Question 13 Marks
Write the formula for the area of a segment in a circle of radius r given that the sector angle is $\theta$ (in degrees).
Answer
In this figure, centre of the circle is $O,$ radius $OA = r$ and $\angle\text{AOB}=\theta$
We are going to find the area of the segment $AXB.$
Area of the segment $AXB =$ Area of the sector $OAXB - $ Area of $\triangle\text{AOB}\ \dots(1)$
We know that area of the sector $\text{OAXB}=\frac{\theta}{360}\times\pi\text{r}^2$
We also know that area of $\triangle\text{AOB}=\text{r}^2$ $\triangle\text{AOB}=\text{r}^2\sin \frac{\theta}{2}\cos\frac{\theta}{2}$
Substituting these values in equation $(1)$ we get,
Area of the segment $AXB =\frac{\theta}{360}\times\pi\text{r}^2-\text{r}^2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$
Area of the segment $AXB=\Big(\frac{\theta}{360}\times\pi-\sin\frac{\theta}{2}\cos\frac{\theta}{2}\Big)\text{r}^2$
So Area of the segment $AXB =\Big(\frac{\pi\theta}{360}-\sin\frac{\theta}{2}\cos\frac{\theta}{2}\Big)\text{r}^2$
Therefore, area of the segment is $=\Big(\frac{\pi\theta}{360}-\sin\frac{\theta}{2}\cos\frac{\theta}{2}\Big)\text{r}^2.$ View full question & answer→Question 23 Marks
In the following figure, $ABCD$ is a rectangle with $AB = 14\ cm$ and $BC = 7\ cm.$ Taking $DC, BC$ and $AD$ as diameters, three semi-circles are drawn as shown in the figure. Find the area of the shaded region.
AnswerArea of the shaded region can be calculated as shown below,
Area of the shaded region $=$ Area of rectangle $−$ area of the semi-circle with diameter $DC$ triangle $+ 2 ×$ area of two semicircles with diameters $AD$ and $BC$
$\therefore$ Area of the shaded region $=7\times14-\frac{\pi\times7^2}{2}+2\times\frac{\pi\times3.5^2}{2}$
$\therefore$ Area of the shaded region $=98-\frac{\pi\times49}{2}+\pi\times12.25$
Substituting $\pi=\frac{22}{7}$ we get,
$\therefore$ Area of the shaded region $=98-\frac{\frac{22}{7}\times49}{2}+\frac{22}{7}\times12.25$
$\therefore$ Area of the shaded region $=98-\frac{22\times7}{2}+22\times1.75$
$\therefore$ Area of the shaded region $=98-77+22\times1.75$
$\therefore$ Area of the shaded region $=21+38.5$
$\therefore$ Area of the shaded region $=59.5$
Therefore, area of the shaded region is $59.5\text{cm}^2$
View full question & answer→Question 33 Marks
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 card board.$\Big(\text{Use }\pi=\frac{22}{7}\Big).$
AnswerLength of rectangle $= 14\ cm$
and breadth $= 7\ cm$
$\therefore$ Total area $= l × b$
$= 14 × 7 = 98cm^2$
Radiud of each circle $= \frac{7}{2} \text{cm}$
$\therefore$ Area of two circle $=2\pi\text{r}^2$
$=2\times\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}=77\text{cm}^2$
Area of the remaining portion
$= 98 - 77 = 21cm^2$ View full question & answer→Question 43 Marks
The perimeter of a certain sector of a circle of radius $5.6\ m$ is $27.2\ m.$ Find the area of the sector.
Answer
$\theta=$ angle subtended at centre
Radius $(r) = 5.6m =\text{OA}\pm\text{OB}$
Perimeter of sector $= 27.2m$
$(AB$ arc length$) + OA + OB = 27.2$
$\Rightarrow\Big(\frac{\theta}{360^\circ}\times2\pi\text{r}\Big)+5.6+5.6=27.2$
$\Rightarrow\frac{5.6\pi\theta}{180^\circ}+11.2=27.2$
$\Rightarrow5.6\times\frac{22}{7}\times\theta=16\times180$
$\Rightarrow\theta=\frac{16\times180}{0.8\times22}=163.64^\circ$
Area of sector $=\frac{\theta}{360^\circ}\times\pi\text{r}^2=\frac{163.64^\circ}{360^\circ}\times\frac{22}{7}\times5.6\times5.6$
$=\frac{163.64}{180}\times11\times0.8\times5.6$
$=44.8\text{cm}^2$ View full question & answer→Question 53 Marks
A plot is in the form of a rectangle $ABCD$ having semi-circle on $BC$ as shown in the following figure. If $AB = 60\ m$ and $BC = 28\ m,$ find the area of the plot.
Answer
Given $AB = 60m = DC [$length$]$
$BC = 28m = AD [$breadth$]$
Radius of semicircle $\text{r}=\frac{1}{2}\times\text{BC}=14\text{m}$
Area of semicircle $\text{r}=\frac{1}{2}\times\text{BC}=14\text{m}$
Area of plot $= ($Area of rectangle $ABCD) + ($area of semicircle$)$
$=(\text{length}\times\text{breadth})+\frac{1}{2}\pi\text{r}^2$
$=(60\times28)+\Big[\frac{1}{2}\times\frac{22}{7}\times14\times14\Big]$
$= 1680 + 308 = 1988m^2$ View full question & answer→Question 63 Marks
In the following figure, $OE = 20\ cm.$ In sector $OSFT,$ square $OEFG$ is inscribed. Find the area of the shaded region.
AnswerIn the figure $OSFT$ is a quadrant and $OEFG$
is a square inscribed in it
The side of the square is $OE = 20\ cm$
$\therefore\text{Diagonal}=\sqrt{2}\times\text{side}$
$\therefore $ Radius $(r)$ of the sector $=20\sqrt{2}\text{cm}$
Now area of quadrant $OTFS =\frac{1}{4}\times \pi\text{r}^2$
$=\frac{1}{4}(3.14)\times\big(20\sqrt{2}\big)^2\text{cm}^2$
$=\frac{1}{4}\times3.14\times800\text{cm}^2=628\text{cm}^2$
Area of square $\mathrm{OEFG = (side)^2}$
$=(20)^2\text{cm}^2=400\text{cm}^2$
Area of shaded region $= 628 - 400$
$= 228cm^2$ View full question & answer→Question 73 Marks
An archery target has three regions formed by the concentric circles as shown in the figure. If the diameters of the concentric circles are in the ratio $1:2:3,$ then find the ratio of the areas of three regions.
