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MATHS 5 Marks Questions

Areas of Circle, Sector and Segment · MP Board (MPBSE) STD 10 (Madhya Pradesh - English Medium). 64 questions.

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Question 15 Marks
Find the area of a quadrant of a circle hose circumference is 44cm.
Answer
Let r be the radius of the circle,
Then, circumference of a circle $=2\pi\text{r}$
$\Rightarrow2\pi\text{r}=44$
$\Rightarrow2\times\frac{22}{7}\times\text{r}=44$
$\Rightarrow\text{r}=7\text{cm}$
$\therefore$ Area of the quadrant $=\Big(\frac{1}{4}\times\frac{22}{7}\times7\times7\Big)\text{cm}^2=\frac{77}{2}\text{cm}^2=38.5\text{cm}^2$
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Question 25 Marks
Four equal circles, each of radius 5cm, touch each other, as shown in the figure. Find the area included between them. $[\text{Take }\pi\ =3.14]$
Answer

Let A, B, C, D be the centres of these circles
Join AB, BC, and DA
Side of square = 10cm
Area of square ABCD
$=(10\times10)\text{cm}^2$
$=100\text{cm}^2$
Area of each sector $=\Big(\pi\text{r}^2\times\frac{\theta}{360}\Big)=3.14\times5\times5\times\frac{90}{360}$
$=19.625\text{cm}^2$
Required area = [area of sq. ABCD - 4(area of each sector)]
$=(100-419.625)\text{cm}^2$
$=(100-78.5)\text{cm}^2=21.5\text{cm}^2$
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Question 35 Marks
Find the perimeter of the s ded region in the figure, if ABCD is a square of side 14cm and APB and CPD are semicircles.
Answer
Side of a square =14cm
⇒ Diameter of a semicircle = 14cm
⇒ Radius of a semicircle = 7cm
$\therefore$ Perimeter of the shaded region
= Arc of semicircle DPC + Arc of semicircle APB + AD + BC
$=\Big\{\Big(\frac{22}{7}\times7\Big)+\Big(\frac{22}{7}\times7\Big)+14+14\Big\}\text{cm}$
$=(22+22+28)\text{cm}$
$=72\text{cm}$
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Question 45 Marks
The side of a square is 10cm. Find (i) the area of the inscribed circle, and (ii) the area of the circumscribed circle. $[\text{Take }\pi\ =3.14]$
Answer
Diameter of the inscibed circle = side of the square = 10cm
$\therefore$ Radus of the inscribrd circle = 5cm

Diameter of the circumscribed circle
= Diagonal of the square
$=(\sqrt{2}\times10)\text{cm}$
Radius of cicumscribed circle $=5\sqrt{2}\text{cm}$
  1. Area of inscribed circle $=\Big(\frac{22}{7}\times5\times5\Big)=78.57\text{cm}^2$
  2. Area of the circumscribed circle $=\Big(\frac{22}{7}\times5\sqrt{2}\times5\sqrt{2}\Big)=157.14\text{cm}^2$
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Question 55 Marks
In the given figure, PQ and AB are respectively the arcs of two concentric circles of radii 7cm and 3.5cm with centre 0. If $\angle\text{POQ}=30^\circ,$ find the area of the shaded region.
Answer
Area of a shaded region = Area of sector OPQ - Area of sector OAB
$=\Bigg[\Bigg(\frac{\frac{22}{7}\times7^2\times30}{360}\Bigg)-\Bigg(\frac{\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}\times30}{360}\Bigg)\Bigg]\text{cm}^2$
$=\Big(\frac{77}{6}-\frac{77}{24}\Big)\text{cm}^2$
$=77\Big(\frac{1}{6}-\frac{1}{24}\Big)\text{cm}^2$
$=77\Big(\frac{4-1}{24}\Big)\text{cm}^2$
$=77\times\frac{3}{24}\text{cm}^2$
$=\frac{77}{8}\text{cm}^2$
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Question 65 Marks
In the given figure, OPQR is a rhombus, three of whose vertices lie on a circle with centre O. If the area of the rhombus is $32\sqrt{3}\text{cm}^2,$ find the radius of the circle.
Answer

OP = OR = OQ = r
Let OQ and PR intersect at S
We know the diagonals of a rhombus bisect each other at right angle.
Therefore we have
$\text{OS}=\frac{1}{2}\text{r}$ and $\angle\text{OSR}=90^\circ$
$\therefore\text{SR}=\sqrt{\text{OR}^2-\text{OS}^2}$
$=\sqrt{\text{r}^2-\frac{\text{r}^2}{4}}=\frac{\sqrt{3}\text{r}}{2}$
$\therefore\text{PR}=2\times\text{SR}=\sqrt{3}\text{r}$
Area of rhombus $=\frac{1}{2}\times\text{OQ}\times\text{PR}$
$=\frac{1}{2}\times\text{r}\times\sqrt{3}\text{r}=\frac{\sqrt{3}\text{r}^2}{2}$
$\therefore\frac{\sqrt{3}\text{r}^2}{2}=32\sqrt{3}\Rightarrow\text{r}^2=\frac{32\sqrt{3}}{\sqrt{3}}\times2=64\text{cm}$
$\text{r}=8\text{cm}$
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Question 75 Marks
In tha given fingure, ABCD is a trapezium of area $24.5cm^2. 24.5cm^2$​​​​​​​ If $\text{AD}||\text{BC},\ \angle\text{DAB}=90^\circ,\ \text{AD}=10\text{cm},\ \text{BC}=4\text{cm}$ and ABE is quadrant of a circle then find the area of the shaded region. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
Answer
Area of the trapezium ABCD $=\frac{1}{2}(\text{AD}+\text{BC})\times\text{AB}$
$\Rightarrow24.5=\frac{1}{2}\times(10+4)\times\text{AB}$
$\Rightarrow24.5=\frac{1}{2}\times14\times\text{AB}$
$\Rightarrow24.5=7\text{AB}$
$\Rightarrow\text{AB}=3.5\text{cm}$
$\Rightarrow$ Radius of a quadrant ABE = 3.5cm
$\therefore$ Area of a quadrant ABE $=\frac{1}{4}\times\frac{22}{7}\times3.5\times3.5=9.625\text{cm}^2$
Now, Area of the shaded region
= Area of the trapezium ABCD - Area of a quadrant ABE
$= 24.5 - 9.625$
$= 14.875cm^2$​​​​​​​
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Question 85 Marks
The area of a circle inscribed in an equilateral triangle is $154 cm^2$​​​​​​​ Find the perimeter of the triangle. $\big[\text{Take }\sqrt{3}=1.73\big]$
Answer
Let the radius of circle be r cm

