Tag: construction of polygons

Quizzes (4) Questions (37)

Questions Related to construction of polygons

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

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

  1. Square

  2. Kite

  3. Trapezium

  4. rhombus
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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

  1. Yes

  2. No

  3. Sometimes yes

  4. Can't say
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $AB=4\ cm$.

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

We cannot construct a square if:

  1. a side is given

  2. a diagonal is given
  3. one angle is $90^0$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
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.