3 Marks Question - Maths STD 10 Questions - Vidyadip

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3 Marks Question

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28 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks

The circumference of a circle exceeds its diameter by 45cm. find the circumference of the circle. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$

Answer

Let r be the radius of the circle.
⇒ Diameter of a cirde = 2r
And, circumference of a circle $=2\pi\text{r}$
It is given that,
Circumference of a circle - Diameter of a circle = 45cm
$\Rightarrow2\pi\text{r}-2\text{r}=45$
$\Rightarrow2\text{r}(\pi-1)=45$
$\Rightarrow2\text{r}\Big(\frac{22}{7}-1\Big)=45$
$\Rightarrow\text{r}\Big(\frac{22-7}{7}\Big)=\frac{45}{2}$
$\Rightarrow\text{r}\times\frac{15}{7}=\frac{45}{2}$
$\Rightarrow\text{r}=\frac{45\times7}{15\times2}$
$\Rightarrow\text{r}=10.5\text{cm}$
$\therefore$ Circumference of a circle $=2\times\frac{22}{7}\times10.5=66\text{cm}$

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Question 23 Marks

Find the lengths of the arcs cut off from a circle of radius 12cm by a chord 12cm long. Also, find the area of the minor gment. $\big[\text{Take }\pi=3.14\text{ and }\sqrt{3}=1.73.\big]$

Answer

$\triangle\text{OAB}$ is equilateral.
So, $\angle\text{AOB}=60$

Arc $\text{ACB}=\Big(2\pi\times12\times\frac{60}{360}\Big)\text{cm}$
$=4\pi\text{cm}$
$=(4\times3.14)\text{cm}$
$=12.56\text{cm}$
Length of arc $\text{BDA}=(2\pi12-\text{Arc}\text{ACB})\text{cm}$
$=24\pi-4\pi\text{cm}=(20\pi)\text{cm}$
$=(20\ 3.14)\text{cm}=62.8\text{cm}$
Area of the minor segment ACBA
$=\Big[\pi\times(12)^2\times\frac{60}{360}-\frac{\sqrt{3}}{4}\times(12)^2\Big]\text{cm}^2$
$=\Big(3.14\times12\times12\times\frac{60}{360}-\frac{1.73}{4}\times12\times12\Big)\text{cm}^2$
$=(75.36-62.28)\text{cm}^2=13.08\text{cm}^2$

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Question 33 Marks

In the given figure, find the area of the shaded region, if ABCD is a square of side 14cm and APO and BPC are semicircles.

Answer

Side of a square = 14cm
⇒ Diameter of a semicircle = 14cm
⇒ Radius of a semicircle = 7cm
Now,
Area of a shaded region = Area of a square - Area of two semicircles
$=\Big[(14\times14)-\Big(\frac{22}{7}\times7\times7\Big)\Big]\text{cm}^2$
$=(196 - 154)\text{cm}^2$
$=42\text{cm}^2$

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Question 43 Marks

Find the area of the sector of a circle having radius 6 cm and of angle $30^\circ$. $\big[\text{Take }\pi=3.14\big]$

Answer

Radius of a circle = r = 6cm
Central angle $=\theta=30^\circ$
$\therefore$ Area of the sector $=\frac{\pi\text{r}^2\theta}{360}$
$=\Big(\frac{3.14\times6\times6\times30^\circ}{360^\circ}\Big)\text{cm}^2$
$= 9.42cm^2$

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Question 53 Marks

The wheels of a car make 2500 revolutions in covering a distance of 4.95km. Find the diameter of a wheel.$\Big[\text{Use }\pi=\frac{22}{7}\Big]$

Answer

Distance covered by the wheel in 1 revolution
$=\Big(\frac{4.95\times1000\times100}{2500}\Big)\text{cm}=198\text{cm}$
$\therefore$ The circumference of the wheel = 198cm
Let the diameter of the wheel be d cm
Then, $\pi\text{d}=198\Rightarrow\frac{22}{7}\times\text{d}=198$
$\Rightarrow\text{d}=\frac{198\times7}{22}=63\text{cm}$
Hence diameter of the wheel is 63cm

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Question 63 Marks

A chord PQ of a circle of radius 10cm subtends an angle of $60^\circ$ at the centre of the circle. Find the area of major and minor segments of the circle. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$

Answer

Radius of a circle = r = 10.5cm
Area of a sector $= 69.3cm^2$
Now, area of the sector $=\frac{\pi\text{r}^2\theta}{360}$
$\Rightarrow69.3=\frac{\frac{22}{7}\times10.5\times10.5\times\theta}{360^\circ}$
$\Rightarrow69.3=\frac{11\times1.5\times10.5\times\theta}{180}$
$\Rightarrow\theta=\frac{69.3\times180}{11\times1.5\times10.5}$
$\Rightarrow\theta=72^\circ$

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Question 73 Marks

In a circle of radius 7cm, a square ABCD is inscribed. Find the area of the circle which is outside the square.