$\Rightarrow\text{r}^2=\Big(154\times\frac{7}{22}\Big)$
$\Rightarrow\text{r}=7\text{cm}$
Let each side of the triangle be a cm and height be h cm
Then, $\text{r}=\frac{\text{h}}{3}$
$\Rightarrow\text{h}=3\text{r}=21\text{cm}$
$\text{h}=\sqrt{\text{a}^2-\frac{\text{a}^2}{4}}=\frac{\sqrt{3\text{a}^2}}{2}=\frac{\sqrt{3\text{a}}}{2}=21$
$\text{a}=\frac{42}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=14\sqrt{3}\text{cm}$
perimeter $=3\text{a}=(3\times14\times\sqrt{3})=(42\times1.73)\text{cm}$
$=72.66\text{cm}$
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Question 95 Marks
In the given figure, ABCD is a square of side 4cm. A quadrant of a circle of radius 1cm is drawn at each vertex of the square and a circle of diameter 2cm is also drawn. Find the area of the shaded region.
Answer
Side of a square = 4cm
⇒ area of a square $= (4)^2 = 16cm^2$​​​​​​​
Radius of a circle = r = 1cm
⇒ Area of 4 quadrants of circle $=4\times\frac{1}{4}\times3.14\times1\times1=3.14\text{cm}^2$
Area of a circle of diameter 2cm $= 3.14 \times 1 \times 1 = 3.14cm^2​​​​​​​$​​​​​​​
$\therefore$ Area of the shaded region = Area of a square - Area of 4 quadrants of circle - Area of a circle of diameter 2cm
$= (16 - 3.14 - 3.14)cm^2$
$= 9.72cm^2​​​​​​​$​​​​​​​
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Question 105 Marks
Find the area of the major segment APB of a circle of radius 35cm and $\angle\text{AOB}=90^\circ$ as shown in the given figure.
Answer
Area of the minor segment ACBA = Area of sector OACBO - Area of $\triangle\text{OAB}$
$=\Big[\Big(\frac{22}{7}\times35\times35\times\frac{90}{360}\Big)-\Big(\frac{1}{2}\times35\times35\Big)\Big]\text{cm}^2$
$=[962.50-612.50]\text{cm}^2$
$=350\text{cm}^2$
Area of the circle $=\frac{22}{7}\times35\times35=3850\text{cm}^2$
$\therefore$ Area of the major segment = Area of a circle - Area of a minor segment
$= (3850 - 350)cm^2$
$= 3500cm^2$​​​​​​​
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Question 115 Marks
A rope by which a cow is tethered is increased from 16m to 23m. How much additional ground does it haven now to graze? $[\text{Take }\pi\ =3.14]$
Answer
Area of plot which cow can graze when = 16m is $\pi\text{r}^2$
$=\Big(\frac{22}{7}\times16\times16\Big)\text{m}^2$
$=804.5\text{m}^2$
Area of plot which cow can graze when radius is increased to 23m
$=\Big(\frac{22}{7}\times23\times23\Big)\text{m}^2$
$=1662.57\text{m}^2$
Additinal ground = Area covered by increased rope - old area
$=(1662.57 - 804.5)\text{m}^2=858\text{m}^2$
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Question 125 Marks
A chord 10cm long is drawn in a circle whose radius $5\sqrt{2}\text{cm}.$ find the areas of both the segments. $[\text{Take }\pi=3.14]$
Answer
Let $\text{OA}=5\sqrt{2}\text{cm},\ \text{ob}=5\sqrt{2}\text{cm}$
And $\text{AB}=10\text{cm}$

Then, $\text{OA}^2+\text{OB}^2=\text{AB}^2$
$\Rightarrow\angle\text{AOB}=90^\circ$
Area of the sector OACBO
$=\frac{\pi\text{r}^2\theta}{360}\text{cm}^2$
$=\Big(3.14\times(5\sqrt{2})\times(5\sqrt{2})\times\frac{90}{360}\Big)\text{cm}^2$
$=39.25\text{cm}^2$
Area of $\triangle\text{AOB}=\frac{1}{2}\text{r}^2\sin\theta=\Big(\frac{1}{2}\times5\sqrt{2}\times5\sqrt{2}\times\sin90^\circ\Big)$
$=25\text{cm}^2$
Area of minor segment = (Area of sector OACBO) - (area of $\triangle\text{OAB}$)
$=(39.25-25)\text{cm}^2=14.25\text{cm}^2$
Area of the major segment BDAB
= Area of cride - area of minor segment
$=\Big(\frac{22}{7}\times5\sqrt{2}\times5\sqrt{2}-14.25\Big)\text{cm}^2$
$=\Big(\frac{1100}{7}-14.25\Big)\text{cm}^2=(157-14.25)\text{cm}^2$
$=142.75\text{cm}^2$
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Question 135 Marks
In the given figure, three sectors of a circle of radius 7cm, making angles of 60°, 80° and 40° at the centre are shaded. Find the area of the shaded region.
Answer
Radius of a circle = r = 7cm
Area of a sector $=\frac{\pi\text{r}^2\theta}{360}$
$\therefore$ Area of the shaded region $=\frac{\pi\text{r}^2\times60^\circ}{360}+\frac{\pi\text{r}^2\times40^\circ}{360}+\frac{\pi\text{r}^2\times80^\circ}{360}$
$=\pi\text{r}^2\Big(\frac{60^\circ+40^\circ+80^\circ}{360}\Big)$
$=\frac{22}{7}\times7\times7\times\frac{180^\circ}{360^\circ}$
$=77\text{cm}^2$
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Question 145 Marks
In the given figure, a square OABC is inscribed in a quadrant OPBQ of a circle. if OA = 20cm, find the area of the shaded region. $\big[\text{Use }\pi=3.14\big]$
Answer
Side of a square = 20cm.
$\therefore$ Area of the square $= (20 \times 20)cm^2 = 400cm^2$​​​​​​​
Diagonal of square $=\sqrt{(20)^2+(20)^2}=\sqrt{800}=20\sqrt{2}\text{cm}$
⇒ Radius of the quadrant $=20\sqrt{2}\text{cm}$
$\therefore$ Area of a quadrant $=\frac{1}{4}\times3.14\times(20\sqrt{2})^2=628\text{cm}^2$
Thus, area of the shaded region = Area of a quadrant - Area of the square
$= (628 - 400)cm^2$
$= 228cm^2​​​​​​​$​​​​​​​
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Question 155 Marks
Find the area of both the segments of a circle of radius 42cm with central angle $120^\circ$. $\Big[\text{Given }\sin120^\circ=\frac{\sqrt{3}}{2}\text{and}\sqrt{3}=1.73\Big]$
Answer
Area of sector OACBO
$=\frac{\pi\text{r}^2\theta}{360}\text{cm}^2=\Big(\frac{22}{7}\times42\times42\times\frac{120}{360}\Big)\text{cm}^2=1848\text{cm}^2$

