Tag: construction of polygons

Questions Related to construction of polygons

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

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

  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) $

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Correct Option: B

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 

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

  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 $(8, - 5)$

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Correct Option: A

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.

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

  1. Rectangle.

  2. Trapezium.

  3. Obtuse triangle.

  4. non-rectangle parallelogram

  5. Accute isosceles triangle.

  6. A quadrilateral PQRS with $\overline {PR} $ as diameter.

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Correct Option: A

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

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

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

  2. r

  3. $\displaystyle \frac{\pi r}{6}$

  4. $\displaystyle \frac{r}{2}$

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Correct Option: B
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 $r$.

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

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

  1. $1m$

  2. $2m$

  3. $3m$

  4. $4m$

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Correct Option: C
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.

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

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

  1. $600$

  2. $300$

  3. $200\sqrt { 2 } $

  4. $200\sqrt { 3 } $

  5. $200\sqrt { 5 } $

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Correct Option: D
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 $6\left ( \dfrac {\sqrt{3}}{4}a^{2} \right )=6*\dfrac {\sqrt{3}}{4}\dfrac {20\sqrt{3}}{3}\dfrac {20\sqrt{3}}{3}=200\sqrt{3}$

Two line segments, each $9\ cm$ long, bisect each other at right angles. Their end points are joined together. The shape formed is a:

maths construction of polygons construction of parallelograms and rectangles construction of special quadrilaterals constructions related to a quadrilateral

  1. Square

  2. Kite

  3. Trapezium

  4. rhombus

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Correct Option: A
Explanation:

Image result for Two line segments, each 9 cm long, bisect each other at right angles. Their end points are joined together. The shape formed is a:

Let $PS$ and $QR$ are the two line segments, each of $9$cm, and bisect each other at right angles.
By joining the end points of these line, we get a shape given in the figure.
In $\triangle POQ$, $\angle POQ=90^{o}$, $OP=OQ=4.5$
By using Pythagoras theorem,
$PQ^{2}=OP^{2}+OQ^{2}$
         $=(4.5)^2+(4.5)^2=40.5$
$\therefore\ PQ=6.36$
Similarly, $QS=6.36=RS=PS$
Thus, length of all sides is same and all angles are right angle.
Hence, the shape formed is square.

A square with side given can be constructed by using the property of its diagonals.

maths construction of polygons construction of parallelograms and rectangles construction of special quadrilaterals constructions related to a quadrilateral

  1. True

  2. False

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Correct Option: A
Explanation:

This statement is true 

We can use property that diagonals are at 45 degree with side and diagonals bisect each other at 90 degree.for construction of square.

Can we construct a rhombus $ABCD$ with $AB=4\ cm$? Its diagonal intersect at the point $O$ and $\angle OAB = 60^0$.

maths construction of polygons construction of parallelograms and rectangles construction of special quadrilaterals constructions related to a quadrilateral

  1. Yes

  2. No

  3. Sometimes yes

  4. Can't say

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Correct Option: A
Explanation:

Given : $AB=4$cm

Diagonal intersect at $O$ and $\angle OAB=60^{o}$ ....... $(1)$
Draw side $AB$ of $4$cm.
In a rhombus, all sides are equal and diagonals bisect the opposite angles
From $(1)$ we get, $\angle A=120^{o}$
$\implies \angle B=60^{o}$ ........... (Adjacent angles are supplementary)
Draw a side $AD$ from A of $4$cm such that $\angle BAD=120^{o}$
Now, from $D$, draw side $DC = 4$cm such that $\angle ADC=60^{o}$
And then join $B-C$ such that $BC=4$cm and $\angle DCB=120^{o}$.
At last we get a rhombus $ABCD$ with length of each side is $4$ cm and diagonals $AC$ and $BD$.
Hence, we can construct a rhombus with $AB=4\ cm$.

We cannot construct a square if:

maths construction of polygons construction of parallelograms and rectangles construction of special quadrilaterals constructions related to a quadrilateral

  1. a side is given

  2. a diagonal is given

  3. one angle is $90^0$

  4. None of these

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Correct Option: C
Explanation:

If a side is given then we can draw a square with the same side as given.

If diagonals are given, by joining the endpoints we can draw the square.
In square, all angles are of $90^{o}$.
If one angle is $90^{o}$ is given, we can't directly conclude that all the angles are $90^{o}$.

Hence, if one angle is $90^{o}$ then we cannot construct a square.