CBSE Questions for Class 10 Maths Areas Related To Cricles Quiz 2 - MCQExams.com

$$\textbf{Step - 1: Verify, If we join any points on a circle we get a diameter of the circle}$$

                   $$\text{A Chord is a line segment that joins any two points of the circle.}$$

                   $$\text{The endpoints of this line segments lie on the circumference of the circle.}$$

                   $$\therefore \text{Option (A) is false}$$

$$\textbf{Step - 2: Verify, A diameter of a circle contains the center of the circle}$$

                   $$\text{Any interval joining two points on the circle and passing through the center is called a}$$

                   $$\text{diameter of the circle.}$$

                   $$\therefore \text{Option (B) is True}$$

$$\textbf{Step - 3: Verify, A semicircle is an arc}$$

                   $$\text{The arc of a circle consists of two points on the circle and all of the points on the circle that lie}$$

                   $$\text{between those two points.}$$

                   $$\text{It's like a segment that was wrapped partway around a circle.}$$

                   $$\text{An arc whose measure equals 180 degrees is called a semicircle since it divides the circle in two}$$

                   $$\therefore \text{Option (C) is True}$$

$$\textbf{Step - 4: Verify, the length of a circle is called its circumference}$$

                   $$\text{A Circle is a round closed figure where all its boundary points are equidistant from a fixed point}$$

                   $$\text{called the center.}$$ 

                   $$\text{The two important metrics of a circle is the area of a circle and the circumference of a circle.}$$

                   $$\therefore \text{Option (D) is True}$$

$$\textbf{Hence, option A is correct as it is false}$$        

Consider a circle with centre at O, radius 10 cm. AB is the chord which subtends $$90^{\circ}$$ at the centre. 
Now, In $$\triangle OAB$$, 
$$OA^2 + OB^2 = AB^2$$ (Using Pythagoras Theorem)
$$10^2 + 10^2 = AB^2$$
$$AB^2 = 200$$
$$AB = 10\sqrt{2}$$

$$\displaystyle 2\pi r=\frac{30}{\pi }$$
$$\displaystyle 2 r=\frac{30}{\pi^{2} }$$

$$r = 14 $$m
Circumference $$ = 2\displaystyle \pi r$$
$$\displaystyle =2\times \frac{22}{7}\times 14= 88m$$

Area of sector $$\displaystyle =\frac{42}{360}\times \pi r^{2}$$
$$\displaystyle =\frac{42}{360}\times \frac{22}{7}\times6\times6=13.2m^{2}$$

Given, 
$$AB=8$$ cm and $$BC=6$$ cm
The diagonal of the inscribed rectangle will be the diameter of the circle.
Thus, by Pythagoras Theorem, we get
     $$(AC)^2=(AB)^2+(AC)^2$$
                  $$=8^2+6^2$$
                 $$=64+36$$
                 $$=100$$
 $$=>AC=\sqrt{100}$$
 $$=>AC=10$$
Thus, radius $$= \frac{10}{2}$$
                     $$=5$$ cm
Area of the rectangle ABCD $$=8\times 6$$
                                                $$=48cm^2$$
Area of the circle $$=\pi (5)^2$$
                               $$=3.14\times 25$$
                               $$=78.5 cm^2$$
Thus, the area of the shaded portion is $$=(78.5-48)cm^2$$
                                                                   $$=30.5 cm^2$$

Area of sector $$OAB$$ $$=\, \displaystyle \frac {x}{360}\, \times\, \pi r^2$$
$$= \displaystyle \frac {60}{360}\, \times\, \displaystyle \frac {22}{7}\, \times\, 21\, \times\, 21$$
$$=231$$ sq. cm
Here, the triangle ABC is an equilateral triangle, so the area of $$\triangle\, OAB\, =\, \displaystyle \frac {\sqrt 3}{4}\, r^2$$
$$= \displaystyle \frac {\sqrt 3}{4}\,\times\, 21\, \times\, 21$$
$$=190.73$$ sq. cm
Therefore, area of shaded region $$=\, 231\, -\, 190.73$$
$$=\, 40.27$$ sq. cm

By Pythagoras theorem, we find that the given triangle is a right angled triangle with $$12$$ as height and $$5$$ as base. 
$$\displaystyle \therefore $$ Area of the triangle $$\displaystyle =\frac{1}{2}\times12\times5sq. units$$
$$=30$$ sq. units
$$\displaystyle \therefore $$ Area of the rectangle $$= length \times breadth = 30$$
$$\displaystyle \Rightarrow Length =\frac{30}{breadth}=\frac{30}{10}=3\, units$$
$$\displaystyle \therefore$$ Perimeter of the rectangle $$ = 2 \times ( 10 + 3 ) $$Units 
$$= 26$$ units.

