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
599.7k+ 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 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Trending doubts
Explain the Treaty of Vienna of 1815 class 10 social science CBSE

Which Indian city is known as the "City of Victory"?

Which instrument is used to measure the Blood Pressure?

Who Won 36 Oscar Awards? Record Holder Revealed

What is the median of the first 10 natural numbers class 10 maths CBSE

Why is it 530 pm in india when it is 1200 afternoon class 10 social science CBSE