AnswerLet the diameters of concentric circles be $k,$
$\therefore$ Radius of concentric circles are $\frac{\text{k}}{2},\text{k}\text{ and }\frac{3\text{k}}{2}.$
$\therefore$ Area of inner circle , $\text{A}_1=\pi\Big(\frac{\text{k}}{2}\Big)^2=\frac{\text{k}^2\pi}{4}$
$\therefore$ Area of middle region,
$\text{A}_2=\pi(\text{k})^2-\frac{\text{k}^2\pi}{4}=\frac{3\text{k}^2\pi}{4}$
$[\therefore\text{area of ring}=\pi(\text{R}^2-\text{r}^2),$ where R is radius of outer and r is radius of inner ring]
and area of outer region, $\text{A}_3=\pi\Big(\frac{3\text{k}}{2}\Big)-\pi\text{k}^2$
$=\frac{9\pi\text{k}^2}{4}-\pi\text{k}^2=\frac{5\pi\text{k}^2}{4}$
$\therefore\text{Required ratio}=\text{A}_1:\text{A}_2:\text{A}_3$
$=\frac{\text{k}^2\pi}{4}:\frac{3\text{k}^2\pi}{4}:\frac{5\pi\text{k}^2}{4}=1:3:5$
View full question & answer→Question 83 Marks
In the following 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:
$(i)$ Quadrant $OACB$
$(ii)$ Shaded region.
AnswerIt is given that $OACB$ is a quadrant of circle with centre at $O$ and radius $3.5\ cm.$
- We know that the area of quadrant of circle of radius $r$ is,
$\text{A}=\frac{1}{4}\pi\text{r}^2$
Substituting the value of radius $\text{r}=3.5\text{cm},$
$\text{A}=\frac{1}{4}\times\frac{22}{7}\times3.5\times3.5$
$=9.625\text{cm}^2$
Hence, the area of $OACB$ is $=9.625\text{cm}^2.$
- It is given that radius of quadrant of small circle is $cm.$
Let the area of quadrant of small circle be. $A'.$
$\text{A}'=\frac{1}{4}\pi\text{r}^2$
$=\frac{1}{4}\times\frac{22}{7}\times2\times2$
$=3.14\text{cm}^2$
It is clear from the above figure that area of shaded region is the difference of larger quadrant and the smaller one. Hence,
Area of shaded region $=\text{A}-\text{A}'$
$=9.625-3.14$
$=6.485\text{cm}^2$ View full question & answer→Question 93 Marks
A circular field has a perimeter of $650\ m.$ A square plot having its vertices on the circumference of the field is marked in the field. Calculate the area of the square plot.
AnswerPerimeter of the circular field $= 650m$
$\therefore$ Radius $ (r) = \frac{\text{Circumference}}{2\pi} $
$=\frac{650\times7}{2\times22}=\frac{2275}{22}\text{m}$
$\therefore$ Diagonal of the inscribed square $=$ diameter of the circle
=$2\text{r}=2\times\frac{2275}{22}=\frac{2275}{11}\text{m}$
$\therefore\text{side}=\frac{\text{diagonal}}{\sqrt{2}}=\frac{2275}{\sqrt{2}\times11}$
and area of the squrae field $= a^2$
$=\Big(\frac{2275}{11\sqrt{2}}\Big)^2\text{m}^2=\frac{5175625}{121\times2}\text{m}^2$
$= 21386.88\text{m}^2$
$=21387\text{m}^2\text{(approx)}$
View full question & answer→Question 103 Marks
A circular pond is of diameter $17.5\ m.$ It is surrounded by a $2\ m$ wide path. Find the cost of constructing the path at the rate of $₹\ 25$ per square metre. $(\text{Use }\pi=3.14)$
AnswerDiameter, $d = 17.5m$
Radius, $r =\frac{17.5}{2}\text{m}$
Radius of the pond with the $2\ m$ wide path $=2+\frac{17.5}{2}\text{m}$
Area of the circular path $=$ Area of the pond with the path $-$ area of the pond
$=\pi\Big(2+\frac{17.5}{2}\Big)-\pi\Big(\frac{17.5}{2}\Big)^2$
$=122.57\text{m}^2$
Cost of constructing the path $= 25 × 122.57 = Rs 3064.2$
View full question & answer→Question 113 Marks
The radii of two circles are $19\ cm$ and $9\ cm$ respectively. Find the radius and area of the circles which has it circumference equal to the sum of the circumferences of the two circles.
AnswerLet the radius of circles be $r\ cm $, $r_1\ cm$ and $r_2\ cm$ respectively. Then their circumferences are $\text{C}=2\pi\text{r}\text{ cm},\text{C}_1=2\pi\text{r}_1\text{cm}$ and $\text{C}_2=2\pi\text{r}_2\text{cm}$ resprctively it is given that,
Circumference C of circle = Circumference $C_1$ of circle + circumference $C_2$ of circle
$2\pi\text{r}=2\pi\text{r}_1+2\pi\text{r}_2$
$2\pi\text{r}=2\pi(\text{r}_1+\text{r}_2)$
$\text{r}=\text{r}_1+\text{r}_2$
We have, $r_1= 19cm$ and $r_2= 9cm$
Substituting the values of $r_1, r_2$
$r = 19 + 9$
$r = 28\ cm$
Hence the radius of the circle is $28\ cm$
We know that the area $A$ of circle is
$\text{A}=\pi\text{r}^2$
Substituting the value of $r$
$\text{A}=\frac{22}{7}\times28\times28$
$=2464\text{cm}^2$
Hence the area of the circle is $= 2464\ cm^2$
View full question & answer→Question 123 Marks
A field is in the form of a circle. A fence is to be erected around the field. The cost fencing would be $Rs.2640$ at the rate of $Rs.12$ per metre. The, the field is to be thoroughly ploughed at the cost of $Re. 0.50$ per $m^2$. What is the amount required to plough the field$?$
$\big[\text{Take }\pi=\frac{22}{7}\big].$
AnswerGiven
Total cost of fencing the circular field $= Rs.\ 2640$
Cost per metre fencing $= Rs\ 12$
Total cost of fencing $=$ circumference $×$ cost per fencing
$⇒ 2640 =$ circumference$× 12$
$\Rightarrow\text{circumference}=\frac{2640}{12}=220\text{m}$
Let radius of field be $r\ m$
Circumference $=2\pi\text{r }\text{m}$
$2\pi\text{r}=220$
$2\times\frac{22}{7}\times\text{r}=220$
$\text{r}=\frac{70}{2}=35\text{m}$
Area of field $=\pi\text{r}^2$
$=\frac{22}{7}\times35\times35$
$= 3850m^2$
Cost of ploughing per $m^2$ land $= Rs.\ 0.50$
Cost of ploughing 3850 $m^2$ land $=\frac{1}{2}\times3850$
$= Rs.\ 1925$
View full question & answer→Question 133 Marks
Find the ratio of the area of the circle circumscribing a square to the area of the circle inscribed in the square.
AnswerLet the side of the square inscribed in a square be a units.