Answer

Radius of a circle = 7cm
⇒ Diagonal of the square = 2 × 7 = 14cm
Now,
Area of the square $=\frac{1}{2}\times(\text{diagonal})^2=\Big(\frac{1}{2}\times14\times14\Big)\text{cm}^2=98\text{cm}^2$
Area of the circle $=\Big(\frac{22}{7}\times7\times7\Big)\text{cm}^2=154\text{cm}^2$
$\therefore$ Required area = Area of the circle - Area of the square
$= (154 - 98)cm^2$
$= 56cm^2$

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Question 83 Marks

Find the radius of a circle whose perimeterter and are are numerically equal.

Answer

let r be the radius of a circle.Then, area of a circle $=\pi\text{r}^2$
Perimeter of a circle $=2\pi\text{r}$
It is given that,
Area of a circle = Perimeter of a circle
$\Rightarrow\pi\text{r}^2=2\pi\text{r}$
$\Rightarrow \text{r}= 2 \text{ units}$

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Question 93 Marks

A park is in the form of a rectangle 120m by 90m. At the centre of the park, there is a circular lawn as shown in the figure. The area of the park excluding the lawn is $2950m^2$ Find the radius of the circular lawn. $\big[\text{Given }\pi = 3.14.\big]$

Answer

Area of rectangle $= (120 \times 90)m^2$
$= 10800m^2$
Area of circular lawn = [Area of rectangle - Area of park excluding circular lawn]
$= [10800 - 2950]m^2 = 7850m^2$
Area of circular lawn = 7850 $\Rightarrow\pi\text{r}^2=7850\text{m}^2$
$3.14 \times r^2 = 7850m^2$
$\text{r}^2=\Big(\frac{7850}{3.14}\Big)\text{m}^2$
$=2500\text{m}^2$
$\text{r}=\sqrt{2500}\text{m}$
or $\text{r}=50\text{m}$
Hence, radius of the circular lawn = 50m

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Question 103 Marks

In the given figure, PQSR represents a flower bed. If OP = 21m and OR = 14m, find the area of the flower bed.

Answer

Area of flower bed = (area of quadrant OPQ) - (area of the quadrant ORS)
$=\Big[\frac{1}{4}\pi\text{r}^2_1-\frac{1}{4}\pi\text{r}^2_2\Big]$
$=\Big[\frac{1}{4}\times\frac{22}{7}\times21\times21-\frac{1}{4}\times\frac{22}{7}\times14\times14\Big]\text{m}^2$
$=[346.5-154]\text{m}^2=192.5\text{m}^2$

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Question 113 Marks

A circular disc of radius 6cm is divided into three sectors with central angles 90°, 120° and 150°. what part of the whole circle is the secto with central angle 150°? Also, calculate the ratio of the areas of the three sectors.

Answer

$\frac{\text{Area of sector with} \ \theta \ = \ 150^\circ}{\text{Area of the circle}}=\frac{\pi\times(6)^2\times\frac{150}{360}}{\pi\times(6)^2}$
$=\frac{150}{360}=\frac{5}{12}$
Required ratio $=\Big(36\pi\times\frac{90}{360}\Big):\Big(36\pi\times\frac{120}{360}\Big):\Big(36\pi\times\frac{150}{360}\Big)$
$=\frac{1}{4}:\frac{1}{3}:\frac{5}{12}=3:4:5$

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Question 123 Marks

The length of a chain used as the bounary of a semicircular park is 108m. Find the area of the park $\Big[\text{Use }\pi=\frac{22}{7}\Big]$

Answer

The length of a chain used as the bounary of a semicircular park is 108m. Find the area of the park.
Let r be the radius of the semicircular park.
Now, perimeter of a semicircular park = 108m
$\Rightarrow\pi\text{r}+2\text{r}=108$
$\Rightarrow\Big(\frac{22}{7}+2\Big)\text{r}=108$
$\Rightarrow\Big(\frac{22+14}{7}\Big)\text{r}=108$
$\Rightarrow\frac{36}{7}\text{r}=108$
$\Rightarrow\text{r}=\frac{108\times7}{36}=21\text{cm}$
$\therefore$ Area of the park $=\frac{1}{2}\times\frac{22}{7}\times21\times21=693\text{m}^2$

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Question 133 Marks

Two circular pieces of equal radii and maximim area, touching each other are cut out from a rectangular cardboard of dimensions 14cm × 7cm. Find the area of the r maining cardboard.

