Tag: 2d and 3d figures

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

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 $=a^2$
                       $=(7\sqrt 2 cm)^2$
                       $=(7\sqrt 2 cm)(7\sqrt 2 cm)$
                       $=98: cm^2$

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 $22\times 7=154$.

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

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

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

$ 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.\  $

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\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\  $