Diameter of the circle outside the square $=$ Diagonal of the square $= \sqrt{2}\text{a}$
Radius $=\frac{\sqrt{2}\text{a}}{2}=\frac{\text{a}}{\sqrt{2}}$
So, the area of the circle circumscribing the square $=\pi\Big(\frac{\text{a}}{\sqrt{2}}\Big)^2$
Now, the radius of the circle inscribed in a square $=\frac{\text{a}}{2}$
Hence, area of the circle inscribed in a square $=\pi\Big(\frac{\text{a}}{2}\Big)^2$
From $(i)$ and $(ii)$
$\frac{\text{Area or circle circumscribing a square}}{\text{Area of circle inscribed in a square }}$$=\frac{\pi\Big(\frac{\text{a}}{\sqrt{2}}\Big)^2}{\pi\Big(\frac{\text{a}}{2}\Big)^2}$
$=\frac{\frac{1}{2}}{\frac{1}{4}}$
$=\frac{2}{1}$
Hence, the required ratio is $2 : 1.$ View full question & answer→Question 143 Marks
In the following figure, shows a kite in which $BCD$ is the shape of a quadrant of a circle of radius $42\ cm.\ ABCD$ is a square and $= \triangle\text{CEF}$ is an isosceles right angled triangle whose equal sides are 6cm long. Find the area of the shaded region.
Answer$ABCD$ is a square with side $= 42\ cm$
$BCD$ is a quadrant in which $\triangle\text{BCD}=90^\circ$ and radius $= 42\ cm$
$\triangle\text{CEF}$ is an isosceles right triangle in which $CE = CF = 6\ cm$
$\therefore$ Area of shaded portion $=$ area of quadrant $BCD +$ area of $\triangle\text{CEF}$
$= \frac{1}{4}\pi\text{r}^2+\frac{1}{2}\text{CF}\times\text{CE}$
$=\frac{1}{4}\times\frac{22}{7}\times42\times42+\frac{1}{2}\times6\times6\text{cm}^2$
$=1386+18=1404\text{cm}^2$ View full question & answer→Question 153 Marks
A circle is inscribed in an equilateral triangle $ABC$ is side $12\ cm,$ touching its sides $($the following figure$).$ Find the radius of the inscribed circle and the area of the shaded part.
AnswerEach side of the equilateral triangle $ABC\ (a) = 12\ cm$
$\therefore\text{Area}=\frac{\sqrt{3}}{4}\text{a}^2=\frac{\sqrt{3}}{4}(12)^2\text{cm}^2$
$=\frac{1.732\times12\times12}{4}=62.352\text{cm}^2$
In $\triangle\text{ABC},$ draw $\text{AD}\perp\text{BC}, O$ will fall on $AD$
and $\text{OD}=\frac{1}{3}\text{AD} ( \because O$ is centroid also$)$
$=\frac{1}{3}\times\frac{\sqrt{3}}{2}\text{(side)}$
$=\frac{\sqrt{3}}{6}\times12=2\sqrt{3}$
$\therefore$ Radius of incircle $ (r) =\text{OD}=2\sqrt{3}\text{cm}$
and area of incircle $=\pi\text{r}^2=\frac{22}{7}\times\big(2\sqrt{3}\big)^2\text{cm}^2$
$=\frac{22}{7}\times12=\frac{264}{7}\text{cm}^2=37.714\text{cm}^2$
$\therefore$ Area of shaded portion
$=62.352-37.714=24.638\text{cm}^2$
View full question & answer→Question 163 Marks
The circumference of a circle exceeds the diameter by $16.8\ cm.$ Find the circumference of the circle.
AnswerLet radius of circle $=\text{r}\text{ cms}$
Diameter $(d) = 2 ×$ radius $= 2r$
Circumference $(c) =2\pi\text{r}$
Given circumference exceeds diameter by $16.8\ cm$
$\text{C}=\text{d}+16.8$
$\Rightarrow2\pi\text{r}=2\text{r}+16.8$
$\Rightarrow2\text{r}(\pi-1)=16.8$
$\Rightarrow2\text{r}\times\Big(\frac{22}{7}-1\Big)=16.8$
$\Rightarrow2\text{r}\times\frac{15}{7}=16.8$
$\Rightarrow\text{r}=\frac{16.8\times7}{30}=5.6\times0.7$
$\Rightarrow\text{r}=3.92\text{cm}$
Circumference $=2\pi\text{r}=2\times\frac{22}{7}\times3.92$
$=\frac{2464}{100}=24.64\text{cm}$
View full question & answer→Question 173 Marks
Find the area of the circle in which a square of area $64cm^2$ is inscribed.$[\text{Use }\pi=3.14]$
AnswerArea of square $= 64cm^2$
$\therefore$ Side of square $=\sqrt{\text{Area}}=\sqrt{64}=8\text{cm}$
$\because$ The square in inscribed in the circle
$\therefore$ Radius of the circle will be $=\frac{1}{2}$ diagonal of
the square $(r) =\frac{1}{2}\times\sqrt{2 }\text{a}=\frac{1}{2}\times\sqrt{2}\times8\text{cm}=4\sqrt{2}$
$\therefore$ Area of the circle $=\pi\text{r}^2$
$=3.14\times(4\frac{1}{2}\times\sqrt{2}\times8\text{cm}=4\sqrt{2})^2\text{cm}^2$
$=3.14\times32=100.48\text{cm}^2$
View full question & answer→Question 183 Marks
The diameter of a coin is $1\ cm ($in the following figure$).$ If four such coins be placed on a table so that the rim of each touches that of the other two, find the area of the shaded region $\big(\text{Take }\pi = 3.1416\big). $
AnswerLook at the figure carefully shaded region is bounded between four sectorsof the circle withsame radius and a square of side $1\ cm.$
Therefore, the area of the shaded region is nothing but the differencethearea of the square andarea of one circle.
$\therefore$ Area of the shaded region $=$ Area of square $-$ Area of a circle
$\therefore$ Area of the shaded region $=1^2-\pi(0.5^2)$
$\therefore$ Area of the shaded region $=1-0.25\pi$
Substituting $\pi=3.1416$ we get,
$\therefore $ Area of the shaded region $= 1-3.1416 × 0.25$
$\therefore$ Area of the shaded region $= 1- 0.7854$
$\therefore$ Area of the shaded rigion $= 0.2146$
Therefore, area of the shaded region is. $0.2146cm^2$
View full question & answer→Question 193 Marks
A rectangular piece is $20m$ long and $15m$ wide. From its four corners, quadrants of radii $3.5m$ have been cut. Find the area of the remaining part.
Answer
Length of rectangular piece $l = 20m$
Breadth of rectangular piece $b = 15m$
Radius of each quadrant $r = 3.5m$
Area of rectangular piece $= ($length $×$ breadth$) = 20 × 15 = 300m^2$.