Answer

Since the dimensions of a rectangular cardboard are 14cm × 7cm,
the diameter of each circle is 7cm,
Now,
Area of the rectangular cardboard $= 14 \times 7 = 98cm^2$
Area of two circle $=2\times\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}=77\text{cm}^2$
$\therefore$ Area of the reamaining cardboard
= Area of the rectangular cardboard - Area of two circles
$= (98 - 77)cm^2$
$= 21cm^2$

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Question 143 Marks

In a circle of radius 10.5cm, the minor arc is one-fifth of the major arc. find the area of the sector corresponding to the major are. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$

Answer

Let the major arc be x cm long
Then, length of the minor arc $=\frac{1}{5}\times\text{cm}$
Circumference $=\Big(\text{x}+\frac{1}{5}\text{x}\Big)\text{cm}=\frac{6\text{x}}{5}\text{cm}$
$\frac{6\text{x}}{5}=2\times\frac{22}{7}\times\frac{21}{2}\Rightarrow\text{x}=55\text{cm}$
Required area $=\Big(\frac{1}{2}\times55\times\frac{21}{2}\Big)\text{cm}^2$
$\Big[\text{Area}=\frac{1}{2}\text{rl}\Big]$
$=2988.75\text{cm}^2$

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Question 153 Marks

In the given figure, find the area of the shaded region, where ABCD is a square of side 14cm and all circles are of the same diameter.

Answer

Side of the square ABCD = 14cm
Area of square $ABCD = 14 14 = 196cm^2$
Radius of each circle $=\frac{14}{4}=3.5\text{cm}$
Area of the circles = 4 area of one circle
$=4\times\pi(3.5)^2$
$=4\times\frac{22}{7}\times3.5\times3.5$
$=154\text{cm}^2$
Area of shaded region = Area of square - area of 4 circles
$= 196 - 154 = 42cm^2$

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Question 163 Marks

If the area of a circle is numerically equal to twice its circumference then what is the diameter of the circle?

Answer

Let r be the radius of the circle.
It is given that,
Area of a circle = 2 × Circumference of a circle
$\Rightarrow\pi\text{r}^2=2\times2\pi\text{r}$
$\Rightarrow r^2 = 4r$
$\Rightarrow r =4$
⇒ Diameter = 2r = 2 × 4 = 8cm

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Question 173 Marks

The sum of the radii of two circles is 7cm, and the differece of their circumferences is 8cm. Find the circumferences of the circleas.$\Big[\text{Use }\pi=\frac{22}{7}\Big]$

Answer

Let the radii of cricles be x cm and (7 - x)cm
Then,
$2\pi\text{x}-[2\pi(7-\text{x})]=8$
$2\pi\text{x}-[14\pi-2\pi\text{x}]=8$
$2\pi\text{x}-14\pi+2\pi\text{x}=8$
$4\pi\text{x}-14\pi=8$
$2\pi\text{x}=4+7\pi$
$2\pi\text{x}=4+22$
$2\pi\text{x}=26$
Substitute the value of $2\pi\text{x}$ in $2\pi(7-\text{x})$
$=14\pi-2\pi\text{x}=14\times\frac{22}{7}-26$
$=44-26=18\text{cm}$
Circumference of the circles are 26cm and 18cm

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Question 183 Marks

A racetrack is in the form of a ring whose inner circumfarence is 352m and outer circumference is 396m. Find the width and the area of the track. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$

