC=2r=2\times14=28\text{ cm}$
$\displaystyle \text{(ii) Circumference of the circle}=2\pi r$
$\displaystyle =2\times\frac{22}{7}\times14=88\text{ cm}$
$\displaystyle \therefore AC=28\text{ cm and circumference}=88\text{ cm}.$
$\\$

$\displaystyle \textbf{Question 5: }\text{A doorway is decorated as shown in the figure.}$
$\displaystyle \text{There are four semi-circles. }BC,\text{ the diameter of the larger semi-circle,}$
$\displaystyle \text{is of length }84\text{ cm. Centers of the three equal semi-circles lie on }BC.$
$\displaystyle ABC\text{ is an isosceles triangle with }AB=AC.\text{ If }BO=OC,$
$\displaystyle \text{find the area of the shaded region.}\hfill \text{[ICSE 2010]}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Since angle in a semi-circle is }90^\circ,\ \angle A=90^\circ$
$\displaystyle \text{Let }AB=AC=x\text{ cm}$
$\displaystyle \text{In right-angled }\triangle ABC,\text{ by Pythagoras theorem,}$
$\displaystyle AB^2+AC^2=BC^2$
$\displaystyle x^2+x^2=84^2$
$\displaystyle 2x^2=84\times84$
$\displaystyle x^2=84\times42$
$\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\times AB\times AC$
$\displaystyle =\frac{1}{2}\times x^2=\frac{1}{2}\times84\times42=1764\text{ cm}^2$
$\displaystyle \text{Radius of the larger semi-circle}=\frac{84}{2}=42\text{ cm}$
$\displaystyle \text{Area of the larger semi-circle}=\frac{1}{2}\pi r^2$
$\displaystyle =\frac{1}{2}\times\frac{22}{7}\times42\times42=2772\text{ cm}^2$
$\displaystyle \text{Diameter of each of the three equal semi-circles}=\frac{84}{3}=28\text{ cm}$
$\displaystyle \text{Radius of each of the three equal semi-circles}=14\text{ cm}$
$\displaystyle \text{Area of the three equal semi-circles}=3\times\frac{1}{2}\pi(14)^2$
$\displaystyle =3\times\frac{1}{2}\times\frac{22}{7}\times14\times14=924\text{ cm}^2$
$\displaystyle \text{Area of the shaded region}$
$\displaystyle =\text{Area of larger semi-circle}+\text{Area of three equal semi-circles}$
$\displaystyle \quad-\text{Area of }\triangle ABC$
$\displaystyle =2772+924-1764=1932\text{ cm}^2$
$\displaystyle \therefore \text{The area of the shaded region is }1932\text{ cm}^2.$
$\\$

$\displaystyle \textbf{Question 6: }\text{The shaded part of the given figure shows the shape}$
$\displaystyle \text{of the top of a table in a restaurant, which is a segment of a circle}$
$\displaystyle \text{with center }O,\ \angle BOD=90^\circ\text{ and }BO=OD=60\text{ cm.}$
$\displaystyle \text{Find (i) the area of the top of the table and (ii) the perimeter}$
$\displaystyle \text{of the table. Take }\pi=3.14.\hfill \text{[ICSE 2002]}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Angle of the shaded major sector}=360^\circ-90^\circ=270^\circ$
$\displaystyle \text{Area of the table}=\frac{270^\circ}{360^\circ}\times\pi r^2$
$\displaystyle =\frac{3}{4}\times3.14\times(60)^2=8478\text{ cm}^2$
$\displaystyle \text{Perimeter of the table}=\text{length of major arc }BD+OB+OD$
$\displaystyle =\frac{270^\circ}{360^\circ}\times2\pi r+60+60$
$\displaystyle =\frac{3}{4}\times2\times3.14\times60+120$
$\displaystyle =282.6+120=402.6\text{ cm}$
$\displaystyle \therefore \text{Area of the table}=8478\text{ cm}^2$
$\displaystyle \text{and perimeter of the table}=402.6\text{ cm}.$
$\\$

