Question

Draw a circle of radius $3.2{\rm{ cm}}$. Take a point P on this circle and draw a tangent at P. (using the centre).

Answer 1

Hint: This question is based on Geometry construction and we have to construct a circle of given radius and then draw a tangent on the circle using the centre of the circle. Now in order to draw the tangent on the circle we draw another circle with the same radius by taking a point on the circumference of the first circle and then by using the theorem that “the tangent on a circle makes an angle of $90^\circ $ between the tangent and the line connecting the point at the circumference to the centre” we draw the tangent on the first circle.

Complete step-by-step answer:
The radius of the circle $OP = 3.2{\rm{ cm}}$

The step by step construction is given below:
1) First take a radius of $3.2{\rm{ cm}}$ using the ruler scale and draw a circle taking O as the centre of the circle.
2) Now take a point P on the circumference of the circle.
3) Now, draw another circle having a radius of $3.2{\rm{ cm}}$ taking point P as the centre.
4) This new drawn circle intersects another circle at two points.
4) Now join point P to any of the two point where it cuts the circle and name that point Q
5) Join PQ with a line segment, where $PQ = 3.2{\rm{ cm}}$ is the same as the radius of the circle.
6) Now draw an angle $\angle QPX = 30^\circ $ using the protractor and then draw a perpendicular bisector of line segment PQ.
7) The point where the angle and the perpendicular bisector intersect each other name that point R.
Therefore, PR is the required tangent on the circle having a radius of $3.2{\rm{ cm}}$.

Note: It should be noted that the $\Delta OPQ$ is an Equilateral triangle so the value of each angle of this triangle is $60^\circ $.
Which means that-
$
\angle OPQ = \angle PQO\\
 = \angle QOP\\
 = 60^\circ
$