# The radii of two concentric circles with centre O are 5 cm and 13 cm respectively. The line drawn from the point A of the outer circle touches the inner circle at point M and line AE intersects the inner circle at the points C and D. If AE=25 cm, then find AD(A-C-D-E).

Last updated date: 27th Mar 2023

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Answer

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Hint: We will use the concept of perpendicular bisector and Pythagoras theorem to find the value of AD in this question.

Complete step-by-step answer:

We have been given two concentric circles with centre O and radius 13 cm and 5 cm. It is also given in the question that the line drawn from the point A of the outer circle touches the inner circle at point M and line AE intersects the inner circle at the points C and D.

Let us take OE as the radius of the bigger circle with length 13 cm and OD as the radius of the smaller circle with length 5 cm.

Now, we can mark a point F on line AE such that we get OF as the perpendicular bisector to line AE. The figure below shows all the details.

OF is a perpendicular bisector to AE, that means OF is perpendicular to AE. So, $\angle OFE=90{}^\circ $ and $\vartriangle OFE$ is a right-angled triangle.

OF is perpendicular bisector to AE, this means that we can write EF=AF and DF=CF.

AE=EF+AF=2AF

2EF=25

$\Rightarrow EF=\dfrac{25}{2}$

We can apply the Pythagoras theorem in $\vartriangle OFE$ and find the value of OF.

We know that by Pythagoras theorem,

$O{{E}^{2}}=O{{F}^{2}}+E{{F}^{2}}$

Substituting OE=13 cm, $EF=\dfrac{25}{2}$ we get,

$\Rightarrow {{13}^{2}}={{\left( \dfrac{25}{2} \right)}^{2}}+O{{F}^{2}}$

$\begin{align}

& \Rightarrow O{{F}^{2}}=\dfrac{676-625}{4} \\

& \Rightarrow O{{F}^{2}}=\dfrac{51}{4} \\

& \Rightarrow OF=\sqrt{\dfrac{51}{4}} \\

& \Rightarrow OF=\dfrac{\sqrt{51}}{2} \\

\end{align}$

Now we can consider the right-angled triangle, $\vartriangle OFD$.

By Pythagoras theorem,

$O{{D}^{2}}=O{{F}^{2}}+D{{F}^{2}}$

Substituting the values of OD=5 cm, OF=$\dfrac{\sqrt{51}}{2}$, we get

$\begin{align}

& D{{F}^{2}}=\dfrac{100-51}{4} \\

& DF=\sqrt{\dfrac{49}{4}}=\pm \dfrac{7}{2} \\

\end{align}$

The length of DF cannot be negative. So, we can take DF= $\dfrac{7}{2}$.

$\begin{align}

& AD=DF+AF \\

& \Rightarrow \dfrac{25}{2}+\dfrac{7}{2}=\dfrac{32}{2}=16cm \\

\end{align}$

Hence, the length of AD is obtained as 16 cm.

Note: The length of a side cannot be negative and perpendicular bisector bisects the line into two equal parts. If this question is asked as an objective question, we can eliminate the options greater than 25 because here AE is given as 25 cm and AD is a part of AE so the length cannot be greater than 25.

Complete step-by-step answer:

We have been given two concentric circles with centre O and radius 13 cm and 5 cm. It is also given in the question that the line drawn from the point A of the outer circle touches the inner circle at point M and line AE intersects the inner circle at the points C and D.

Let us take OE as the radius of the bigger circle with length 13 cm and OD as the radius of the smaller circle with length 5 cm.

Now, we can mark a point F on line AE such that we get OF as the perpendicular bisector to line AE. The figure below shows all the details.

OF is a perpendicular bisector to AE, that means OF is perpendicular to AE. So, $\angle OFE=90{}^\circ $ and $\vartriangle OFE$ is a right-angled triangle.

OF is perpendicular bisector to AE, this means that we can write EF=AF and DF=CF.

AE=EF+AF=2AF

2EF=25

$\Rightarrow EF=\dfrac{25}{2}$

We can apply the Pythagoras theorem in $\vartriangle OFE$ and find the value of OF.

We know that by Pythagoras theorem,

$O{{E}^{2}}=O{{F}^{2}}+E{{F}^{2}}$

Substituting OE=13 cm, $EF=\dfrac{25}{2}$ we get,

$\Rightarrow {{13}^{2}}={{\left( \dfrac{25}{2} \right)}^{2}}+O{{F}^{2}}$

$\begin{align}

& \Rightarrow O{{F}^{2}}=\dfrac{676-625}{4} \\

& \Rightarrow O{{F}^{2}}=\dfrac{51}{4} \\

& \Rightarrow OF=\sqrt{\dfrac{51}{4}} \\

& \Rightarrow OF=\dfrac{\sqrt{51}}{2} \\

\end{align}$

Now we can consider the right-angled triangle, $\vartriangle OFD$.

By Pythagoras theorem,

$O{{D}^{2}}=O{{F}^{2}}+D{{F}^{2}}$

Substituting the values of OD=5 cm, OF=$\dfrac{\sqrt{51}}{2}$, we get

$\begin{align}

& D{{F}^{2}}=\dfrac{100-51}{4} \\

& DF=\sqrt{\dfrac{49}{4}}=\pm \dfrac{7}{2} \\

\end{align}$

The length of DF cannot be negative. So, we can take DF= $\dfrac{7}{2}$.

$\begin{align}

& AD=DF+AF \\

& \Rightarrow \dfrac{25}{2}+\dfrac{7}{2}=\dfrac{32}{2}=16cm \\

\end{align}$

Hence, the length of AD is obtained as 16 cm.

Note: The length of a side cannot be negative and perpendicular bisector bisects the line into two equal parts. If this question is asked as an objective question, we can eliminate the options greater than 25 because here AE is given as 25 cm and AD is a part of AE so the length cannot be greater than 25.

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