Area of $\triangle\text{OAB}=\frac{1}{2}\text{r}^2\sin\theta$
$=\Big(\frac{1}{2}\times42\times42\times\sin120^\circ\Big)$
$=\Big(21\times42\times\frac{\sqrt{3}}{2}\Big)\text{cm}^2$
$=(21\times21\times1.73)\text{cm}^2=762.93\text{cm}^2$
Area of monor segment ACBA
= (Area of sector OACBO) - $\big($Area of the $\triangle\text{OAB}\big)$
$= (1848 - 762.93)cm^2 = 1085.07$
Area major segment BADB
= (area of the cirde) - (area of minor segment)
$=\frac{22}{7}\times42\times42-1085.07$
$=(5544-1085.07)\text{cm}^2=4458.93\text{cm}^2$
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Question 165 Marks
A square ABCD is inscribed in a circle of radius r. Find the area of the square.
Answer
Since square ABCD is inscribed in a circle of radius r,
diagonal of a square = AC = 2r
$\therefore$ Area of a square ABCD $=\frac{1}{2}\times(\text{diagonal})^2=\frac{1}{2}\times(2\text{r})^2=2\text{r}^2\text{sq. units}$
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Question 175 Marks
Find the area of the shaded region in the given figure, where a circular arc of radius 6cm has been drawn with vertex of an equilateral triangle of side 12cm as centre and a sector of circle of radius 6cm with centre B is made. $\big[\text{Use }\sqrt{3}=1.73\text{ and }\pi=3.14\big]$
Answer
Since $\triangle\text{OAD}$ is in equilateral triangle,
$\angle\text{O}=\angle\text{A}=\angle\text{B}=60^\circ$ and $\text{OA}=\text{OB}=\text{AB}=12\text{cm}$
$\Rightarrow$ Radius of the ciecle = r = 6cm
Area of the shaded circular part
$=\pi\text{r}^2-\frac{\pi\text{r}^2\times60}{360}$
$=\pi\text{r}^2\Big(1-\frac{1}{6}\Big)$
$=3.14\times6\times6\times\frac{5}{6}$
$=94.2\text{cm}^2$
Area of shaded tringular region
$=\frac{\sqrt{3}}{4}\times(\text{side})^2-\frac{\pi\text{r}^2\times60}{360}$
$=\frac{\sqrt{3}}{4}\times12\times12-\frac{\frac{22}{7}\times6\times6\times60}{360}$
$=36\sqrt{3}-18.86$
$=36\times1.73-18.86$
$=62.28-18.86$
$=43.42\text{cm}^2$
$\therefore$ Area of the shaded region
= Area of the shaded circular part + Area of shaded triangular region
$=94.2\text{cm}^2+43.42\text{cm}^2$
$=137.62\text{cm}^2$
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Question 185 Marks
In the given figure, PSR, RTQ and PAQ are three semicircles of diameter 10cm, 3cm, and 7cm respectively. Find the perimeter of shaded region. $\big[\text{Use }\pi=3.14\big]$
Answer
Perimeter of shaded region
= Perimeter of semicircle PSR + Perimeter of semicircle RTQ + Perimeter of semicircle PAQ
⇒ Perimeter $=(5\pi+1.5\pi+3.5\pi)\text{cm}$
$=10\pi\text{cm}$
$=10\times3.14\text{cm}$
$=31.4\text{cm}$
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Question 195 Marks
In the given figure, OABC is a square of side 7cm. If COPB is a quadrant of a circle with centre C find the area of the shaded region.
Answer
Side of a square = 7cm
⇒ Area of a square $= (7)^2 = 49cm^2$​​​​​​​
Now, radius of a circle r = side of a square = 7cm
⇒ Area of a quadrant of a circle $=\frac{1}{4}\times\frac{22}{7}\times7\times7=\frac{77}{2}=38.5\text{cm}^2$
$\therefore$ Area of the shaded region = Area of a square - Area of a quadrants of circle
$= (49 - 38.5)cm^2$
$= 10.5cm^2​​​​​​​$​​​​​​​
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Question 205 Marks
The diameters of the front and rear wheels of a tractor are 80cm and 2m respectively. Find the number of volutions that a rear wheel makes to cover the distance which the front wheel covers in 800 revolutions,
Answer
Radius of the front wheel $=40\text{cm}=\frac{2}{5}\text{m}$
Circumference of the front wheel $=\Big(2\pi\times\frac{2}{5}\Big)\text{m}=\frac{4\pi}{5}\text{m}$
Distance moved by it in 800 revolution
$=\Big(\frac{4\pi}{5}\times800\Big)\text{m}(640\pi)\text{m}$
Circumference of rear wheel $=(2\pi\ 1)\text{m}=(2\pi)\text{m}$
Required number of revolutions $=\Big(\frac{640\pi}{2\pi}\Big)=320$
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Question 215 Marks
In the given figure, O is the centre of the bigger circle, and AC is its diameter. Another circle with AB as diameter is drawn. If AC = 54cm and BC = 10cm, find the area of the shaded region.
Answer