Area of quadrant each $=\frac{1}{4} ($area of circle with radius $3.5m)$
$=\frac{1}{4}\times\pi\text{r}^2$
$=\frac{1}{4}\times\frac{22}{7}\times3.5\times3.5=\frac{38.5}{4}\text{m}^2$
Area of remaining part $= [$area of rectangular piece$] - 4 [$area of each quadrant$]$
$=300-4\Big[\frac{38.5}{4}\Big]=300-38.5$
$=261.5\text{m}^2$ View full question & answer→Question 203 Marks
The area of a circle inscribed in an equilateral triangle is $154cm^2$. Find the perimeter of the triangle.$\Big[\text{use } \pi = \frac{22}{7} \text{and } \sqrt{3} = 1.73\Big]$
AnswerArea of the inscribed circle of $\triangle\text{ABC}=154\text{cm}^2$
Let $r$ be the radius, then
$\pi\text{r}^2=154$
$\Rightarrow\frac{22}{7}\text{r}^2=154\Rightarrow\text{r}^2=\frac{154\times7}{22}$
$\Rightarrow\text{r}^2=49=(7)^2$
$\therefore r = 7\ cm$
$\therefore OP = 7\ cm$
$\therefore\text{In the equilateral }\triangle\text{ABC},$
$AP$ is $\perp$ BC which bisects $BC$ at $P$
$\therefore AP = 3OP = 3 × 7 = 21\ cm$
Let $a$ be side of the triangle $ABC$
$\therefore\frac{\sqrt{3}}{2}\text{a}=\text{AP}=21\Rightarrow\text{a}=\frac{21\times2}{\sqrt{3}}$
$\Rightarrow\text{a}=\frac{21\times2\times\sqrt{3}}{\sqrt{3}+\sqrt{3}}=\frac{21\times2\sqrt{3}}{3}=7\times2\sqrt{3}$
$= 7 × 2(1.73) = 24.22\ cm$
Perimeter of the triangle $= 3a$
$= 3 × 24.22 = 72.7\ cm$ View full question & answer→Question 213 Marks
Find the area of the following figure, in square $cm,$ correct to one place of decimal.$\Big(\text{Take }\pi=\frac{22}{7}\Big).$
AnswerJoin $AD$
$ABCD$ is a square whose each side $= 10\ cm$
Area of square $= a^2= (10)^2= 100cm^2$
Area of half semicircle whose radius is $\frac{10}{2}=5$
$=\frac{1}{2}\pi\text{r}^2=\frac{1}{2}\times\frac{22}{7}\times5\times5$
$=\frac{275}{7}=39.28\text{cm}^2$
$\therefore$ Total area of the figure $= 100 - 24 + 39.28$
$= 76 + 39.28 = 115.28cm^2$
$= 115.28cm^2$ View full question & answer→Question 223 Marks
In the following figure, the square $ABCD$ is divided into five equal parts, all having same area. The central part is circular and the lines $AE, GC, BF$ and $HD$ lie along the diagonals $AC$ and $BD$ of the square. If $AB = 22\ cm,$ find:
The perimeter of the part $ABEF.$
AnswerWe have a square $ABCD.$
We have,
$AB = 22cm$
We have to find the perimeter of $ABEF.$ Let $O$ be the centre of the circular region.Use Pythagoras theorem to get,
$2 (AE + r)^2= 22^2$
$AE + r = 15.56$
$AE = (15.56 - 5.56)cm$
$=10c,$
Similarly,
$BF = 10\ cm$
Now length of arc $EF,$
$=\frac{\text{Perimeter of circular region}}{4}$
$=\frac{34.88}{4}\text{cm}$
$= 8.64\ cm$
So, perimeter of $ABFE,$
$= AB + BF + EF + AE$
$= 50.64\ cm$ View full question & answer→Question 233 Marks
Prove that the area of a circular path of uniform width surrounding a circular region of radius r is $\pi\text{h}(2\text{r}+\text{h}).$
Answer
The width of the circular path $= h$
Let the inner circle be region A and the outer circle be region $B$
Radius of region $A = r$
Radius of region $B = r + h$
Area of the circular path = Area of region $B −$ Area of region $A$
$=\pi(\text{r}+\text{h})^2-\pi\text{r}^2$
$=\pi(\text{r}^2+\text{h}^2+2\text{rh}-\text{r}^2)$
$=\pi\text{h}(\text{h}+2\text{r})$
Hence proved View full question & answer→Question 243 Marks
In the the following figure, $PSR, RTQ$ and $PAQ$ are three semicircles of diameter $10\ cm, 3\ cm$ and $7\ cm$ respectively. Find the perimeter of shaded region.
AnswerRadius of larger semicircle $(\text{r}_1)=\frac{10}{2}=5\text{cm}$
Radius of large semicircle $=(\text{r}_2)=\frac{7}{2}\text{cm}$
Radius of small semicircle $(\text{r}_2)=\frac{3}{2}\text{cm}$
$\therefore$ perimeter of the shaded portion
$=\pi\text{r}_1+\pi\text{r}_2+\pi\text{r}_3$
$=\pi\Big(5+\frac{7}{2}+\frac{3}{2}\Big)\text{cm}=\pi\times10\text{cm}$
$=\frac{22}{7}\times10=\frac{220}{7}=31.41\text{cm}$
View full question & answer→Question 253 Marks
In the following figure, two circles with centres $A$ and $B$ touch each other at the point $C$. If $AC = 8\ cm$ and $AB = 3\ cm,$ find the area of the shaded region.
AnswerArea of the shaded region can be calculated as shown below,
Area of the shaded region = Area of circle with radius $AC -$ area of circlewith radius radius $BC$
We have given radius of the outer circle that is 8cm but we don’t know theradius of theinner circle.
We can calculate the radius of the inner circle as shown below,
$BC = AC - AB$
$\therefore BC = 8 - 3$
$\therefore BC = 5$
$\therefore$ Area of the shaded region $=\pi\times8\times8-\pi\times5\times5$
$\therefore$ Area of the shaded region $=\pi\times64-\pi\times25$
$\therefore$ Area of the shaded region $=\pi\times39$
Substituting $\pi=\frac{22}{7}$ we get,
$\therefore$ Area of the shaded region $=\frac{22}{7}\times39$
$\therefore$ Area of the shaded region $=122.57$
Therefore, area of the shaded region is $122.57cm^2$.
View full question & answer→Question 263 Marks
Four equal circles, each of radius a, touch each other. Show that the area between them is $\frac{6}{7}\text{a}^2$ $\Big(\text{Take }\pi=\frac{22}{7}\Big).$
AnswerIt is given that four equal circles of radius a touches each other.
So,
Area of circle $= \pi\text{a}^2$
Since circles touches each other, the lines joining their centre make a square $ABCD.$
The side of square is $2a.$
Area of quadrant inside squqre $=\frac{1}{4}\pi\text{a}^2$
Area of shaded region $=$ Area of square $-\ 4\ ×$ Area of quadrant
$=(2\text{a})^2-4\times\frac{\pi\text{a}^2}{4}$
$=4\text{a}^2-\frac{22}{7}\text{a}^2$
$= \frac{6}{7}\text{a}^2$ View full question & answer→Question 273 Marks
The perimeter of a sector of a circle of radius $5.7\ m$ is $27.2\ m.$ Find the area of the sector.