Let length of each side of the square $$=a$$

 Area = $$(3\displaystyle \times 1) + (1\displaystyle \times 1) + (3\displaystyle \times 1) + (1\displaystyle \times 1) + (3\displaystyle \times 1)$$
$$= 3 + 1 + 3 + 3 + 1$$
$$= 11$$ sq cm

Given that, AC is the diameter of a circle with radius $$5 \ cm$$
and $$AB=BC, CD=8\ cm$$

$$ Let\quad the\quad given\quad perimeter\quad be\quad P.\\ We\quad calculate\quad the\quad areas\quad of\quad the\quad given\quad figures\quad \\ and\quad compare.\\ Option\quad A\longrightarrow circle.\\ The\quad perimeter(circumference)=P\\ \Longrightarrow 2\pi r=P\quad (when\quad r=radius\quad of\quad the\quad circle.)\\ \Longrightarrow r=\frac { P }{ 2\pi  } \\ \therefore \quad ar.circle=\pi { r }^{ 2 }=\pi \times { \left( \frac { P }{ 2\pi  }  \right)  }^{ 2 }=\frac { { P }^{ 2 } }{ 4\times 3.14 } =\frac { { P }^{ 2 } }{ 12.56 } .\\ Option\quad B\longrightarrow Equilateral\quad triangle.\\ One\quad side=\frac { P }{ 3 } .\\ \therefore \quad area=\frac { \sqrt { 3 }  }{ 4 } { side }^{ 2 }=\frac { \sqrt { 3 }  }{ 4 } \times { \left( \frac { P }{ 3 }  \right)  }^{ 2 }=\frac { { P }^{ 2 } }{ 20.77 } .\\ Option\quad C\longrightarrow Square\\ side=\frac { P }{ 4 } \\ \therefore \quad area=\frac { P }{ 4 } \times \frac { P }{ 4 } =\frac { { P }^{ 2 } }{ 16 } .\\ Option\quad D\longrightarrow Regular\quad pentagon.\\ side=\frac { P }{ 5 } .\\ \therefore \quad ar.pentagon=\frac { 1 }{ 4 } \sqrt { 5\left( 5+2\sqrt { 5 }  \right)  } \times { side }^{ 2 }\\ =\frac { 1 }{ 4 } \sqrt { 5\left( 5+2\sqrt { 5 }  \right)  } \times { \left( \frac { P }{ 5 }  \right)  }^{ 2 }=\frac { { P }^{ 2 } }{ 19.95 } .\\ \therefore \quad comparing\quad the\quad areas\quad we\quad get\\ the\quad ar.circle\quad is\quad the\quad greatest.\\ Ans-\quad Option\quad A\\  $$ 

$$\Rightarrow \displaystyle \frac {\theta}{360}\, \times\, \pi r^2\, =\, 9\, \displaystyle \frac {3}{7}$$
$$\Rightarrow \displaystyle \frac {120}{360}\, \times\, \displaystyle \frac {22}{7}\, \times\, r^2\, =\, \displaystyle \frac {66}{7}$$
$$\Rightarrow r^2\, =\, \displaystyle \frac {66}{7}\, \times\, \displaystyle \frac {360}{120}\, \times\, \displaystyle \frac {7}{22}\, =\, 9$$
$$\Rightarrow r\, =\, \sqrt 9\, =\, 3$$ cm