Diameter of bigger circle = AC = 54cm
Radius of bigger circle $=\frac{\text{AC}}{2}$
$=\Big(\frac{54}{2}\Big)\text{cm}$
$=27\text{cm}$
Diameter AB smaller circle
$= AC - BC$
$= (54 - 10)cm = 44cm$
Radius of smaller circle $=\frac{44}{2}\text{cm}=22\text{cm}$
Area of bigger circle $=\pi\text{r}^2=\Big(\frac{22}{7}\times27\times27\Big)\text{cm}^2$
$= 2291.14cm^2$​​​​​​​
Area of smaller circle $=\pi\text{r}^2=\Big(\frac{22}{7}\times22\times22\Big)\text{cm}^2$
$= 1521.11cm^2​​​​​​​$​​​​​​​
Area of shaded region = area of bigger circle - area of smaller circle
$= (2291.14 - 1521.14)cm^2$
$= 770cm^2​​​​​​​$​​​​​​​
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Question 225 Marks
In the given figure, ABCD is a square of side 7cm, DPBA and DQBC are quadrants of circles each of the radius 7cm. Find the area of shaded region.
Answer
Area of the shaded region
= Area of quadrant DPBA + Area of quadrant DQBC - Area of a square ABCD
$=\Big\{\Big(\frac{1}{4}\times\frac{22}{7}\times7\times7\Big)+\Big(\frac{1}{4}\times\frac{22}{7}\times7\times7\Big)-(7\times7)\Big\}\text{cm}^2$
$=\Big(\frac{77}{2}+\frac{77}{2}-49\Big)\text{cm}^2$
$=(77-49)\text{cm}^2$
$=28\text{cm}^2$
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Question 235 Marks
The length of an arc of a circle, subteding an angle of 54° at the centre is 16.5cm. Calculate the radius, circumference and area of the circle. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
Answer
Length of arc $=\frac{2\pi\text{r}\theta}{360}=16.5\text{cm}$
$2\times\frac{22}{7}\times\text{r}\times\frac{54^\circ}{360^\circ}=16.5$
$\text{r}=\frac{16.5\times7\times360}{2\times22\times54}=17.5\text{cm}$
Circumference of circle $=2\pi\text{r}$
$\Big(2\times\frac{22}{7}\times17.5\Big)=110\text{cm}$
Area of circle $=\pi\text{r}^2=\Big(\frac{22}{7}\times17.5\times17.5\Big)\text{cm}^2$
$=962.5\text{cm}^2$
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Question 245 Marks
Four cows are tethered at the four comers of a square field of side 50m such that each can graze the maximum unshared area. What area will be left ungrazed? $[\text{Tale }\pi=3.14]$
Answer

Ungrazed area
= Shadwd area
$=\Big[(50\times50)-\frac{4\times\pi\times(25)^2\times90}{360}\Big]\text{m}^2$
$=[2500-3.14\times25\times25]\text{m}^2$
$=[2500-1962.5]\text{m}^2$
$=537.5\text{m}^2$
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Question 255 Marks
In the given figure, APB and AQO ar semicircles and AO = OB. if the perimeter of the figure is 40cm, find the area of the shaded region.
Answer
Let AO = OB = r
Perimeter of the given figure = Perimeter of arc APB + OB + Perimeter of arc OQA
$\Rightarrow40=\pi\text{r}+\text{r}+\frac{\pi\text{r}}{2}$
$=40=\Big(\frac{3\pi}{2}+1\Big)\text{r}$
$=40=\Big(\frac{3\times22}{2\times7}+1\Big)\text{r}$
$=40=\Big(\frac{33}{7}+1\Big)\text{r}$
$\Rightarrow40=\frac{40}{7}\text{r}$
$\Rightarrow\text{r}=7\text{cm}$
Now, Area of the shaded region
Area of semicircle APB + Area of semicircle AQO
$=\Big[\Big(\frac{1}{2}\times\frac{22}{7}\times7\times7\Big)+\Big(\frac{1}{2}\times\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}\Big)\Big]\text{cm}^2$
$=\Big(77+\frac{77}{4}\Big)\text{cm}^2$
$=\frac{384}{4}\text{cm}^2$
$=96.25\text{cm}^2$
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Question 265 Marks
The cost of fencing a circular field at the rate of Rs. 25 per metre is t Rs. 5500. The field is to be ploughed at the rate of 50 paise per $m^2$​​​​​​​ Find the cos of ploughing the field. $\Big[\text{Take }\pi=\frac{22}{7}.\Big]$
Answer
Cost of fencing a circular field = Rs. 5500
Rate of fencing per metre = Rs. 25
$\therefore$ Perimeter of a circular field $=\frac{\text{Cost of fencing}}{\text{Rate Per metre}}=\Big(\frac{5500}{25}\Big)\text{m}=220\text{m}$
Let r be the radius of the circular field.
Then, $2\pi\text{r}=220$
$\Rightarrow2\times\frac{22}{7}\times\text{r}=220$
$\Rightarrow\text{r}=220\times\frac{7}{44}$
$\Rightarrow\text{r}=35\text{m}$
$\therefore$ Area of the circular field $=\pi\text{r}^2=\Big(\frac{22}{7}\times35\times35\Big)=3850\text{m}^2$
Cost of ploughing per $m^2$​​​​​​​​​​​​​​ = 50 paise
$\therefore$ Cost of ploughing 3850 $m^2​​​​​​​$​​​​​​​ $= \text{Rs.}\frac{50}{100}\times3850=\text{Rs.}1925$
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Question 275 Marks
If a square is inscribed in a circle, find the ratio of the area of the circle and the square.
Answer
Let the radius of circle be r cm