$\displaystyle \textbf{Question 7: }\text{Calculate the area of the shaded portion.}$
$\displaystyle \text{The quadrants shown in the figure are each of radius }7\text{ cm.}$
$\displaystyle \hfill \text{[ICSE 2000]}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Side of the square}=2r=2\times7=14\text{ cm}$
$\displaystyle \text{Area of the square}=14\times14=196\text{ cm}^2$
$\displaystyle \text{Area of four quadrants}=4\times\frac{1}{4}\pi r^2=\pi r^2$
$\displaystyle =\frac{22}{7}\times7\times7=154\text{ cm}^2$
$\displaystyle \text{Area of the shaded portion}$
$\displaystyle =\text{Area of the square}-\text{Area of four quadrants}$
$\displaystyle =196-154=42\text{ cm}^2$
$\displaystyle \therefore \text{The area of the shaded portion is }42\text{ cm}^2.$
$\\$

$\displaystyle \textbf{Question 8: }\text{In the figure given below, }ABCD\text{ is a rectangle}$
$\displaystyle \text{with }AB=14\text{ cm and }BC=7\text{ cm. From the rectangle,}$
$\displaystyle \text{a quarter circle }BFEC\text{ and a semicircle }DGE\text{ are removed.}$
$\displaystyle \text{Calculate the area of the remaining piece of the rectangle.}$
$\displaystyle \text{Take }\pi=\frac{22}{7}.\hfill \text{[ICSE 2014]}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Area of rectangle }ABCD=14\times7=98\text{ cm}^2$
$\displaystyle \text{Radius of quarter circle }BFEC=7\text{ cm}$
$\displaystyle \text{Area of quarter circle }BFEC=\frac{1}{4}\pi(7)^2=\frac{49\pi}{4}$
$\displaystyle DE=DC-EC=14-7=7\text{ cm}$
$\displaystyle \text{Radius of semicircle }DGE=\frac{7}{2}\text{ cm}$
$\displaystyle \text{Area of semicircle }DGE=\frac{1}{2}\pi\left(\frac{7}{2}\right)^2$
$\displaystyle =\frac{1}{2}\times\frac{49\pi}{4}=\frac{49\pi}{8}$
$\displaystyle \text{Area of remaining piece}$
$\displaystyle =98-\left(\frac{49\pi}{4}+\frac{49\pi}{8}\right)$
$\displaystyle =98-\frac{147\pi}{8}$
$\displaystyle =98-\frac{147}{8}\times\frac{22}{7}$
$\displaystyle =98-57.75=40.25\text{ cm}^2$
$\displaystyle \therefore \text{The area of the remaining piece is }40.25\text{ cm}^2.$
$\\$

$\displaystyle \textbf{Question 9: }\text{The given figure shows a running track surrounding}$
$\displaystyle \text{a grass enclosure }PQRSTU.\text{ The enclosure consists of a rectangle }PQST$
$\displaystyle \text{with a semicircular region at each end. }PQ=200\text{ m and }PT=70\text{ m}.$
$\displaystyle \text{(i) Calculate the area of the grass enclosure in m}^2\text{.}$
$\displaystyle \text{(ii) If the track is of constant width }7\text{ m, calculate the outer}$
$\displaystyle \text{perimeter }ABCDEF\text{ of the track.}\hfill \text{[ICSE 1999]}$ $\displaystyle \text{Answer:}$
$\displaystyle PQ=200\text{ m},\quad PT=70\text{ m}$
$\displaystyle \text{Radius of each inner semicircle}=\frac{70}{2}=35\text{ m}$
$\displaystyle \text{Radius of each outer semicircle}=35+7=42\text{ m}$
$\displaystyle \text{(i) Area of the grass enclosure}$
$\displaystyle =\pi(35)^2+70\times200$
$\displaystyle =\frac{22}{7}\times35\times35+14000$
$\displaystyle =3850+14000=17850\text{ m}^2$
$\displaystyle \text{(ii) Outer perimeter of the track}$
$\displaystyle =2\pi(42)+200+200$
$\displaystyle =2\times\frac{22}{7}\times42+400$
$\displaystyle =264+400=664\text{ m}$
$\displaystyle \therefore \text{(i) Area of the grass enclosure}=17850\text{ m}^2$
$\displaystyle \text{and (ii) outer perimeter of the track}=664\text{ m}.$
$\\$