AnswerRadius of the circle $(r) = 5.7m$
Perimeter of the sector $= 27.2m$
Length of the arc $=$ Perimeter $- 2r$
$= (27.2 - 2 \times 5.7)m$
$= 27.2 - 11.4 = 15.8m$
Let $\theta$ be the central angle, then
$2\pi\text{r}\times\frac{\theta}{360^\circ}=15.8$
$\Rightarrow2\times\frac{22}{7}\times5.7\times\frac{\theta}{360^\circ}=15.8$
$\Rightarrow\frac{\theta}{360^\circ}=\frac{15.8\times7}{2\times22\times5.7}=\frac{110.6}{250.8}\dots(\text{i})$
$\therefore$ Area of the sector $=\pi\text{r}^2\times\frac{\theta}{360^\circ}$
$=\frac{22}{7}\times(5.7)^2\times\frac{110.6}{250.8} [$From $(i)]$
$=\frac{22}{7}\times\frac{32.49\times110.6}{250.8}=\frac{32.49\times15.8}{11.4}$
$=45.03\text{m}^2$
View full question & answer→Question 283 Marks
In a hospital used water is collected in a cylindrical tank of diameter $2\ m$ and height $5\ m.$ After recycling, this water is used to irrigate a park of hospital whose length is $25\ m$ and breadth is $20\ m.$ If tank is filled completely then what will be the height of standing water used for irrigating the park.
AnswerDiameter of cylinder $(d) = 2m$
Radius of cylinder $(r) = 1m$
Height of cylinder $(H) = 5m$
Volume of cylinderical tank, $\text{V}_\text{c}=\pi\text{r}^2\text{H}=\pi\times(1)^2\times5=5\pi\text{ m}^3$
Length of the park $(l) = 25m$
Breadth of park $(b) = 20m$
height of standing water in the park $= h$
Volume of water in the park $= lbh =25\times20\times\text{h m}^3$
Now water from the tank is used to irrigate the park. So,
Volume of cylinderical tank $=$ Volume of water in the park
$\Rightarrow5\pi=25\times20\times\text{h}$
$\Rightarrow\frac{5\pi}{25\times20}=\text{h}$
$\Rightarrow\text{h}=\frac{\pi}{100}\text{m}$
$\Rightarrow\text{h}=0.0314\text{m}$
View full question & answer→Question 293 Marks
In the following figure, an equilateral triangle $ABC$ of side $6\ cm$ has been inscribed a circle. Find the area of the shaded region. $\big(\text{Take }\pi=3.14\big).$
AnswerWe have to find the area of the shaded portion. We have $\triangle\text{ABC}$ which is an equilateral triangle and $\text{AB}=6\text{cm}$ .Let $r$ be the radius of the circle.
We have $O$ as the circumcentre.
$\angle\text{OBA}=30^\circ$
So,
$\cos (30^\circ)=\frac{3}{\text{r}}$
Thus, $\text{r}=2\sqrt{3}$
So area of the shaded region,
$=$ Area of the circle $-$ar $ (\triangle\text{ABC})$
$=(3.14)(2\sqrt{3})^2-\frac{\sqrt{3}}{4}(6)^2$
$=(37.68-15.59)\text{cm}^2$
$=22.126\text{cm}^2$ View full question & answer→Question 303 Marks
Find the area of the sector of a circle of radius $5\ cm,$ if the corresponding arc length is $3.5\ cm.$
AnswerLet the central angle of the sector be $\theta.$
Given that, radius of the sector of a circle $(r) = 5\ cm$
and arc length $(l) = 3.5\ cm$
$\therefore$ Central angle of the sector,
$\theta=\frac{\text{are length(l)}}{\text{radius}}$
$\Rightarrow\theta=\frac{3.5}{5}=0.7\text{ Radian}$ $\Big[\therefore\theta=\frac{\text{l}}{\text{r}}\Big]$
$\Rightarrow\theta=\Big(0.7\times\frac{180^\circ}{\pi}\Big)$ $\Big[\therefore1\text{R}=\frac{180^\circ}{\pi}\text{D}^\circ\Big]$
Now, area of sector with angle $\theta=0.7$
$=\frac{\pi\text{r}^2}{360^\circ}\times(0.7)\times\frac{180^\circ}{\pi}$
$=\frac{(5)^2}{2}\times0.7=\frac{25\times7}{2\times10}=\frac{175}{20}=8.75\text{cm}^2$
Hence, required area of the sector of a circle is $8.75cm^2$. View full question & answer→Question 313 Marks
In a circle of radius $35\ cm,$ an arc subtends an angle of $72^\circ $ at the centre. Find the length of the arc and area of the sector.
AnswerWe know that the arc length $l$ and area $A$ of a sector of an angle $\theta$ in the circle of radius $r$ is given.
by $\text{l}=\frac{\theta}{360^\circ}\times2\pi\text{r}$ and $\text{A}=\frac{\theta}{360^\circ}\times\pi\text{r}^2$ respectively.
It is given that, $r = 35\ cm$ and $\theta=72^\circ$
We will calculate the arc length using the value of r and $\theta,$
$\text{l}=\frac{72^\circ}{360^\circ}\times2\pi\times35\text{cm}$
$=\frac{75^\circ}{360^\circ}\times2\times\frac{22}{7}\times35\text{cm}$
$=44\text{cm}$
Now, we will find the value of area $A$ of the sector
$\text{A}=\frac{72^\circ}{360^\circ}\times\pi\times35\times35\text{cm}^2$
$=770\text{cm}^2$
View full question & answer→Question 323 Marks
The sum of the radii of two circles is $140\ cm$ and the difference of their circumferences is $88\ cm.$ Find the diameters of the circles.
AnswerLet the radius of two circles be $r_1cm$ and $r_2cm$ respectively. Then their circumferences are $\text{C}_1=2\pi\text{r}_1\text{cm}$ and $\text{C}_2=2\pi\text{r}_2\text{cm}$ respectively and their areas are $\text{A}_1=\pi\text{r}^2_1\text{cm}^2$ and $\text{A}_2=\pi\text{r}^2_2\text{cm}^2$ respectively.
It is given that the sum of the radii of two circles is $140\ cm$ and difference of their circumferences is $88\ cm.$
So,
$r_1+ r_2= 140\ cm ....(1)$
$C_1- C_2= 88\ cm$
$2\pi\text{r}_1-2\pi\text{r}_2=88\text{cm}$
$2\pi(\text{r}_1-\text{r}_2)=88\text{cm}$
$\text{r}_1-\text{r}_2=\frac{88}{2\pi}\text{cm}$
$\text{r}_1-\text{r}_2=\frac{88}{2\times\frac{22}{7}}\text{cm}$
$\text{r}_1-\text{r}_2=\frac{88\times7}{44}\text{cm}$
$r_1-r_2 = 14\ cm ....(2)$
Now, solving $(1)$ and $(2)$
$\text{r}_1=77\text{cm}$
$\text{r}_2=63$
Thus diameters of circles are,
$2\text{r}_1=154\text{cm}$
$2\text{r}_2=126\text{cm}$
View full question & answer→Question 333 Marks
$AB$ is a chord of a circle with centre $O$ and radius $4\ cm. AB$ is of length $4\ cm$ and divides the circle into two segments. Find the area of the minor segment.
