
Construct a tangent to a circle of radius $4cm$ from a point on the concentric circle of radius $6cm$ and measure its length and also verify the obtained measurements by actual calculation.
Answer
485.1k+ views
Hint: First, we have to construct two concentric circles as given in the question of the give radii. Taking the theorem regarding the circles: Radius is perpendicular to the tangent at a given point on a circle, draw the perpendicular bisector of the radius because the perpendicular bisector gives us an angle of ${90^ \circ }$.
Complete step-by-step solution:
A point $O$ is taken as the centre of the circle. Mark the point on the paper, label it as $O$ as the centre.
Taking $O$ as the centre, draw two concentric circles with radii $4cm$ and $6cm$.
Now, mark a point on the outer circle and join $OP$. Here, $OP = 6cm$.
A perpendicular bisector $AB$ for $OP$ is drawn, cutting $OP$ at $Q$.
With $Q$ as the centre and $OQ$ as the radius, a circle is drawn to intersect the inner circle at points ${T_1}$ and ${T_2}$.
$P{T_1}$ and $P{T_2}$ are joined – which are the required tangents which are equal.
The approximate value of $OP$ when measured with a ruler is around $4.5cm$.
Verification by actual Calculation:
In the above figure:
$O{T_1}$ is joined to form a right-angled triangle, $\Delta O{T_1}P$, since radius is perpendicular to the tangent at the point of contact on the circle.
In $\Delta O{T_1}P$,
By Pythagoras theorem:
$O{P^2} = OT_1^2 + P{T^2}$
$ \Rightarrow PT = \sqrt {O{P^2} - PT_1^2} $
Substituting the values,
$ \Rightarrow PT = \sqrt {{{(6)}^2} + {{(4)}^2}} $
Squaring we get,
$ \Rightarrow PT = \sqrt {36 - 16} $
Simplifying to solve,
$ = \sqrt {20} = \sqrt {5 \times 4} = \sqrt 5 \times \sqrt 4 $
Hence, the value of $\sqrt 5 = 2.236$,
$ = 2 \times 2.236$
Hence,
$ = 4.47 \approx 4.5cm$
$\therefore $ The measure of the tangent is $4.5cm$
Note: We have to remember that a tangent line to a circle is a line that touches the circle at exactly one point, never entering the circle’s interior. The radius of a circle is perpendicular to the tangent line through its endpoint on the circle’s circumference. Conversely, the perpendicular to a radius through the same endpoint is a tangent line. Take care of the approximate measurements while measuring the radius using the compass. If there is an error, the final value might be distracted. You have to take careful measures while approximating the final value, the rounding off factors should be taken care of.
Complete step-by-step solution:
A point $O$ is taken as the centre of the circle. Mark the point on the paper, label it as $O$ as the centre.

Taking $O$ as the centre, draw two concentric circles with radii $4cm$ and $6cm$.
Now, mark a point on the outer circle and join $OP$. Here, $OP = 6cm$.
A perpendicular bisector $AB$ for $OP$ is drawn, cutting $OP$ at $Q$.

With $Q$ as the centre and $OQ$ as the radius, a circle is drawn to intersect the inner circle at points ${T_1}$ and ${T_2}$.
$P{T_1}$ and $P{T_2}$ are joined – which are the required tangents which are equal.
The approximate value of $OP$ when measured with a ruler is around $4.5cm$.
Verification by actual Calculation:
In the above figure:
$O{T_1}$ is joined to form a right-angled triangle, $\Delta O{T_1}P$, since radius is perpendicular to the tangent at the point of contact on the circle.
In $\Delta O{T_1}P$,
By Pythagoras theorem:
$O{P^2} = OT_1^2 + P{T^2}$
$ \Rightarrow PT = \sqrt {O{P^2} - PT_1^2} $
Substituting the values,
$ \Rightarrow PT = \sqrt {{{(6)}^2} + {{(4)}^2}} $
Squaring we get,
$ \Rightarrow PT = \sqrt {36 - 16} $
Simplifying to solve,
$ = \sqrt {20} = \sqrt {5 \times 4} = \sqrt 5 \times \sqrt 4 $
Hence, the value of $\sqrt 5 = 2.236$,
$ = 2 \times 2.236$
Hence,
$ = 4.47 \approx 4.5cm$
$\therefore $ The measure of the tangent is $4.5cm$
Note: We have to remember that a tangent line to a circle is a line that touches the circle at exactly one point, never entering the circle’s interior. The radius of a circle is perpendicular to the tangent line through its endpoint on the circle’s circumference. Conversely, the perpendicular to a radius through the same endpoint is a tangent line. Take care of the approximate measurements while measuring the radius using the compass. If there is an error, the final value might be distracted. You have to take careful measures while approximating the final value, the rounding off factors should be taken care of.
Recently Updated Pages
Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Earth rotates from West to east ATrue BFalse class 6 social science CBSE

The easternmost longitude of India is A 97circ 25E class 6 social science CBSE

Write the given sentence in the passive voice Ann cant class 6 CBSE

Convert 1 foot into meters A030 meter B03048 meter-class-6-maths-CBSE

What is the LCM of 30 and 40 class 6 maths CBSE

Trending doubts
Which one is a true fish A Jellyfish B Starfish C Dogfish class 10 biology CBSE

Dr BR Ambedkars fathers name was Ramaji Sakpal and class 10 social science CBSE

A boat goes 24 km upstream and 28 km downstream in class 10 maths CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

What is the full form of POSCO class 10 social science CBSE
