Tag: circumscribing and inscribing a circle on a regular hexagon

Questions Related to circumscribing and inscribing a circle on a regular hexagon

What are the tools required for constructing a tangent to a circle?

maths geometric constructions 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

  1. ruler

  2. compass

  3. pencil

  4. all the above

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Correct Option: D
Explanation:

The tools required for constructing a tangent to a circle is ruler, compass and pencil.

Let C be the circle with centre at $(1, 1)$ and radius $=1$. If T is the circle centred at $(0, y)$, passing through origin and touching the circle C externally, then the radius of T is equal to?

maths geometric constructions 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

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

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

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

  4. $\dfrac{1}{4}$

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Correct Option: A

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

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

  1. $\cfrac { 133 }{ 3 } $

  2. $\cfrac { 125 }{ 3 } $

  3. $42$

  4. $41$

  5. $40$

Show answer
Correct Option: B
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.

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

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

  1. $88^\circ$

  2. $78^\circ$

  3. $68^\circ$

  4. $58^\circ$

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Correct Option: C
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 ${{68}^{\circ }}$.

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

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

  1. $6\ cm$

  2. $7\ cm$

  3. $8\ cm$

  4. $9\ cm$

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Correct Option: B
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 $\triangle OAP$

$\angle OPA = 90$

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

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

$AP = 0.6 \times 6 = 3.6$

$AB = 2 \times 3.6 = 7cm$

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

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

  1. $1$

  2. $2$

  3. $3$

  4. $4$

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Correct Option: A
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.

The number of independent measurement required to construct a $\Delta$ le is 

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

  1. $3$

  2. $4$

  3. $2$

  4. $5$

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Correct Option: A
Explanation:

Triangle has $3$ sides.
So, number of measurements required to construct a triangle is $3$.

The minimum number of dimensions needed to construct a rectangle is:

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

  1. $1$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: B
Explanation:

We can construct a rectangle when:

(i) two adjacent sides are given
(ii) one side and the diagonal is given
(iii) both diagonals are given
In the above cases the number of dimensions needed to construct a rectangle is $2$.

State True or False
There is a triangle whose sides have lengths 10.2 cm, 5.8 cm and 4.5 cm 

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

  1. True

  2. False

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Correct Option: A
Explanation:

Suppose such a triangle is possible Then the sum of the lengths of any two side would be greater than the length of the third side  Let us check this
Is 4.5+5.8>10.2  Yes 
Is 5.8+10.2>4.5  Yes
Is 10.2+4.5>5.8  Yes
Therefore the triangle is possible

The number of independent measurements required to construct a $\Delta$ is

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

  1. 3

  2. 4

  3. 2

  4. 5

Show answer
Correct Option: A
Explanation:

We have three measurements to construct a $\Delta$ le,