Answer
Radius of circle $r = 4\ cm= OA = OB$
Length of chord $AB = 4\ cm$
$OA$ is equilateral triangle $\angle\text{AOB}=60^\circ\rightarrow\theta$
Angle subtended at centre $\theta=60^\circ$
Area of segment $($shaded region$) =(\text{area of sector})-(\text{area of }\triangle\text{AOB})$
$=\frac{\theta}{360^\circ}\times\pi\text{r}^2-\frac{\sqrt{3}}{4}(\text{Side})^2$
$=\frac{60}{360}\times\frac{22}{7}\times4\times4-\frac{\sqrt{3}}{4}\times4\times4$
$=\frac{176}{3}-4\sqrt{3}=58.67-6.92=51.75\text{cm}^2$ View full question & answer→Question 343 Marks
In the following figure, $ABCD$ is a trapezium of area $24.5cm^2$, If $\text{AD}||\text{BC}, \angle\text{DAB} = 90^\circ$ $AD = 10\ cm, BC = 4\ cm$ and $ABE$ is quadrant of a circle, then find the area of the shaded region.
AnswerArea of trapezium $=\frac{1}{2}(\text{AD}+\text{BC})\times\text{AB}$
$\Rightarrow 24.5=\frac{1}{2}(10+4)\times\text{AB}$
$\Rightarrow \text{AB}=3.5\text{cm}$
Area of shaded region $=$ Area of trapezium $ABCD -$ Area of quadrant $ABE$
$=24.5-\frac{1}{4}\times\frac{22}{7}\times(3.5)^2$
$=24.5-9.625$
$=14.875\text{cm}^2$
Hence, the area of shaded region is $14.875cm^2$
View full question & answer→Question 353 Marks
In the given figure, $ABCD$ is a trapezium with $\text{AB}||\text{DC},\text{AB}=18\text{cm} DC = 32\ cm$ and the distance between $AB$ and $DC$ is $14\ cm.$ Circles of equal radii $7\ cm$ with centres $A, B, C$ and $D$ have been drawn. Then find the area of the shaded region.$\Big(\text{Use }\pi=\frac{22}{7}\Big)$
AnswerIn trapezium $ABCD$
$\text{AB } || \text{ DC}$
$AB = 18\ cm, DC = 32\ cm$ Height $= 14\ cm$
Radius of each at the corner of trapezium $= 7\ cm$
$\therefore$ Angle of a quadrilateral $=360^\circ$
$\therefore 4$ sectrors compete a circle
$\therefore$ Angle of a circle $=\pi \text{r}^2=\pi\cdot(7)^2=\frac{22}{7}\times49\text{cm}^2$
$=154\text{cm}^2$
and area of trapezium
$=\frac{1}{2}(\text{AB}+\text{DC)}\times\text{h}$
$=\frac{1}{2}(18+32)\times14\text{cm}^2$
$=\frac{1}{2}\times50\times14=350\text{cm}^2$
$\therefore$ Area of shaded portion
$= 350 - 154 = 196\text{cm}^2$ View full question & answer→Question 363 Marks
A play ground has the shape of a rectangle, with two semi-circles on its smaller sides as diameters, added to its outside. If the sides of the rectangle are $36\ m$ and $24.5\ m,$ find the area of the playground. $\Big(\text{Take }\pi=\frac{22}{7}\Big).$
AnswerIt is given that a play ground has a shape of rectangle, with two semicircles on its smaller sides as diameter, added to its outside. So,
Area of play ground = Area of rectangle $+\ 2\ ×$ Area of semicircle
We have, sides of rectangle $l = 36 m$ and $b = 24.5m.$
Since, the diameter of semicircle is $2r = b. $ then,
$r = 24.52 = 12.25mr = 24.52 = 12.25m$
Area of semicircle $=\frac{\pi\text{r}^2}{2}$
$=\frac{1}{2}\times\frac{22}{7}\times12.25.\times12.25$
$= 235.81m^2$
Area of rectangle $= l × b$
$= 36 × 24.5$
$= 882m^2$
Thus, the area of playground is
Area of play ground $=$ Area of rectangle $+\ 2\ ×$ Area of semicircle
$= 882 + 2 × 235.81$
$= 882 + 471.62$
$= 1353.62m^2$
View full question & answer→Question 373 Marks
Four equal circles, each of radius $5\ cm,$ touch each other as shown in the following figure. Find the area included between them$\big(\text{Take } \pi = 3.14).$
AnswerRadius of each circle $= 5\ cm$
$\because$ The four circles touch eachother externally
$\therefore$ By joining their centres, we get a square whose side will be $5 + 5 = 10\ cm$
Now area of square so formed $=a^2=(10)^2=100 \mathrm{~cm}^2$
and area of 4 quadrants $=4 \times \frac{1}{4} \pi \mathrm{r}^2=\pi \mathrm{r}^2$
$=3.14 \times(5)^2 \mathrm{~cm}^2=3.14 \times 25 \mathrm{~cm}^2=78.5 \mathrm{~cm}^2$
$\therefore$ Area of the part included between the circles
$ =100-78.5 $
$ =21.5 \mathrm{~cm}^2$ View full question & answer→Question 383 Marks
In the following figure, shows a sector of a circle, centre $O,$ containing an angle $\theta^\circ.$ Prove that:
Area of the shaded region is $\frac{\text{r}^2}{2}\Big(\tan\theta-\frac{\pi\theta}{180}\Big)$
AnswerIt is given that the radius of circle is $r$ and the angle $\angle\text{AOC}=\theta^\circ$
In $\triangle\text{AOB},$
It is given that $OA = r .$
$\cos\theta=\frac{\text{OA}}{\text{OB}}$
$\text{OB}=\frac{\text{OA}}{\cos\theta}$
$\text{OB}=\text{r}\sec\theta$
$\tan\theta=\frac{\text{AB}}{\text{OA}}$
$\text{AB}=\text{OA}\tan\theta$
$\text{AB}=\text{r}\tan\theta$
We know that area $A$ of the sector at an angle $\theta$ in the circle of radius $r$ is
$\text{A}=\frac{\theta}{360^\circ}\times\pi\text{r}^2$
Thus
Area of sector $\text{AOC}=\frac{\theta}{360^\circ}\pi\text{r}^2$
Area of $\triangle\text{AOB}=\frac{1}{2}\times\text{OA}\times\text{AB}$
$=\frac{1}{2}\times\text{r}\times\text{r}\tan\theta$
$=\frac{1}{2}\times\text{r}^2\tan\theta$
Area of shaded region $ABC =$ Area of $\triangle\text{AOB }-$ Area of sector $AOC$
$=\frac{1}{2}\text{r}^2\tan\theta-\frac{\theta}{360^\circ}\times\pi\text{r}^2$
$=\frac{\text{r}^2}{2}\Big(\tan\theta-\frac{\pi\theta}{180^\circ}\Big)$
Hence, Area of shaded region $\text{ABC}=\frac{\text{r}^2}{2}\Big(\tan\theta-\frac{\pi\theta}{180^\circ}\Big)$ View full question & answer→Question 393 Marks
Figure shows a sector of a circle of radius $r\ cm$ containing an angle $\theta.$ The area of the sector is $A \ cm^2$ and perimeter of the sector is $50\ cm.$ Prove that.