Than diagonal of square = diameter of circle = 2r cm
Area of the circle $=(\pi\text{r}^2)=\text{cm}^2$
Area of square $=\frac{1}{2}\times(\text{diagonal})^2$
$=\frac{1}{2}\times4\text{r}^2=2\text{r}^2\text{cm}$
Ratio $=\frac{\text{Area of circle}}{\text{Area of square}}=\frac{\pi\text{r}^2}{2\text{r}^2}=\frac{\pi}{2}=(\pi:2)$
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Question 285 Marks
From a thin metallic piece in the shpe of a trapezium ABCD in which $\text{AB}||\text{CD}$ and $\angle\text{BCD}=90^\circ$ a quarter circle BFEC is removed. Given, AB = BC = 3.5cm and DE = 2cm, calculate the area of remaining (shaded) part of metals sheet.
Answer
Clearly, AB = BC = CE = 3.5cm and DE = 2cm
⇒ CD = DE + EC = 2 + 3.5 = 5.5cm
$\therefore$ Area of the shaded part = Area of trapezium ABCD - Area of quadrant BCE
$=\Big[\Big\{\frac{1}{2}(\text{AB+CD}\times\text{BC})\Big\}-\Big\{\frac{1}{4}\times\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}\Big\}\Big]\text{cm}^2$
$=\Big[\Big\{\frac{1}{2}(3.5+5.5)\times3.5\Big\}-\frac{77}{8}\Big]\text{cm}^2$
$=\Big[\Big\{\frac{1}{2}\times9\times3.5\Big\}-\frac{77}{8}\Big]\text{cm}^2$
$=[15.75-9.625]\text{cm}^2$
$=6.125\text{cm}^2$
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Question 295 Marks
Find the area of a quadrant of a circle whose circumference is 88cm. $[\text{Take }\pi\ =3.14]$
Answer
Let r be the radius of a circle.
Circumference of a circle = 88cm
$\Rightarrow2\pi\text{r}=88$
$\Rightarrow2\times\frac{22}{7}\times\text{r}=88$
$\Rightarrow\text{r}=\frac{88\times7}{2\times22}$
$\Rightarrow\text{r}=14\text{cm}$
$\therefore$ Area of a quadrant $=\frac{1}{4}\pi\text{r}^2=\frac{1}{4}\times\frac{22}{7}\times14\times14=154\text{cm}^2$
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Question 305 Marks
In the given figure, OABC is a quadrant of a circle with centre O and radius 3.5cm. If OD = 2cm, find the area of the shaded region.
Answer

Shaded area = (area of quadrant) - (area of DAOD)
$=\Big[\frac{1}{4}\pi\text{r}^2-\frac{1}{2}\times\text{h}\times\text{b}\Big]$
$=\Big[\frac{1}{4}\times\frac{22}{7}\times3.5\times3.5-\frac{1}{2}\times2\times3.5\Big]\text{cm}^2$
$=(9.625-3.5)\text{cm}^2=6.125\text{cm}^2$
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Question 315 Marks
A chord of a circle of radius 30cm makes an angle of 60° at the centre of the circle. Find the areas of the minor and major segments. $[\text{Take }\pi=3.14\text{ and }\sqrt{3}=1.732]$
Answer
Let AB be the chord of circle of centre O and radius = 30cm such that $\angle\text{AOB}=60^\circ$

Area of the sector OACBO
$=\Big(3.14\times30\times30\times\frac{60}{360}\Big)\text{cm}^2$
$=471\text{cm}^2$
Area of $\triangle\text{OAB}=\frac{1}{2}\text{r}^2\sin\theta=\Big(\frac{1}{2}\times30\times30\times\sin60^\circ\Big)\text{cm}^2$
$=\Big(\frac{1}{2}\times30\times30\times\frac{\sqrt{3}}{2}\Big)\text{cm}^2=(225\sqrt{3})\text{cm}^2$
$=(225\times1.73)\text{cm}^2=389.25\text{cm}^2$
Area of the monor segment ACBA
= (area of the sector OACBO) - $($area of the $\triangle\text{OAB})$
$=(471-389.25)\text{cm}^2=81.75\text{cm}^2$
Area of the major segment BADB
= (area of circle) - (area of the minor sement)
$=[(3.14\times30\times30)-81.75)]\text{cm}^2=2744.25\text{cm}^2$
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Question 325 Marks
Find the area of the shaded region in the given figure, if ABCD is a rectangle with sides 8cm and 6cm and O is the centre of the circle.
Answer
Length of a rectangle = 8cm
Breadth of a rectangle = 6cm
$\therefore$ Area of rectangle $ABCD = 8 \times 6 = 48cm^2$​​​​​​​
Now,
$AC^2 = AB^2 + BC^2 = 8^2 + 6^2 = 64 + 36 = 100$
$\Rightarrow\text{AC}=\sqrt{100}=10\text{cm}$
⇒ Diameter of a circle = 10cm
⇒ Radius of a circle = 5cm
$\therefore$ Area of a circle $=\frac{22}{7}\times5\times5=78.57\text{cm}^2$
Thus, area of shaded region = Area of a circle - Area of rectangle ABCD
$= (78.57 - 48)cm^2$
$= 30.57cm^2​​​​​​​$​​​​​​​
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Question 335 Marks
A wire is bent to form a square enclosing an area of $484m^2$​​​​​​​. Using the same wire, a circle is formed. Find the area of the circle.
Answer
Area of a square formed $= 484m^2$​​​​​​​
$\Rightarrow \text{(Side)}^2 = 484$
$\Rightarrow \text{Side}=\sqrt{484}\text{m}=22\text{m}$
$\therefore$ Perimeter of a square = 4 × side = 4 × 22 = 88m
Let r be the radius of the circle formed
Now,
circumference of a circle = Perimeter of square
$\Rightarrow2\times\frac{22}{7}\times\text{r}=88$
$\Rightarrow\text{r}=\Big(88\times\frac{7}{44}\Big)=14\text{m}$
$\therefore$ Area of the circle $=\pi\text{r}^2=\Big(\frac{22}{7}\times14\times14\Big)\text{m}^2=616\text{m}^2$
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Question 345 Marks
Four equal circles, each of radius a units, touch each other. show that area between them is $\Big(\frac{6}{7}\text{a}^2\Big)$ sq units. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
Answer

