Tag: construction

Quizzes (3) Questions (26)

Questions Related to construction

Multiple choice maths construction circumscribing and inscribing a circle on a regular hexagon constructions related to a circle construction of polygons construction of tangent to a circle construction of tangents construction of line segment and circle of given radius construction related to lines

The centre of the circle circumscribing the square whose three sides are $3x+y=22,x-3y=14$ and $3x=y=62$ is:

  1. $\left( \dfrac { 3 }{ 2 } ,\dfrac { 27 }{ 2 } \right) $
  2. $\left( \dfrac { 27 }{ 2 } ,\dfrac { 3 }{ 2 } \right) $
  3. $(27,3)$
  4. $\left( 1,\dfrac { 2 }{ 3 } \right) $
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice maths construction circumscribing and inscribing a circle on a regular hexagon constructions related to a circle construction of polygons construction of tangent to a circle construction of tangents construction of line segment and circle of given radius construction related to lines

A square is inscribed in the circle $x^2 + y^2 -2x +4y - 93 = 0$ with its sides parallel to the coordinates axes. The coordinates of its vertices are 

  1. $( - 6, - 9), \, ( - 6, 5), \, (8, - 9)$ and $(8, 5)$
  2. $( - 6, 9), \, ( - 6, - 5), \, (8, - 9)$ and $(8, 5)$
  3. $( - 6, - 9), \, ( - 6, 5), \, (8, 9)$ and $(8, 5)$
  4. $( - 6, - 9), \, ( - 6, 5), \, (8, - 9)$ and $(8, - 5)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The circle equation is (x-1)^2 + (y+2)^2 = 93 + 1 + 4 = 98. The radius is sqrt(98) = 7*sqrt(2). For a square with sides parallel to axes, the distance from center (1, -2) to vertices is the radius.

Multiple choice maths construction circumscribing and inscribing a circle on a regular hexagon constructions related to a circle construction of polygons construction of tangent to a circle construction of tangents construction of line segment and circle of given radius construction related to lines

For each of the following, drawn a circle and inscribe the figure given.If a polygon of the given type can't be inscribed,write not possible.

  1. Rectangle.

  2. Trapezium.

  3. Obtuse triangle.

  4. non-rectangle parallelogram

  5. Accute isosceles triangle.
  6. A quadrilateral PQRS with $\overline {PR} $ as diameter.
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice maths construction circumscribing and inscribing a circle on a regular hexagon constructions related to a circle construction of polygons construction of tangent to a circle construction of tangents construction of line segment and circle of given radius construction related to lines

In regular hexagon, if the radius of circle through vertices is r, then length of the side will be

  1. $\displaystyle \frac{2\pi r}{6}$

  2. r
  3. $\displaystyle \frac{\pi r}{6}$
  4. $\displaystyle \frac{r}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\Rightarrow$   Radius of a circle is $r$.

$\Rightarrow$   In regular hexagon all sides are equal.
$\Rightarrow$   The regular hexagon has 6 equilateral triangles. The diameter of the circle is $2r$ in this case, will coincide with 2 equilateral triangles. So the side of the hexagon will be $r$.
$\therefore$   Length of side of hexagon is $r$.

Multiple choice maths construction circumscribing and inscribing a circle on a regular hexagon constructions related to a circle construction of polygons construction of tangent to a circle construction of tangents construction of line segment and circle of given radius construction related to lines

When constructing the circles circumscribing and inscribing a regular hexagon with radius $3$ m, then inscribing hexagon length of each side is

  1. $1m$
  2. $2m$
  3. $3m$
  4. $4m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When constructing the circles circumscribing and inscribing a regular hexagon with radius $3$ m, then inscribing hexagon length of each side is $3$ m.

Multiple choice maths construction circumscribing and inscribing a circle on a regular hexagon constructions related to a circle construction of polygons construction of tangent to a circle construction of tangents construction of line segment and circle of given radius construction related to lines

The area of a circle inscribed in a regular hexagon is $100\pi$. The area of the hexagon is:

  1. $600$
  2. $300$
  3. $200\sqrt { 2 } $
  4. $200\sqrt { 3 } $
  5. $200\sqrt { 5 } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Area of circle $=100\pi $
$\pi r^{2}=100\pi $
$r^{2}=100$
$r=10$
Now, a regular hexagon is made up of 6 equilateral $\bigtriangleup s $ of equal areas. Now, height of equilateral $\bigtriangleup  $ is equal to radius of circle.Therefore, ar. of 1 equilateral $\bigtriangleup=\dfrac {1}{2} $ x base x height
$\Rightarrow \dfrac {\sqrt{3}}{4}a^{2}=\dfrac {1}{2}a*10\Rightarrow a=\dfrac {4*10}{2\sqrt{3}}=\dfrac {20\sqrt{3}}{3} $
Area of hexagon $6\left ( \dfrac {\sqrt{3}}{4}a^{2} \right )=6*\dfrac {\sqrt{3}}{4}\dfrac {20\sqrt{3}}{3}\dfrac {20\sqrt{3}}{3}=200\sqrt{3}$