$\theta=\frac{360}{\pi}\Big(\frac{25}{\text{r}}-1\Big)$
AnswerRadius of the sector of the circle $= r\ cm$
and angle at the centre $= 0$
Area of sector $OAB = Acm^2$
and perimeter of sector $OAB = 50cm$
Area of the sector $=\pi\text{r}^2\times\frac{\theta}{360^\circ}$
$\Rightarrow\text{A}=\pi\text{r}^2\times\Big(\frac{\theta}{360^\circ}\Big)$
Perimeter $= 2OA + arc\ AB$
$\Rightarrow50=2\text{r}+2\pi\text{r}\times\Big(\frac{\theta}{360^\circ}\Big)$
$\Rightarrow50-2\text{r}=2\pi\text{r}\Big(\frac{\theta}{360^\circ}\Big)$
$\Rightarrow\frac{\theta}{360^\circ}=\frac{50-2\text{r}}{2\pi\text{r}}=\frac{50}{2\pi\text{r}}-\frac{2\text{r}}{2\pi\text{r}}$
$\Rightarrow\frac{\theta}{360^\circ}=\frac{25}{\pi\text{r}}-\frac{1}{\pi}\ ....(\text{i})$
$\Rightarrow\theta=360\Big(\frac{25}{\pi\text{r}}-\frac{1}{\pi}\Big)=\frac{360^\circ}{\pi}\Big(\frac{25}{\text{r}}-1\Big)$ View full question & answer→Question 403 Marks
In the following figure find the area of the shaded region. $\big(\text{Use }\pi=3.14\big)$
AnswerSide of large square $= 14cm$
Radius of each semicircle $=\frac{4}{2}=2\text{cm}$
Side of square $= 4cm$
Area of square $=4 \times 4=16 \mathrm{~cm}^2$
$\therefore$ Area of semicircles $=4 \times \frac{1}{2} \pi \mathrm{r}^2$
$=2 \times 3.14 \times 2 \times 2 $
$ =8 \times 3.14 $
$ =25.12 \mathrm{~cm}^2$
$\therefore$ Area of shaded region $=$ Area of large square $-$ Area of central portion
$=(14) 2-(16+25.12) \mathrm{cm}^2 $
$ =154.88 \mathrm{~cm}^2$
View full question & answer→Question 413 Marks
If the circumference of two circles are in the ratio $2 : 3, $ what is the ratio of their areas$?$
AnswerWe are given ratio of circumferences of two circles. If $\text{c}=2\pi\text{r}$ and $\text{c}' =2\pi\text{r}'$are circumferences of two circles such that
$\frac{\text{c}}{\text{c}'}=\frac{2}{3}$
$\Rightarrow\frac{2\pi\text{r}}{2\pi\text{r}'}=\frac{2}{3}\ \dots(1)$
Simplifying equation $(1)$ we get,
$\frac{\text{r}}{\text{r}'}=\frac{2}{3}$
Let $\text{A}= \pi\text{r}^2$and $\text{A}'= \pi\text{r}^2$ are the areas of the respective circles and we are asked to find their ratio.
$\frac{\text{A}}{\text{A}'}=\frac{\pi\text{r}^2}{\pi\text{r}'^2}$
$\frac{\text{A}}{\text{A}'}=\frac{\text{r}^2}{\text{r}'^2}$
$\frac{\text{A}}{\text{A}'}=\Big(\frac{\text{r}}{\text{r}'}\Big)^2\ \dots(2)$
We know that $\frac{\text{r}}{\text{r}'}=\frac{2}{3}$ substituting this value in equation $(2) $ we get,$\frac{\text{A}}{\text{A}'}=\Big(\frac{2}{3}\Big)^2$
$\Rightarrow\frac{\text{A}}{\text{A}'}=\frac{4}{9}$
Therefore, ratio of their areas is $4:9.$
View full question & answer→Question 423 Marks
From each of the two opposite corners of a square of side $8\ cm,$ a quadrant of a circle of radius $1.4\ cm$ is cut. Another circle of radius $4.2\ cm$ is also cut from the centre as shown in the following figure. Find the area of the remaining (Shaded) portion of the square.$\Big(\text{Use }\pi=\frac{22}{7}\Big)$
AnswerSide of a square $ABCD = 8\ cm$
$\therefore$ Area $=a^2=(8)^2=64 \mathrm{~cm}^2$
Radius of each quadrants $= 1.4\ cm$
$\therefore$ Area of $2$ quadrants
$=2\times\frac{1}{4}\pi\text{r}^2=\frac{1}{2}\times\frac{22}{7}(1.4)^2\text{cm}^2$
$=\frac{1}{2}\times\frac{22}{7}\times1.4\times1.4=4.312\text{cm}^2$
Radius of center circle $= 4.2\ cm$
Area of center circle $=\pi\text{r}^2$
$=\frac{22}{7}\times4.2\times4.2$
$=22\times0.6\times4.2$
$=55.44\text{cm}^2$
Area of shaded region $=$ Area of square $-$ Area of two quadrant $-$ Area of circle
$=64-4.312-55.44$
$=4.23$ View full question & answer→Question 433 Marks
A rectangular park is $100m$ by $50m$. It is surrounding by semi-circular flower beds all round. Find the cost of levelling the semi-circular flower beds at $60$ paise per square metre $\big(\text{Take }\pi=3.14\big).$
AnswerSince four semicircular flower beds rounds the rectangular park. Then, diameters of semicircular.
plots are $2r_1= l$ and $2r_2= w$
So, the radius of semicircle at larger side of rectangle
$\text{r}_1=\frac{\text{l}}{2}$
$=\frac{100}{2}$
$=50\text{m}$
Area of semicicluar plot at larger side of rectangle $=\frac{1}{2}\pi\text{r}^2$
$=\frac{1}{2}\times3.14\times50\times50$
$=3925\text{m}^2$
And the radius of semicircle at smaller side of rectangle
$\text{r}_2=\frac{\text{l}}{2}$
$=\frac{50}{2}$
$=25\text{m}$
Area of semicicluar plot at samller side of rectangle $=\frac{1}{2}\pi\text{r}^2$
$=\frac{1}{2}\times3.14\times25\times25$
$=981.25\text{m}^2$
Now, the total area of semicircular plot is sum of area of four semicircular plots.
