### Area of complex plane figures - class-X

 Description: area of complex plane figures Number of Questions: 14 Created by: Amit Pandey Tags: perimeter, area and volume perimeter and area area area and volume figures (two and three dimensional) measurements maths circumference and area of a circle
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Find the area of equilateral  triangle inscribed in a circle of unit radius.

1. 3/4

2. $\dfrac {3\sqrt { 3 } }{4}$

3. 3

4. $\frac { 3\sqrt { 3 } }{ 2 }$

Correct Option: B
Explanation:

The radius of circumcircle  of equilateral triangle is  $\dfrac 23h=1\h=\dfrac 32$

The side of equilateral triangle is given as $\dfrac{4h}{\sqrt 3} \\dfrac{4}{\sqrt 3}\times \dfrac 32=2\sqrt 3$
The area of triangle is given as $\dfrac {\sqrt 3}{4}(2\sqrt 3)^2=3\sqrt 3$

A square is inscribed in a circle of radius $7: cm$. Find area of the square.

1. $98 : cm^{2}$

2. $97 : cm^{2}$

3. $91 : cm^{2}$

4. $90 : cm^{2}$

Correct Option: A
Explanation:

Given,
Radius of the circle $=7:cm$
Let the side of the square be $a:cm$.
A square when inscribed in a circle then the diameter of the circle must be diagonal of the square.
Therefore,
Diagonal of square $=\sqrt {a^2+a^2}$
$=a\sqrt 2$
Now,
Diameter of the circle $=2\times 7$
$=14:cm$
$=>\sqrt 2 a=14$
$=>a=\dfrac{14}{\sqrt 2}$
$=>a=7\sqrt 2: cm$
Therefore,
Area of square $=a^2$
$=(7\sqrt 2 cm)^2$
$=(7\sqrt 2 cm)(7\sqrt 2 cm)$
$=98: cm^2$

The ratio of areas of square and circle is given n : 1 where n is a natural number. If the ratio of side of square and radius of circle is k :1, where k is a natural number, then n will be multiple of

1. $77$

2. $22$

3. $154$

4. Data insufficient

Correct Option: C
Explanation:

Let a be the side of the square & r be the radius of the circle, then $\dfrac {a^2}{\pi r^2}=n$
Now, $\dfrac {a}{r}=k$
$k=\dfrac {a}{r}=\sqrt {\dfrac {22\times n}{7}},n$ has to be multiple of $22\times 7=154$.

A rectangular sheet of acrylic is 50 cm by 25 cm . From it 60 circular buttons, each of diameter 2.8 cm have been cut out. The area of the remaining sheet is

1. 1260.82 $\displaystyle cm^{2}$

2. 880.4 $\displaystyle cm^{2}$

3. 630.4 $\displaystyle cm^{2}$

4. None of these

Correct Option: B
Explanation:

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

If one side of a square is 2.4 m. Then what will be the area of the circle inscribed in the square?

1. $1.44 \displaystyle\, m^{2}$

2. $\displaystyle 1\frac{11}{25}\pi$ $\displaystyle m^{2}$

3. $\displaystyle \frac{11}{25}\pi$ $\displaystyle m^{2}$

4. None of these

Correct Option: B
Explanation:

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
$\displaystyle =\pi \times 1.2 m\times 1.2 m$
$\displaystyle =1.44 \pi m^{2}$

$\displaystyle =1\dfrac{11}{25}\pi m^{2}$
$\displaystyle \therefore$ The required area $\displaystyle =1\frac{11}{25}\pi m^{2}$

Size of a tile is $9$ inches by $9$ inches. The number of tiles needed to cover a floor of $12$ feet by $18$ feet is

1. $384$

2. $32$

3. $24$

4. $216$

Correct Option: A
Explanation:

Number of tiles$\displaystyle =\dfrac{Area\, of\, floor}{Area\, of\, 1 tile}$

$\displaystyle =\dfrac{12\times 12\times 18\times 12}{9\times 9}=384$

A pentagon is made up of an equilateral $\triangle ABC$ of side length $2cm$ on top of a square $BCDE$. Circumscribe a circle through points, $A, D$ and $E$. The radius of the circle is

1. $1+\dfrac {\sqrt {3}}{2}$

2. $5-2\sqrt {3}$

3. $2$

4. $1+\sqrt {3}$

Correct Option: B

A copper wire when bent in the form of an equilateral triangle has an area of $121\, \sqrt3\, cm^2$. If the same wire is bent into the form of a circle, then the area enclosed by the wire is

1. 110.25 $cm^2$

2. 346.5 $cm^2$

3. 121.5 $cm^2$

4. 336.5 $cm^2$

Correct Option: B
Explanation:

$Given-\ A\quad wire\quad is\quad bent\quad into\quad an\quad equilateral\quad triangle.\ Its\quad area=121\sqrt { 3 } { cm }^{ 2 }.\ The\quad same\quad wire\quad is\quad bent\quad into\quad a\quad circle.\ To\quad find\quad out-\ ar.circle=?\ Solution-\ The\quad same\quad wire\quad is\quad bent\quad into\quad an\quad equilateral\quad triangle\quad \ and\quad a\quad circle.\ \therefore \quad Perimeter\quad P\quad of\quad the\quad triangle\ =circumference\quad C\quad of\quad the\quad circle.\ Let\quad the\quad side\quad of\quad the\quad equilataral\quad triangle\quad be\quad x.\ i.e\quad P=3x=C.........(i)\ Then\quad ar.triangle=\frac { \sqrt { 3 } }{ 4 } { x }^{ 2 }{ cm }^{ 2 }.=121\sqrt { 3 } { cm }^{ 2 }\quad (given)\ \Longrightarrow x=22cm.\ So\quad P=3x=3\times 22cm=66cm=C\quad (by\quad i)........(ii)\ Again\quad let\quad us\quad assume\quad that\quad the\quad radius\quad of\quad the\quad circle=r.\ Then\quad the\quad circumference=C=2\pi r\Longrightarrow r=\frac { C }{ 2\pi } =\frac { 66 }{ 2\pi } (from\quad ii)\ \Longrightarrow r=\frac { 66 }{ 2\times \frac { 22 }{ 7 } } cm=10.5cm.\ \therefore \quad ar.circle=\pi { r }^{ 2 }=\frac { 22 }{ 7 } \times { \left( 10.5 \right) }^{ 2 }{ cm }^{ 2 }=346.5{ cm }^{ 2 }.\ Ans-\quad Option\quad B.\$

The sides of a triangle are $5$, $12$ and$13$ units. A rectangle of width $10$ units is constructed equal in area to the area of the triangle. Then the perimeter of the rectangle is

1. 30 units

2. 26 units

3. 13 units

4. 15 units

Correct Option: B
Explanation:

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.

Which of the following shapes of equal perimeter the one having the largest areas is

1. circle

2. equilateral triangle

3. square

4. regular pentagon

Correct Option: A
Explanation:

$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\$

A square sheet of paper is converted into a cylinder by rolling it along its length. What is the ratio of the base radius to side of the square ?

1. $\displaystyle \frac{1}{2\pi}$

2. $\displaystyle \frac{\sqrt{2}}{\pi}$

3. $\displaystyle \frac{1}{\sqrt{2\pi }}$

4. $\displaystyle \frac{1}{\pi}$

Correct Option: A
Explanation:

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

Base radius of cylinder $=r$
Surface area of sheet $={ a }^{ 2 }$
Surface area of cylinder $=2\pi rh$
But height $h=a$, because the square sheet is rolled it along its length.
Thus, surface area of cylinder $=2\pi ra$
Therefore, ${ a }^{ 2 }=2\pi ra\Rightarrow a=2\pi r\Rightarrow \dfrac { r }{ a } =\dfrac { 1 }{ 2\pi }$

A polygon has 44 diagonals, The number of its sides is

1. 11

2. 10

3. 8

4. 7

Correct Option: A
Explanation:
Number of diagonals in a polygon $=\cfrac{n(n-3)}2$
$\implies 44=\cfrac{n(n-3)}2$
$\implies n^2-3n-88=0$
$\implies (n-11)(n-8)=0$
$\implies n=11$ or $n=-8$
Therefore, number of sides in a polygon $=11.$
Hence, A is the correct option.

Four circular cardboard pieces of radii $7 cm$ are placed on a paper in such a way that each piece touches other two pieces. The area of the  region enclosed between these pieces   is

1. $42$ $cm^2$

2. $21$  $cm^2$

3. $84$ $cm^2$

4. $96$ $cm^2$

Correct Option: A
Explanation:

The diameter of circle =$2\times 7=14$ cm

The circles together formed a shape  square diameter of 2 circle to get from a side =$2\times 14=28$ cm
Then area of square =$a^{2}=(28)^{2}=784 cm^{2}$
And area of each  circle =$\pi r^{2}=\frac{22}{7}\times (7)^{2}=154 cm^{2}$
So area of four circles =$4\times 154=616 cm^{2}$
Then area of region enclosed between these pieces $= 784-616=168$ sq cm
Then area of region enclosed between one  pieces=$\frac{168}{4}=42 cm^{2}$

The ratio of the slant height of two right cones of equal base is 3 : 2 then the ratio of their volumes is

1. $1:4$

2. $9:4$

3. $3:2$

4. None

Correct Option: B
Explanation:

Let the radius of both the cones be r, slant height of first cone $= 3a$; slant height of second cone $= 2a$

For a cone, l $= \sqrt { { h }^{ 2 }+ {r}^{ 2 } }$ where h is the height

Hence, for cone 1,  $3a = \sqrt { { h 1 }^{ 2 }+ {r}^{ 2 } }$

$9{a}^{2} = { h 1 }^{ 2 } + {r}^{ 2 }$

${ h 1 }^{ 2 } = 9{a}^{2} - {r}^{ 2 }$

$h 1 = \sqrt {9{a}^{2} - {r}^{ 2 }}$

For cone 1,  $2a = \sqrt { { h 2 }^{ 2 }+ {r}^{ 2 } }$

$4{a}^{2} = { h 2 }^{ 2 } + {r}^{ 2 }$

${ h 2 }^{ 2 } =4{a}^{2} - {r}^{ 2 }$

$h 2 = \sqrt {4{a}^{2} - {r}^{ 2 }}$
Now, ratio of their volumes is calculated as:
$V _{ 1 } : V _{ 2 }$
$\frac { 1 }{ 3 }\pi {r }^{ 2 }h 1 = \frac { 1 }{ 3 } \pi { r }^{2 }h 2$

$h 1 : h 2$
$\sqrt {9{a}^{2} - {r}^{ 2 }} : \sqrt {4{a}^{2} - {r}^{ 2 }}$
$9{a}^{2} - {r}^{ 2 }: 4{a}^{2} - {r}^{ 2 }$
$9{a}^{2} : 4{a}^{2}$
$9:4$

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