Area of shaded region $$=$$ Area of trapezium $$-2$$ $$\times $$ Area of circle

Here $$r=12cm$$, $$\theta=60^o$$
Area of the corresponding segment  $$ABXA$$
$$=\displaystyle\left[\frac{\pi r^2\theta}{360}-\frac{1}{2}r^2\sin\theta\right]$$
$$=\displaystyle\left[\frac{3.14\times 12\times 12\times 60^o}{360^o}-\frac{1}{2}\times 12\times \sin 60^o\right]$$
$$=\displaystyle\left[\frac{314}{100}\times 12\times 2-6\times 12\times \displaystyle\frac{\sqrt 3}{2}\right]cm^2$$.    $$\displaystyle\left[\because \sin 60^o=\frac{\sqrt 3}{2}\right]$$
$$=\displaystyle\left[\frac{7536}{100}-36\times 1.73\right]cm^2$$
$$=[75.36-62.28]cm^2=13.08cm^2$$

Required area
= Area of sheet - 60 $$\displaystyle \times $$ area of 1 button
$$\displaystyle=\left ( 50\times 25-60\times \dfrac{22}{7}\times 1.4\times 1.4 \right )cm^{2}$$
$$\displaystyle =\left ( 1250-369.6 \right )cm^{2}$$
$$\displaystyle =880.4cm^{2}$$

We have,
$$OA=R=21m$$ and $$OC=r=14m$$
$$\therefore$$ Area of the flower bed
=Area of a quadrant of a circle of radius R - Area of a quadrant of a circle of radius r
$$=\displaystyle\frac{1}{4}\pi R^2-\frac{1}{4}\pi r^2$$
$$=\displaystyle\frac{\pi}{4}(R^2-r^2)$$
$$=\displaystyle\frac{1}{4}\times \frac{22}{7}(21^2-14^2)cm^2$$     $$[\therefore R=12m$$ and $$r=14m]$$
$$=\displaystyle\left(\frac{1}{4}\times \frac{22}{7}\times(21+14)(21-14)\right)m^2$$
$$=\displaystyle\left(\frac{1}{4}\times \frac{22}{7}\times 35\times 7\right)m^2=192.5m^2$$

Let ABCD be a square with side $$58$$ metres
Let its diagonals intersect at O

Then $$OA=OB=OC=OD$$ and

$$\displaystyle \angle AOD=\angle BOC=90^{\circ}$$

With radius$$= \displaystyle \dfrac{1}{2}AC\text{ arcs have been drawn namely } AED \ and \ CBF$$

$$\displaystyle \therefore \text{Radius of each arc}=\dfrac{1}{2}\times diagonal$$

$$\displaystyle =\dfrac{1}{2}\sqrt{2}\times 58$$

$$=1.41 \times 29 =40.89 metres.$$

$$\displaystyle 58\times 58+2\times \left [ \dfrac{22}{7}\times \left ( 40.89 \right )^{2}\times \dfrac{90}{360}-\dfrac{1}{2}\times \left ( 40.89 \right )^{2}90^{\circ} \right ]$$

$$\displaystyle 33.64+\dfrac{11\times \left ( 40.89 \right )^{2}}{7}-\left ( 40.89 \right )^{2}$$

$$=(3364+2627.4-1671.9)$$ sq metres

$$=4319.5$$ sq metres

Area of equilateral $$\Delta ABC=49\sqrt { 3 } { cm }^{ 2 }$$

Area of remaining park $$=$$ area of park $$- 4$$$$\times $$area of $$1$$ quadrant 

Given that, diameter of the circle $$\displaystyle =20 \ cm$$
$$\displaystyle \therefore \ $$ Radius of the circle $$\displaystyle \left ( r \right )=10 \ cm$$

length of arc = $$\dfrac { \theta  }{ { 360 }^{ 0 } } \times 2\pi r=\dfrac { { 120 }^{ 0 } }{ { 360 }^{ 0 } } \times 2\times \dfrac { 22 }{ 7 } \times 21=44cm$$

The radius of the circle inscribed in the square
of side 2.4m
$$\displaystyle r=\dfrac{2.4}{4}m=1.2m$$
$$\displaystyle \therefore$$ Area of the circle $$\displaystyle =\pi r^{2}$$ square units
$$\displaystyle =\pi \times 1.2 m\times 1.2 m$$
$$\displaystyle =1.44 \pi m^{2}$$

Area of the shaded region $$=$$ Area of square $$-$$ Area of 1 Circle (4 quarter  = 1 circle)