Total Area of plot $= 2 × 3925 + 2 × 981.25$
$= 7850 + 1962.5m^2$
$= 9812.5m^2$
Since, The cost of levelling semicircular flower bed per square meter $= Rs. 0.60$
So, The cost of levelling $9812.5$ square meter flower bed $= Rs. 0.60 × 9812.5$
$= Rs. 5887.50$
View full question & answer→Question 443 Marks
A car travels $1$ kilometre distance in which each wheel makes $450$ complete revolutions. Find the radius of the its wheels.
AnswerDistance covered by the car in $450$ revolutions $= 1\ km = 1000\ m$
$\therefore$ Distance covered in $1$ revolution $=\big(\frac{1000}{450}\big)$
$=\big(\frac{20}{9}\big)\text{m}$
Perimeter of wheel $=2\pi\text{r}$
$\Rightarrow\frac{2\times22}{7}\text{r}=220\Rightarrow\text{r}=\frac{220\times7}{2\times22}$
$⇒ r = 35\ m.$
View full question & answer→Question 453 Marks
Find the area of a shaded region in the the following figure,where a circular arc of radius $7\ cm$ has been drawn with vertex $A$ of an equilateral triangle $ABC$ of side $14\ cm$ as centre. $\big(\text{Use }\pi=\frac{22}{7}$ and $\sqrt{3}=1.73\big)$
AnswerIn equilateral traingle all the angles are of $60^\circ $
$\therefore\angle\text{BAC}=60^\circ$
Area of the shaded region $= ($Area of triangle $ABC -$ Area of sector having central angle $60^\circ ) +$ Area of sector having central angle $(360^\circ - 60^\circ )$
$=\frac{\sqrt{3}}{4}\text{(AB)}^2-\frac{60^\circ}{360^\circ}\pi(7)^2+\frac{300^\circ}{360^\circ}\pi(7)^2$
$=\frac{\sqrt{3}}{4}(14)^2-\frac{1}{6}\times\frac{22}{7}(7)^2+\frac{5}{6}\times\frac{22}{7}(7)^2$
$=84.77-25.67+128.35$
$=187.45\text{cm}^2$
Hence, the area of shaded region is $187.45cm^2$
View full question & answer→Question 463 Marks
The side of a square is $10\ cm.$ Find the area of circumscribed and inscribed circles.
AnswerCircumscribed circle
Radius $=\frac{1}{2} ($diagonal of square$)$
$=\frac{1}{2}\times\sqrt{2}\text{ side}$
$=\frac{1}{2}\times\sqrt{2}\times10$
$=5\sqrt{2}\text{cm}$
$\text{Area}=\pi\text{r}^2$
$=\frac{22}{7}\times25\times2$
$=\frac{1100}{7}\text{cm}^2$
Inscribed circle
$\text{Radius}=\frac{1}{2}(\text{sides})$
$=\frac{1}{2}\times10$
$=5\text{cm}$
$\text{Area}=\pi\text{r}^2$
$=\frac{22}{7}\times5\times5$
$=\frac{550}{7}\text{cm}^2$ View full question & answer→Question 473 Marks
In the given figure, the side of square is $28\ cm$ and radius of each circle is half of the length of the side of the square where O and $O'$ are centres of the circles. Find the area of shaded region.
AnswerSide $= 28\ cm,$ Radius $=\frac{28}{2}\text{cm}=14\text{cm}$
The area of the shaded region
$=$ Area of square $+\frac{3}{4} ($Area of circle$)$
$+\frac{3}{4} ($Area of circle$)$
$=$ Area of square $+\frac{3}{2} ($Area of circle$)$
[Area of square = (Side)$^2$; Area of circle $=\pi\text{r}^2$]
$=(28)^2+\frac{3}{2}\times\frac{22}{7}\times14\times14$
$=784+924=1708\text{cm}^2$
View full question & answer→Question 483 Marks
The length of the minute hand of a clock is $14\ cm.$ Find the area swept by the minute hand in $5$ minutes.
AnswerAngle make by the minute hand in $1$ minute $= 6^\circ $
Angle make by the minute hand in $5$ minute $= 5 ⨯ 6^\circ = 30^\circ $
Area of the sector having central angle is given by
$\frac{30^\circ}{360^\circ}\pi(14)^2$
$=\frac{1}{12}\times\frac{22}{7}(14)^2$
$=51.33\text{cm}^2$
Hence, the area swept by minute hand in $5$ minutes is $51.33cm^2$
View full question & answer→Question 493 Marks
Find the area enclosed between two concentric circles of radii $3.5\ cm$ and $7\ cm.$ A third concentric circle is drawn outside the $7\ cm$ circle, such that the area enclosed between it and the $7\ cm$ circle is same as that between the two inner circles. Find the radius of the third circle correct to one decimal place.
Answer
The area enclosed between the two circles of radii $3.5\ cm$ and $7\ cm$
$=\pi(7^2-3.5^2)$
$=115.\text{cm}^2$
Let the radius of the outemost circle be $r\ cm.$
Area betweent the circles with radius $r$ and $7\ cm = $Area betweent the circles with radius $7\ cm$ and $3.5\ cm$
$\pi(\text{r}^2-7^2)=115.5$
$\Rightarrow(\text{r}^2-7^2)=\frac{115.5}{\pi}$
$\Rightarrow\text{r}^2=36.75+49=85.75\text{cm}$
$\Rightarrow\text{r}=9.26\text{cm}$ View full question & answer→Question 503 Marks
Find the area of the shaded region in the following figure, if $AC = 24\ cm, BC = 10\ cm$ and $O$ is the centre of the circle.
AnswerIt is given a triangle $ABC$ is cut from a circle.
$\text{AC}=24\text{cm}$
$\text{BC}=10\text{cm}$
Area of $\triangle\text{ABC} =\frac{1}{2}\ \text{AC}\ \times\text{BC}$
$=\frac{1}{2}\times24\times10$
$=120\text{cm}^2$
In $\triangle\text{ABC},$
$\angle\text{ACB}=90^\circ ,$ Since any angle inscribed in semicircle is always right angle
By applying Pythagoras theorem,
$ \text{AB}^2=\text{AC}^2+\text{BC}^2$
$=24\times24+10\times10$
$=576+100$
$=676\text{cm}^2$
$\text{OA}=\frac{\text{AB}}{2}$
$=\frac{26}{2}\text{cm}$
$=13\text{cm}$
We know that the area $A$ of circle of radius $r$ is
$\text{A}=\pi\text{r}^2$
Substituting the value of radius $r,$
$\text{A}=3.14\times13\times13$
$=530.66\text{cm}^2$
Area of semicircle $=\frac{1}{2}\pi\text{r}^2$
$=\frac{530.66}{2}\text{cm}^2$
$=265.33\text{cm}^2$
Area of sharded region $=$ Area of circle of circle $-$ Area of semicircle $-$ Area of triangle.
$=530.66-265.33-120$
$=145.33\text{cm}^2$ View full question & answer→