Required area = [area of square - areas of quadrants of circles]
Let the side = 2a unit and radius = a units
Area of square = (side side) = (2a 2a) sq. units
$=4\text{a}^2\text{sq. units}$
Area of quadrant $=\frac{1}{4}\pi\text{r}^2$
Area of 4 quadrants $=4\times\frac{1}{4}\pi\text{r}^2=\pi\text{r}^2=\frac{22}{7}\times\text{a}\times\text{a}=\frac{22}{7}\text{a}^2\text{sq. units}$
Required area $=\Big(4\text{a}^2-\frac{22}{7}\text{a}^2\Big)\text{sq. units}=\frac{6\text{a}^2}{7}$
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Question 355 Marks
Four equal circles are described about the four corners of a square so that each touches two of the others, as shown in the figure. find the area of the shaded region, if each side of the square measures 14cm. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
Answer


Each side of the square is 14cm
Then, area of square $= (14 \times 14)cm^2$
$= 196cm^2$​​​​​​​
Thus, radius of each circle 7cm
Required area = area of square ABCD
-4(Area of sector with r = 7cm, $\theta$ = 90°)
$=\Big[196-4\times\frac{22}{7}\times7\times7\times\frac{90}{360}\Big]\text{cm}^2$
$=[196-154]\text{cm}^2$
$=42\text{cm}^2$
Area of the shaded region $= 42cm^2​​​​​​​$​​​​​​​
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Question 365 Marks
The area of an equilateral triangle is $49\sqrt{3}\text{cm}^2.$ 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 the triangle not included in the circles. $\big[\text{Take }\sqrt{3}=1.73\big]$
Answer
Area of equilateral triangle $\text{ABC}=49\sqrt{3}\text{cm}^2$
Let a be its side $\therefore\frac{\sqrt{3}}{4}\text{a}^2=49\sqrt{3}$ Or $\text{a}^2=49\times4$ $\therefore\text{a}=7\times2$ $\Rightarrow\text{a}=14\text{cm}$ Area of sector $\text{BDF}=\pi\text{r}^2\times\frac{\theta}{360^\circ}$ $=\frac{22}{7}\times7\times7\times\frac{60}{360}\text{cm}$ $=\frac{11\times7}{3}\text{cm}^2=\frac{77}{3}\text{cm}^2$ Area pf sector BDF = Area of sector CDE = Area of sector AEF Sum of area of all the sectors $=\frac{77}{3}\times3\text{cm}^2=77\text{cm}^2$ $\therefore$ Shaded area = Area of $\triangle\text{ABC}$ - Sum of area of all sectors $=49\sqrt{3}-77\text{cm}^2=(84.77-77.00)\text{cm}^2$ $=7.77\text{cm}^2$
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Question 375 Marks
A racetrack is in the form of a ring whose inner circumference is 352m and outer circumference is 396m. Find the Width and the area of the track.
Answer
Let r m and R m be the inner and outer boundaries, respectively.
Thus, we have:
$2\pi\text{r}=352$
$\Rightarrow\text{r}=\frac{352}{2\pi}$
Also,
$2\pi\text{R}=396$
$\Rightarrow\text{R}=\frac{396}{2\pi}$
Width of the track $=(\text{R}-\text{r})$
$=\Big(\frac{396}{2\pi}-\frac{352}{2\pi}\Big)\text{m}$
$=\frac{1}{2\pi}(396-352)\text{m}$
$=\Big(\frac{1}{2}\times\frac{7}{22}\times44\Big)\text{m}$
$=7\text{m}$
Area of the track $=\pi\big(\text{R}^2-\text{r}^2\big)$
$=\pi\big(\text{R}+\text{r}\big)\big(\text{R}-\text{r}\big)$
$=\Big[\pi\Big(\frac{396}{2\pi}+\frac{352}{2\pi}\Big)\times\Big(\frac{396}{2\pi}-\frac{352}{2\pi}\Big)\Big]\text{m}^2$
$=\Big(\pi\times\frac{748}{2\pi}\times7\Big)\text{m}^2$
$=\frac{748}{2}\times7\text{m}^2$
$=2618\text{m}^2$
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Question 385 Marks
With the vertices A, B and C of a tringle ABC as centres, arcs are drawn with radii 5cm each as shown in the given figure. If AB = 14cm, BC = 48cm and CA = 50cm then find the area of the shaded region. $\big[\text{Use }\pi=3.14\big]$
Answer
Area of sector $= \frac{\angle\text{A}}{360^\circ}\times\pi\text{r}^2+\frac{\angle\text{B}}{360^\circ}\times\pi\text{r}^2+\frac{\angle\text{C}}{360^\circ}\times\pi\text{r}^2$
$=\frac{3.14\times25}{360^\circ}(\angle\text{A}+\angle\text{B}+\angle\text{C})$
$=\frac{3.14\times25}{360^\circ}\times180^\circ$
$=39.25\text{cm}^2$
Since, 14cm, 48cm and 50cm is a Pythagorean triplet, it is a right $\triangle$
Area $=\frac{1}{2}\times48\times14=336\text{cm}^2$
$\therefore$ Area of shaded portion $= (336 - 39.25)cm^2$
$= 296.75cm^2$​​​​​​​
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Question 395 Marks
In the given figure, ABCD is a rectangle with $\text{AB}=80\text{cm}$ and $\text{BC}=70\text{cm},$ $\angle\text{AED}=90^\circ$ and $\text{DE}=42\text{cm}$ A semicircle is dn wn, taking BC as diameter. Find the area o the shaded region.
Answer
Length of rectangle ABCD = AB = 80cm
Breadth of rectangle ABCD = BC = 70cm
$\therefore$ Area of rectangle $ABCD = AB \times BC = 80 \times 70 = 5600cm^2$​​​​​​​
In right-angled $\triangle\text{AED},$
$AE^2 = (AD^2 - DE^2) = (70^2 - 42^2) = (70 + 42) = 112 \times 28 = 4 \times 28 \times 28$
$\Rightarrow AE = 2 \times 28 = 56cm$
$\therefore$ Area of $\triangle\text{AED}=\frac12\times\text{DE}\times\text{AE}=\frac12\times42\times56=1176\text{cm}^2$
Area of semi-circle $=\frac12\pi\times\Big(\frac{70}{2}\Big)^2$
$=\Big\{\frac{1}{2}\times\frac{22}{7}\times35\times35\Big\}\text{cm}^2=1925\text{cm}^2$
Thus, Area of the shaded region
= Area of rectangle ABCD $- ($Area of $\triangle\text{AED}$ + Area of semi-circle$)$
$= 5600 - (1176 + 1925)$
$= 5600 - 3101$
$= 2499cm^2​​​​​​​$
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Question 405 Marks
If three circles of radius a each, drawn such that each touches the other two, prove that the area included between included between them is equal to $\frac{4}{25}\text{a}^2.$ $\big[\text{Take } \sqrt{3}=1.73\text{ and}\ \pi=3.14\big]$
Answer