$\displaystyle \textbf{Question 10: }\text{A rectangular playground has two semicircles added}$
$\displaystyle \text{to the outside with its smaller sides as diameters. If the sides of the}$
$\displaystyle \text{rectangle are }120\text{ m and }21\text{ m, find the area of the playground}$
$\displaystyle \left(\pi=\frac{22}{7}\right).\hfill \text{[ICSE 2000]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of each semicircle}=\frac{21}{2}=10.5\text{ m}$
$\displaystyle \text{The two semicircles together form one complete circle.}$
$\displaystyle \text{Area of the circle}=\pi(10.5)^2$
$\displaystyle =\frac{22}{7}\times10.5\times10.5=346.5\text{ m}^2$
$\displaystyle \text{Area of the rectangle}=120\times21=2520\text{ m}^2$
$\displaystyle \text{Area of the playground}=346.5+2520=2866.5\text{ m}^2$
$\displaystyle \therefore \text{The area of the playground is }2866.5\text{ m}^2.$
$\\$

$\displaystyle \textbf{Question 11: }\text{In the figure alongside, }OAB\text{ is a quadrant}$
$\displaystyle \text{of a circle. The radius }OA=3.5\text{ cm and }OD=2\text{ cm.}$
$\displaystyle \text{Calculate the area of the shaded portion. Take }\pi=\frac{22}{7}.$
$\displaystyle \hfill \text{[ICSE 2013]}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of quadrant }OAB,\ r=3.5\text{ cm}$
$\displaystyle \text{Area of quadrant }OAB=\frac{1}{4}\pi r^2$
$\displaystyle =\frac{1}{4}\times\frac{22}{7}\times(3.5)^2=9.625\text{ cm}^2$
$\displaystyle \text{Here, }\angle AOD=90^\circ$
$\displaystyle \text{Base }OA=3.5\text{ cm and height }OD=2\text{ cm}$
$\displaystyle \text{Area of }\triangle AOD=\frac{1}{2}\times3.5\times2=3.5\text{ cm}^2$
$\displaystyle \text{Area of the shaded portion}$
$\displaystyle =\text{Area of quadrant }OAB-\text{Area of }\triangle AOD$
$\displaystyle =9.625-3.5=6.125\text{ cm}^2$
$\displaystyle \therefore \text{The area of the shaded portion is }6.125\text{ cm}^2.$
$\\$

$\displaystyle \textbf{Question 12: }ABC\text{ is an isosceles right-angled triangle}$
$\displaystyle \text{with }\angle ABC=90^\circ.\text{ A semicircle is drawn with }AC\text{ as diameter.}$
$\displaystyle \text{If }AB=BC=7\text{ cm, find the area of the shaded region.}$
$\displaystyle \text{Take }\pi=\frac{22}{7}.\hfill \text{[ICSE 2012]}$ $\displaystyle \text{Answer:}$
$\displaystyle \triangle ABC\text{ is a right-angled triangle.}$
$\displaystyle \therefore AC^2=AB^2+BC^2$
$\displaystyle AC^2=7^2+7^2=49+49=98$
$\displaystyle AC=7\sqrt{2}\text{ cm}$
$\displaystyle \text{Radius of the semicircle}=\frac{AC}{2}=\frac{7\sqrt{2}}{2}\text{ cm}$
$\displaystyle \text{Area of the semicircle}=\frac{1}{2}\pi\left(\frac{7\sqrt{2}}{2}\right)^2$
$\displaystyle =\frac{1}{2}\times\frac{22}{7}\times\frac{98}{4}=38.5\text{ cm}^2$
$\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\times7\times7=24.5\text{ cm}^2$
$\displaystyle \text{Area of the shaded region}$
$\displaystyle =\text{Area of the semicircle}-\text{Area of }\triangle ABC$
$\displaystyle =38.5-24.5=14\text{ cm}^2$
$\displaystyle \therefore \text{The area of the shaded region is }14\text{ cm}^2.$
$\\$