Answer

Let r m and R m be the radii of inner circle and outer boundaries respectively.
Then, $2\pi\text{r}=352$ and $2\pi\text{R}=396$
$\text{r}=\frac{352}{2\pi},\ \text{R}=\frac{396}{2\pi}$
Width of the track = (R - r)m
$=\Big(\frac{396}{2\pi}-\frac{352}{2\pi}\Big)\text{m}=\Big(\frac{44}{2\pi}\Big)\text{m}$
$=\Big(\frac{44}{2}\times\frac{7}{22}\Big)\text{m}=7\text{m}$
Area the track $=\pi(\text{R}^2-\text{r}^2)=\pi(\text{R}+\text{r})(\text{R}-\text{r})$
$=\Big[\pi\Big(\frac{352}{2\pi}+\frac{396}{2\pi}\Big)\times7\Big]\text{m}^2$
$=\Big[\Big(\pi\times\frac{748}{2\pi}\Big)\times7\Big]\text{m}^2=(374\times7)\text{m}^2$
$=2618\text{m}^2$

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Question 193 Marks

The radius of a circle with centre O Is 7cm. Two radii OA and OB are drawn at right angles to each other find the area of minor and major segments. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$

Answer

Radius of the circle = r = 7cm
Central angle $=\theta=90^\circ$
$=\frac{22}{7}\times7\times7\times\frac{90}{360}-\frac{1}{2}\times7\times7\times\sin90^\circ$
$=\frac{77}{2}-\frac{49}{2}$
$=\frac{7-49}{2}$
$=\frac{28}{2}$
$=14\text{cm}^2$
Area of a cirde $=\pi^2=\frac{22}{7}\times7\times7=154\text{cm}^2$
Area of the major segment = Area of a circle - Area of the monor segment
$= (154 - 14)cm^2$
$= 140cm^2$

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Question 203 Marks

The perimeter of a certain sector of a circle of radius 6.5cm is 31cm. Find the area of the sector.

Answer

Let sector of circle is OAB Perimeter of a sector of circle = 31cm OA + OB + length of arc AB = 31cm

6.5 + 6.5 + arc AB = 31cm arc AB = 31 - 13 = 18cm Area of circle $=\frac{1}{2}$ lr $=\frac{1}{2}\times18\times6.5=58.5\text{ cm}^2$

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Question 213 Marks

Find the diameter of the circle whose area is equal to the sum of the areas of two circles having radii 4cm and 3cm.

Answer

Let the r radius of the large circle be R. Then, We have Area of large circle of radius R = Area of a circle of radius 4cm + Area of a circle of radius 3cm $\Rightarrow\pi\text{R}^2=\Big(\pi\times4^2+\pi\times3^2\Big)$ $\Rightarrow\pi\text{R}^2=(16\pi+9\pi)$ $\Rightarrow\text{R}^2=25$$\Rightarrow\text{R}=5\text{cm}$
$\Rightarrow\text{Diameter}=2\text{R}=10\text{cm}$

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Question 223 Marks

A sector is cut from a circle of radius 21cm. The angle of the sector is 150°. Find the length of the arc and the area of the sector. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$

Answer

Length of the are $=\frac{2\pi\text{r}\theta}{360},\ \text{r}=21\text{cm},\theta=150^\circ$
$=\Big(\frac{2\pi\times21\times150}{360}\Big)\text{cm}=(17.5\pi\text{cm})$
Length of are $=\Big(17.5\times\frac{22}{7}\Big)\text{cm}=55\text{cm}$
Area of the sector $=\frac{\pi^2\theta}{360}=\Big(\frac{\pi\times21\times21\times150}{360}\Big)\text{cm}^2$
$=\Big(\frac{22}{7}\times183.75\Big)\text{cm}^2=577.5\text{cm}^2$

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Question 233 Marks

A horse is tethered to one comer of a field which is in the shape of an equilateral triangle of side 12m lf the lrngth of the rope is 7m, find the area of the field which the horse cannot geaze. $\text{Take }\sqrt{3}=1.732$ the answer correct to 2 places of decimal $\Big[\text{Use }\pi=\frac{22}{7}\Big]$

Answer

Each angle of equilateral triangle is 60

Area which cannot be grazed = $($area of equilateral $\triangle\text{ABC})$
= (area of the sector with r = 7m, $\theta$ = 60°)
$=\Big[\frac{\sqrt{3}}{4}\times(12)^2-\frac{22}{7}\times(7)^2\times\frac{60}{360}\Big]\text{m}^2$
$=\Big[(\sqrt{3}\times12\times3)-\frac{(22\times7)}{6}\Big]$
$=62.35-25.66\text{m}^2$
$=36.68\text{m}^2$
Area that the horse connot graze is $36.68m^2$

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Question 243 Marks

From a rectangular sheet of paper ABCD with AB = 40cm and AD = 28cm, a semicircular portion with BC as diameter is cut off. Find the area of the remaining paper.