Multiple choice maths construction circumscribing and inscribing a circle on a regular hexagon construction of tangent to a circle construction of tangents construction of line segment and circle of given radius construction related to lines

The sides of a triangle are $25,39$ and $40$. The diameter of the circumscribed circle is: 

  1. $\cfrac { 133 }{ 3 } $
  2. $\cfrac { 125 }{ 3 } $
  3. $42$
  4. $41$
  5. $40$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Circum radius formula

$R$ $=\cfrac { abc }{ \sqrt { (a+b+c)(b+c-a)(c+a-b)(a+b-c) }  }$ .
Where  $a, b, c$  are sides of triangle 
$\Rightarrow$ $R$ $=\cfrac { 25\times 39\times 40\quad  }{ \sqrt { (140\quad \times (54)\times (26)\quad \times (240) }  } $
$=\cfrac { 25\times 39\times 40\quad  }{ \sqrt { { 2 }^{ 3 } } \times 13\times 2\times { 3 }^{ 3 }\times 2\times 13\times { 2 }^{ 3 }\times 3 } $
$=\cfrac { 25\times 39\times 40\quad  }{ \sqrt { { 2 }^{ 8 } } \times { 3 }^{ 4 }\times { 13 }^{ 2 } } $.
$=\cfrac { 25 \times \ 39 \times 40  }{ { 2 }^{ 4 }\times { 3 }^{ 2 }\times { 13 } } =\quad \cfrac { 25 \times 39 \times40\quad  }{ 16\times 9\times { 13 } }$ 
$=\cfrac { 125 }{ 6 }$ 
$\therefore$   Diameter $=\cfrac { 125\times \ 2 }{ 6 } = \cfrac { 125 }{ 3 } $

$\therefore$ B) Answer.

Multiple choice maths construction circumscribing and inscribing a circle on a regular hexagon construction of tangent to a circle construction of tangents

The angles of a pentagon in degrees are $y^\circ$, $(y+20^\circ)$, $(y+40^\circ)-(y+60^\circ)$ and $(y+80^\circ)$. The smallest angle of the pentagon is

  1. $88^\circ$
  2. $78^\circ$
  3. $68^\circ$
  4. $58^\circ$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Consider the given angles.

${{y}^{\circ }},\left( {{y}^{\circ }}+{{20}^{\circ }} \right),\left( {{y}^{\circ }}+{{40}^{\circ }} \right),\left( {{y}^{\circ }}+{{60}^{\circ }} \right),\left( {{y}^{\circ }}+{{80}^{\circ }} \right)$

 

We know that the sum of all angles of pentagon

$ {{y}^{\circ }}+\left( {{y}^{\circ }}+{{20}^{\circ }} \right)+\left( {{y}^{\circ }}+{{40}^{\circ }} \right)+\left( {{y}^{\circ }}+{{60}^{\circ }} \right)+\left( {{y}^{\circ }}+{{80}^{\circ }} \right)={{540}^{\circ }} $

$ 5{{y}^{\circ }}+{{200}^{\circ }}={{540}^{\circ }} $

$ 5{{y}^{\circ }}={{340}^{\circ }} $

 

Hence, the smallest angle of the pentagon is ${{68}^{\circ }}$.

Multiple choice maths construction circumscribing and inscribing a circle on a regular hexagon construction of tangent to a circle construction of tangents

Construct a regular pentagon inside a circle of radius $6\ cm$. The length of each side of the pentagon is: (approx.)

  1. $6\ cm$
  2. $7\ cm$
  3. $8\ cm$
  4. $9\ cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Each side of the pentagon makes an angle x at the center

$\implies 5x= 360 $

$x = 72$

Now lets consider side AB which is a chord to the circle

Let OP be a perpendicular to AB

$\implies AP = BP \implies AB = 2AP$

IN $\triangle OAP$

$\angle OPA = 90$

$\angle POA = \dfrac{x}{2} = \dfrac{72}{2} = 36$

$\sin 36 = \dfrac{AP}{OA}$

$AP = 0.6 \times 6 = 3.6$

$AB = 2 \times 3.6 = 7cm$

Multiple choice maths construction circumscribing and inscribing a circle on a regular hexagon construction of tangent to a circle construction of tangents construction of line segment and circle of given radius construction related to lines

The minimum number of dimensions needed to construct an equilateral triangle is:

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As we know that all angles in an equilateral triangle measures $60^o$. Hence we need only the length of the side to construct an equilateral triangle.