$\displaystyle \textbf{Question 13: }\text{(i) From a rectangular cardboard }ABCD,$
$\displaystyle \text{two circles and one semicircle of the largest sizes are cut out,}$
$\displaystyle \text{as shown below. Calculate the ratio of the area of the remaining}$
$\displaystyle \text{cardboard and the area of the cardboard cut out.}$
$\displaystyle \text{(ii) If the figure of part (i), given above, has radius of each circle}$
$\displaystyle 3\text{ cm and }\pi=3.14,\text{ find the area of the shaded portion}$
$\displaystyle \text{within the rectangle.}\hfill \text{[ICSE 2008]}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Let the radius of each circle be }r.$
$\displaystyle \therefore BC=2r\text{ and }AB=5r$
$\displaystyle \text{Area of rectangle }ABCD=5r\times2r=10r^2$
$\displaystyle \text{Area of cardboard cut out}$
$\displaystyle =2\pi r^2+\frac{1}{2}\pi r^2=\frac{5}{2}\pi r^2$
$\displaystyle =2.5\times3.14r^2=7.85r^2$
$\displaystyle \text{Area of remaining cardboard}=10r^2-7.85r^2=2.15r^2$
$\displaystyle \text{(i) Required ratio}=\frac{\text{Area of remaining cardboard}}{\text{Area of cardboard cut out}}$
$\displaystyle =\frac{2.15r^2}{7.85r^2}=\frac{2.15}{7.85}=\frac{43}{157}$
$\displaystyle \therefore \text{Required ratio}=43:157$
$\displaystyle \text{(ii) When }r=3\text{ cm, area of shaded portion}$
$\displaystyle =10(3)^2-7.85(3)^2$
$\displaystyle =90-70.65=19.35\text{ cm}^2$
$\displaystyle \therefore \text{The area of the shaded portion is }19.35\text{ cm}^2.$
$\\$

$\displaystyle \textbf{Question 14: }\text{In an equilateral }\triangle ABC\text{ of side }14\text{ cm,}$
$\displaystyle \text{side }BC\text{ is the diameter of a semicircle, as shown in the given figure.}$
$\displaystyle \text{Find the area of the shaded region. Take }\pi=\frac{22}{7}\text{ and }\sqrt{3}=1.732.$
$\displaystyle \hfill \text{[ICSE 2007]}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of the semicircle}=\frac{14}{2}=7\text{ cm}$
$\displaystyle \text{Area of the semicircle}=\frac{1}{2}\pi r^2$
$\displaystyle =\frac{1}{2}\times\frac{22}{7}\times(7)^2=77\text{ cm}^2$
$\displaystyle \text{Height of equilateral }\triangle ABC$
$\displaystyle =\sqrt{14^2-7^2}=\sqrt{196-49}=\sqrt{147}=7\sqrt{3}$
$\displaystyle =7\times1.732=12.124\text{ cm}$
$\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\times14\times12.124$
$\displaystyle =84.868\text{ cm}^2$
$\displaystyle \text{Area of the shaded region}=77+84.868$
$\displaystyle =161.868\text{ cm}^2$
$\displaystyle \therefore \text{The area of the shaded region is }161.868\text{ cm}^2.$
$\\$

$\displaystyle \textbf{Question 15: }\text{Calculate the area of the shaded region,}$
$\displaystyle \text{if the diameter of the semicircle is equal to }14\text{ cm.}$
$\displaystyle \text{Take }\pi=\frac{22}{7}.\hfill \text{[ICSE 2011]}$
$\displaystyle \text{Answer:}$ $\displaystyle \text{Diameter of semicircle }EFD=14\text{ cm}$
$\displaystyle \therefore ED=AC=14\text{ cm}$
$\displaystyle AB=BC=AE=CD=7\text{ cm}$
$\displaystyle \text{Area of rectangle }ACDE=14\times7=98\text{ cm}^2$
$\displaystyle \text{Area of semicircle }EFD=\frac{1}{2}\pi(7)^2=77\text{ cm}^2$
$\displaystyle \text{Area of two quadrants}$
$\displaystyle =2\times\frac{1}{4}\pi(7)^2=\frac{1}{2}\pi(7)^2=77\text{ cm}^2$
$\displaystyle \text{Area of shaded region}$
$\displaystyle =\text{Area of rectangle }ACDE+\text{Area of semicircle }EFD$
$\displaystyle \quad-\text{Area of two quadrants}$
$\displaystyle =98+77-77=98\text{ cm}^2$
$\displaystyle \therefore \text{The area of the shaded region is }98\text{ cm}^2.$
$\\$