Question

Draw each of the shapes given below and fill in the blanks.

Shape	Centre of rotation(intersection of diagonals\ Intersection of axes of symmetry)	Angle of rotation	Order of rotation
Square
Rectangle
Rhombus
Equilateral triangle
Regular Hexagon
Circle
Semicircle

Answer 1

Hint: In order to solve this problem, we need to know the actual meaning of centre of rotation, angle of rotation and order of rotation. The point of intersection where all the axis of symmetry meets is called the centre of rotation. The angle of rotation. It is the minimum angle upon which the shape maps into itself during rotation ${{360}^{\circ }}$ . Order of rotation is the number of times the shape maps onto itself during rotation of ${{360}^{\circ }}$ .

Complete step-by-step answer:
Now we need to find the three terms in each of the following cases
Let’s first understand the meaning of all the three terms
Let's start with the center of rotation.
The point of intersection where all the axis of symmetry meets is called the centre of rotation.
Basically, it is the point from where the whole figure looks symmetric.
Coming on to next, the angle of rotation. It is the minimum angle upon which the shape maps into itself during rotation of ${{360}^{\circ }}$ .
We will find out more when we solve the examples.
Order of rotation is the number of times the shape maps onto itself during rotation of ${{360}^{\circ }}$ .
Now, let's start with a square shape.
 As we can say that the square is symmetric under from all sides and the centre of rotation is the point of intersection of its diagonals.
We can show that from the figure below,

Point A is the intersection of the diagonals and it’s also the centre of rotation.
The square maps itself in the interval of ${{90}^{\circ }}$ . Therefore, the angle of rotation of the square is ${{90}^{\circ }}$ .
During one complete revolution ${{360}^{\circ }}$ , it maps four times onto itself. Therefore, the order of symmetry is 4 ( ${{90}^{\circ }},{{180}^{\circ }},{{270}^{\circ }},{{360}^{\circ }}$ ).
Let's go to the second shape that is a rectangle.
As we can say that the square is symmetric under from all sides and the centre of rotation is the point of intersection of its diagonals.
We can show that from the figure below,

Point B is the intersection of the diagonals and it’s also the centre of rotation.
The rectangle maps itself in the interval of ${{180}^{\circ }}$ . Therefore, the angle of rotation of the rectangle is ${{180}^{\circ }}$ .
During one complete revolution ${{360}^{\circ }}$ , it maps two times onto itself. Therefore, the order of symmetry is 2 ( ${{180}^{\circ }},{{360}^{\circ }}$ ).
Now, moving on the third figure that is a rhombus,
The centre of rotation is the point of intersection of its diagonals.
We can show that from the figure below,

Point C is the intersection of the diagonals and it’s also the centre of rotation.
The rhombus maps itself in the interval of ${{180}^{\circ }}$ . Therefore, the angle of rotation of rhombus is ${{180}^{\circ }}$ .
During one complete revolution ${{360}^{\circ }}$ , it maps two times onto itself. Therefore, the order of symmetry is 2 ( ${{180}^{\circ }},{{360}^{\circ }}$ ).
The next figure is the equilateral triangle,
The centre of rotation is the point of intersection of its medians.
The median is the line joining the point of intersection of the vertex and the midpoint of the line opposite to it.
We can show that from the figure below,

Point D is the intersection of the medians and it’s also the centre of rotation.
The equilateral triangle maps itself in the interval of ${{120}^{\circ }}$ . Therefore, the angle of rotation of the equilateral triangle is ${{120}^{\circ }}$ .
During one complete revolution ${{360}^{\circ }}$ , it maps three times onto itself. Therefore, the order of symmetry is 3 ( ${{120}^{\circ }},{{240}^{\circ }},{{360}^{\circ }}$ ).
The next diagram is of a regular hexagon.
The centre of rotation is the point of intersection of diagonals.
The diagonals in the regular hexagon is the intersection of opposite vertices of a regular hexagon.
We can show that from the figure below,

Point E is the intersection of the diagonals and its also the centre of rotation.
The regular hexagon maps itself in the interval of ${{60}^{\circ }}$ . Therefore, the angle of rotation of a regular hexagon is ${{60}^{\circ }}$ .
During one complete revolution ${{360}^{\circ }}$ , it maps six times onto itself. Therefore, the order of symmetry is 6 ( ${{60}^{\circ }},{{120}^{\circ }},{{180}^{\circ }},{{240}^{\circ }},{{300}^{\circ }},{{360}^{\circ }}$ ).
The next diagram is of a circle.
The centre of rotation is the centre of the circle itself.
The circle is symmetric about the centre of the circle
We can show that from the figure below,

Point F is the centre of the circle and it's also the centre of rotation.
The circle maps itself in all the intervals. Therefore, the angle of rotation of a circle is any angle.
During one complete revolution ${{360}^{\circ }}$ , it maps infinite times onto itself. Therefore, the order of symmetry is infinite.
The next diagram is of a semicircle.
The centre of rotation is the centre of the semi-circle itself.
The semicircle is symmetric about the centre of the semicircle
We can show that from the figure below,

Point G is the centre of the semicircle and it’s also the centre of rotation.
The semicircle maps only once during one complete rotation. Therefore, the angle of rotation of a semicircle is ${{360}^{\circ }}$ .
Therefore, the order of symmetry is 1 ( ${{360}^{\circ }}$).
All the values from the table are known, therefore substituting all the values we get,

Shape	Centre of rotation(intersection of diagonals\ Intersection of axes of symmetry)	Angle of rotation	Order of rotation
Square	Intersection point of diagonals	${{90}^{\circ }}$	4
Rectangle	Intersection point of diagonals	${{180}^{\circ }}$	2
Rhombus	Intersection point of diagonals	${{180}^{\circ }}$	2
Equilateral triangle	Intersection point of medians	${{120}^{\circ }}$	3
Regular Hexagon	Intersection point of diagonals	${{60}^{\circ }}$	6
Circle	Centre of circle	Any angle	Infinite
Semicircle	Centre of semicircle	${{360}^{\circ }}$	1

Note:In a rectangle, the angle of rotation changed from ${{90}^{\circ }}$ to ${{180}^{\circ }}$ with respect to square because only two sides of the rectangle are equal. Similarly, in rhombus, the angle of rotation changed from ${{90}^{\circ }}$ to ${{180}^{\circ }}$ with respect to square because rhombus is tilted to the side which makes it loses the symmetry. Also, we can derive the formula for the order of rotation. It can be shown as follows, $\text{Order of rotation}=\dfrac{\text{Angle of rotation}}{{{360}^{\circ }}}$ .