Let A, B, C, be the centres of these circles. joint AB, BC, CA
Required area = $\big($area of $\triangle\text{ABC}$ with each side 2$\big)$
-3(area of sector with r = a cm, $\theta$ = 60°)
$=\Big[\frac{\sqrt{3}}{4}\times(2\text{a})^2-\frac{3\pi\text{a}^2\times60}{360}\Big]$
$=(1.73\text{a}^2-1.57\text{a}^2)$
$=0.16\text{a}^2$
$=\frac{16}{100}\text{a}^2$
$=\Big(\frac{4}{25}\text{a}^2\Big)\text{sq. units}$
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Question 415 Marks
The inside perimeter of a running track shown in the shown in the figure is 400m. the length of the straigth portions is 90m, and the ends are semicicles. if the track. Also find the length of the outer boundary of the track. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
Answer

Length of the inner curved portion
= (400 - 290)m
= 220m
Let the radius of each inner curved part be r
Then, $\frac{22}{7}\times\text{r}=110\text{m}$
$\text{r}=\Big(110\times\frac{7}{22}\Big)\text{m}=35\text{m}$
Inner radius = 35m, outer radius = (35 + 14) = 49m
$\therefore$ Area of the track = (area of 2 rectangles each 90m 14m) + (area pf circular ring with R = 49m, r = 35m)
$\Big[2\times90\times14+\frac{22}{7}\big\{(49)^2-(35)^2\big\}\Big]\text{m}^2$
$\Big[2520+\frac{22}{7}(49+35)(49-35)\Big]\text{m}^2$
$\big[2520+3696\big]\text{m}^2=6216\text{m}^2$
Length of outer boundary of the track
$\Big[2\times90+2\times\frac{22}{7}\times49\Big]\text{m}=488\text{m}$
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Question 425 Marks
In the given figure, from a rectangular region ABCD with AB = 20cm, a right triangle AED with AE = 9cm and DE = 12cm, is cut off. On the other end, taking BC as diameter, a semicircle is added on outside the region. Find the area of the shaded region. $\big[\text{Use }\pi=3.14\big]$
Answer
In right-angled $\triangle\text{AED},$
$AD^2 = DE^2 + AE^2 = 12^2 + 9^2 = 144 + 81 = 225$
$\Rightarrow\text{AD}=\sqrt{225}=15\text{cm}$
Now, Area of $\triangle\text{AED}=\frac12\times\text{DE}\times\text{AE}=\frac12\times12\times9=54\text{cm}^2$
Length of rectangle ABCD = AB = 20cm
Breadth of rectangle ABCD = AD = 15cm
$\therefore$ Area of rectangle $ABCD = AB \times BC = 20 \times 15 = 300cm^2$​​​​​​​
Area of semi-circle $=\frac{1}{2}\pi\times\Big(\frac{15}{2}\Big)^2=\Big\{\frac12\times3.14\times7.5\times7.5\Big\}\text{cm}^2=88.3125\text{cm}^2$
Thus, Area of rectangle ABCD + Area of semi-circle - Area of $\triangle\text{AED}$
$= 300 + 88.31 - 54$
$= 334.31cm^2​​​​​​​$​​​​​​​
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Question 435 Marks
Three semicircles each of diameter 3cm, a circle of diameter 4.5,cm and a semicircle of radius 4.5cm are drawn in the given figure. Find the area of the shaded region. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
Answer
Area of semi-circle PQR $=\frac{1}{2}\pi\Big(\frac92\Big)^2$
$=\frac{81\pi}{8}\text{cm}^2$
Area of region circle, $\text{A}=\pi\Big(\frac94\Big)^2$
$=\frac{81\pi}{16}\text{cm}^2$
Area of region (B + C) $=\pi\Big(\frac32\Big)^2$
$=\frac{9\pi}{4}\text{cm}^2$
Area of region $\text{D}=\frac{1}{2}\pi\Big(\frac{3}{2}\Big)^2$
$=\frac{9\pi}{8}\text{cm}^2$
Area of shaded region = Area of semicircle - Area of circle - Area of region (B + C) + Area of region D
$=\frac{18\pi}{8}-\frac{81\pi}{16}-\frac{9\pi}{4}+\frac{9\pi}{8}$
$=\frac{63\pi}{16}\text{cm}^2$
$=\frac{99}{8}\text{cm}^2$
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Question 445 Marks
A square tank has an area of $1600cm^2.$ There are four semicircular plots around it. Find the cost of turfing the plots at Rs. 12.50 per $m^2$​​​​​​​. $[\text{Take }\pi=3.14]$
Answer
Let am be the side of the square.
Area of the square
$=\text{a}^2$
Thus. we have
$\text{a}^2=1600$
$\Rightarrow\text{a}=40$
Area of the plots = 4(Area of the semicircle of radius 20m)
$=\Big|4\Big(\frac{1}{4}\pi\text{r}^2\Big)\Big|\text{m}^2$
$=\Big|\Big(\frac{1}{2}\times3.14\times20\times20\Big)\Big|\text{m}^2$
$=2512\text{m}^2$
$\therefore$ Costofturfmgtheplotsat12.50 per $m^2$​​​​​​​ = Rs. (2512 × 12.50)
= Rs. 31400
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Question 455 Marks
A round table cover has six equal designs as shown in the given figure. If the radius of the cover is 35cm then find the total area of the design. $\big[\text{Use }\sqrt{3}=1.732\text{ and }\pi=3.14\big]$
Answer