Answer

Length of a rectangular sheet of paper $= AB = 40cm$
Breadth of a rectangular sheet of paper $= AD = 28cm$
⇒ Area of a rectangular sheet of paper $= AB \times AD = 40 \times 28 = 1120cm^2$
Diameter of a Semicircular portion $= AD = 28cm$
⇒ Radius$ = 14cm$
⇒ Area of a Semicircular portion $=\frac{1}{2}\times\frac{22}{7}\times14\times14=308\text{cm}^2$
$\therefore$ Area of the remaining paper = Area of a rectangular sheet of paper - Area of a Semicircular portion
$= (1120 - 308)cm^2$
$= 812cm^2$

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Question 253 Marks

The short and long hands of a clock are 4cm and 6cm long respectively. find the sum of distances travelled by their tips in 2 days. $[\text{Take }\pi\ =3.14]$

Answer

In 2 days, the shot hand will complete 4 rounds
$\therefore$ Distance travelled by its tip in 2 days
= 4(circumference of the circle with r = 4cm)
$=(4\times2\pi\times4)\text{cm}=32\pi\text{cm}$
In 2 days, the long hand will complete 48 rounds
$\therefore$ length moved by its tip
= 48(circumference of the circle eith r = 6cm)
$=(48\times2\pi\times6)\text{cm}=576\pi\text{cm}$
$\therefore$ Sum of the lengths moved
$=(32\pi+576\pi)=608\pi\text{cm}$
$=(608\times3.14)\text{cm}=1909.12\text{cm}$

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Question 263 Marks

In the given figure, the shape of the top of a table is that of a sector of a circle with centre O and $\angle\text{AOB}=90^\circ.$ If AO = OB = 42cm then find the perimeter of the top of the table.

Answer

$\angle\text{AOB}=90^\circ$
AO = OB = 42cm
⇒ Radius of a circle = 42cm
$\therefore$ Required perimeter = Circumference of a circle - Length of arc AB + (AO + OB)
$=\Big\{\Big(2\times\frac{22}{7}\times42\Big)-\Big(2\times\frac{22}{7}\times42\times\frac{90}{360}\Big)+(42+42)\Big\}\text{cm}$
$=\Big\{\Big(2\times\frac{22}{7}\times42\Big)-\Big(2\times\frac{22}{7}\times42\times\frac{90}{360}\Big)+(42+42)\Big\}\text{cm}$
$=(264-66+84)\text{cm}$
$=282\text{cm}$

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Question 273 Marks

In the given figure, APB and CQD are semicircles of diameter 7cm each, while ARC an BSD are semicircles of diameter 14cm each. Find the,

Perimeter,
Area of the shad d region.

Answer

Perimeter of the shaded region

= Perimeter of semicircles (ARC + BSD) + Perimeter of semicircles (APB + CQD)

$=\Big\{2\Big(\frac{22}{7}\times7\Big)+2\Big(\frac{22}{7}\times\frac{7}{2}\Big)\Big\}\text{cm}$

$=(44+22)\text{cm}$

$=66\text{cm}$

Area of the shaded region

= Area of semicircles (ARC + BSD) - Area of semicircles (APB + CQD)

$=\Big\{2\Big(\frac{1}{2}\times\frac{22}{7}\times7\times7\Big)-2\Big(\frac{1}{2}\times\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}\Big)\Big\}\text{cm}^2$

$=\Big(2\times77-2\times\frac{77}{4}\Big)\text{cm}^2$

$=(154-38.5)\text{cm}^2$

$=115.5\text{cm}^2$

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Question 283 Marks

A horse is placed for grazing inside a reangular field 70m by 2m. It is tethered to one comer by a rope 21m long. On how much area an it graze? How much area is left ungrazed? $\Big[\text{Use }\pi=\frac{22}{7}\Big]$

Answer

Area ehich the horse can graze = Area of the quadrant of radius 21m
$=\Big(\frac{1}{4}\times\frac{22}{7}\times21\times21\Big)\text{m}^2$
$=346.5\text{m}^2$
Area ungrazed $=[(70\times52)-346.5]\text{m}^2$
$=3293.5\text{m}^2$

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