ABCDEF is a hexagon
$\therefore\angle\text{AOB}=60,$ Radius = 35cm
Area of sector AOB
$=\pi\text{r}^2\times\frac{60^\circ}{360^\circ}=\frac{\pi\times35\times35}{6}\text{cm}^2$
$=\frac{3.14\times35\times35}{6}\text{cm}^2$
$=641.083\text{cm}^2$
Area of $\triangle\text{AOB}=\frac{\sqrt{3}}{4}\times\text{r}^2=\frac{\sqrt{3}}{4}\times35\times35\text{cm}^2$
$=530.425\text{cm}^2$
Area of segment $APB = (641.083 = 530.425)cm^2 = 110.658cm^2$​​​​​​​
Area of design (shaded area) $= 6110.658m^2 = 663.948m^2$
$= 663.948m^2​​​​​​​$​​​​​​​
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Question 465 Marks
Three equal circles, each of radius 6cm, touch one another as shown in the figune. find the area enclosed between them. $\big[\text{Take }\pi=3.14\text{ and }\sqrt{3}=1.732.\big]$
Answer


Let A, B, C, be the centres of these circles. joint AB, BC, CA
Required area = $\big($area of $\triangle\text{ABC}$ with each side a = 12cm$\big)$
-3(area of sector with r = 6, $\theta$ = 60°)
$=\Big[\frac{\sqrt{3}}{4}\times(12)^2-3\times\Big(3.14\times(6)^2\times\frac{60}{360}\Big)\Big]$
$=\Big[\frac{\sqrt{3}}{4}\times12\times12-3\times3.14\times6\Big]\text{cm}$
$=(36\times1.73-56.52)\text{cm}^2$
$=(62.28-56.52)\text{cm}^2$
$=5.76\text{cm}^2$
The area enclosed $= 5.76cm^2$​​​​​​​
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Question 475 Marks
In the given figure, O is the centre of the circle with $\text{AC}=24\text{cm},\ \text{AB}=7\text{cm}$ and $\angle\text{BOD}=90^\circ$ Find the area of shaded region. $\big[\text{Use }\pi=3.14\big]$
Answer
In right-angled $\triangle\text{BAC},$
$CB^2 = AC^2 + AB^2 = 24^2 + 7^2 = 576 + 49 = 625$
$\Rightarrow\text{CB}=\sqrt{325}=25\text{cm}$
$\Rightarrow\text{OC}=\frac12\text{CB}=\frac{25}{2}\text{cm}=12.5\text{cm}$ = radius of the circle
Now, area of $\triangle\text{BAC}=\frac12\times\text{AC}\times\text{AB}=\frac{1}{2}\times24\times7=84\text{cm}^2$
Area of the circle $= 3.14 \times 12.5 \times 12.5 = 490.625cm^2$​​​​​​​
Now, area of the shaded region
= Area of the circle - Area of $\triangle\text{BAC}$ - Area of quadrant COD
$= (490.625 - 84 - 122.66)cm^2$
$= 283.96cm^2​​​​​​​$​​​​​​​
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Question 485 Marks
In the given figure, the side of suare is 28cm 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. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
Answer

Side of square = 28cm and radius of each circle $=\frac{28}{2}\text{cm}$
Area of the shaded region = Area of the square + Area of the two circle - Area of the quadrants
$=(28)^2+2\times\pi\times\Big(\frac{28}{2}\Big)^2-2\times\frac14\times\pi\times\Big(\frac{28}{2}\Big)^2$
$=(28)^2+\frac32\times\pi\times\Big(\frac{28}{2}\Big)^2$
$=(28)^2\Big(1+\frac32\times\frac{22}{7}\times\frac12\times\frac{1}{2}\Big)$
$=(28)^2\Big(1+\frac{33}{28}\Big)$
$=(28)^2\times\frac{61}{28}$
$=28\times61$
$=1708\text{cm}^2$
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Question 495 Marks
PQRS is a diameter of a circle of radius 6cm. The lengths PQ, QR an RS are equal. Semicircles are drawn with PQ and QS as diameters, as shown in the given figure. If PS= 12cm, find the perimeter and area of the shaded region. $\big[\text{Take }\pi=3.14\big]$
Answer

PS = 12cm
PQ = QR = RS = 4cm, QS = 8cm
Perimeter = arc PTS + arc PBQ + arc QES
$=(\pi\times6+\pi\times2+\pi\times4)\text{cm}$
$=12\pi \ \text{cm}$
$=12\pi=12\times3.14\text{cm}$
$=37.68\text{cm}$
Area of shaded region = (area of the semi cirlce PBQ) + (area of semicircle PTS) - (area of semicircle QES)
$=\Big[\frac{1}{2}\pi\times(2)^2\times\pi\times(6)^2-\frac{1}{2}\times\pi\times(4)^2\Big]\text{cm}^2$
$=\big[2\pi+18\pi-8\pi\big]=12\pi \ \text{cm}^2=(12\times3.14)\text{cm}^2$
$=37.68 \ \text{cm}^2$
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Question 505 Marks
In the given figure, two concentric circles with centre O have radii 21cm and 42cm. If $\angle\text{AOB}=60^\circ,$ find the area of the shaded region. $\Big[\text{Use }\pi=\frac{22}{7}.\Big]$
Answer
Radii of the concentric circles = 21cm and 42cm
Area between the circles $=\pi\text{R}^2-\text{r}^2=\Big(\frac{22}{7}\Big)\big(42^2-21^2\big)=4158\text{cm}^2$
Angle subtended by the arc in the inner circle $= 60^\circ$
Area of the sector in the inner circle
$=\Big(\frac{60^\circ}{360^\circ}\Big)\times\pi\text{R}^2=\Big(\frac{60^\circ}{360^\circ}\Big)\times\Big(\frac{22}{7}\Big)\times(42)^2=924\text{cm}^2$
Area of the portion of the sector in between the circles $= 924 - 231 = 693cm^2$
Area of the shaded portion = (Area between the circles) - (Area of the portion of the sector in between the circles)
$= 4158 - 693$
$= 3465$
Therefore, the area of the shaded region is $3465cm^2.$
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