Question

Radius of a circle with center ‘O’ is\[4\sqrt{5}\]. AB is the diameter of the circle,\[AE||BC,BC=8cm\]. Line EC is tangent to the circle at point D. Find the length of DE.

A. \[4\sqrt{5}\]
B. \[6\sqrt{5}\]
C. \[8\]
D. \[10\]

Answer 1

Hint:
Find the length of OC by considering the right triangle OBC. \[\angle EOC={{90}^{\circ }}.\]
So consider the triangle EOC and find the length of OE. Then by Pythagora's theorem find the length of ED.

Complete step-by-step answer:
Given is that ‘O’ is the center of the circle. The radius of the circle is given as\[4\sqrt{5}\].
So, OA = OB = OD, they are radii of the circle.
Thus, length of\[OA=OB=OD=4\sqrt{5}\]
AB is the diameter of the circle. We know diameter is equal to twice the radius.
\[\begin{align}
  & \therefore diameter=2\times radius \\
 & d=2r=2\times 4\sqrt{5}=8\sqrt{5} \\
\end{align}\]
It is said that AE is parallel to BC i.e. \[AE||BC\].
It is given in figure that \[\angle OBC={{90}^{\circ }}\]as\[AE||BC\].
Given, now the length of BC = 8 cm.
Let us consider the triangle BOC formed, we know \[OB=4\sqrt{5}cm\]and\[BC=8cm\].
Therefore, by the Pythagoras theorem,
\[\begin{align}
  & O{{C}^{2}}=O{{B}^{2}}+B{{C}^{2}} \\
 & OC=\sqrt{O{{B}^{2}}+B{{C}^{2}}}=\sqrt{{{\left( 4\sqrt{5} \right)}^{2}}+{{8}^{2}}}=\sqrt{80+64}=\sqrt{144}=12 \\
\end{align}\]
Thus we got the length of OC = 12 cm.

From the figure we can see that the angles OAE, OBC and ODC are 90 degrees. Thus we can also say that \[\angle EOC={{90}^{\circ }}\] from the figure.
We also know that the sum of the reciprocals of the squares of the length of the arms of a right angled triangle is equal to the reciprocal of the square of the length of the perpendicular dropped on the hypotenuse from the right vertex.
Thus we can say that from the right\[\Delta EOC\],
\[\begin{align}
  & \dfrac{1}{O{{D}^{2}}}=\dfrac{1}{O{{C}^{2}}}+\dfrac{1}{O{{E}^{2}}} \\
 & \dfrac{1}{{{\left( 4\sqrt{5} \right)}^{2}}}=\dfrac{1}{{{12}^{2}}}+\dfrac{1}{O{{E}^{2}}} \\
 & \Rightarrow \dfrac{1}{O{{E}^{2}}}=\dfrac{1}{80}-\dfrac{1}{144} \\
 & \dfrac{1}{O{{E}^{2}}}=\dfrac{144-80}{80\times 144}=\dfrac{64}{80\times 144} \\
 & \therefore O{{E}^{2}}=\dfrac{80\times 144}{64} \\
 & O{{E}^{2}}=180 \\
 & \therefore OE=\sqrt{180}=\sqrt{6\times 6\times 5}=6\sqrt{5}cm \\
\end{align}\]
Now, let’s consider\[\Delta EOD\], we know that by using Pythagoras theorem, on\[\Delta EOD\],
\[\begin{align}
  & E{{O}^{2}}=O{{D}^{2}}+E{{D}^{2}} \\
 & \Rightarrow E{{D}^{2}}=O{{E}^{2}}-O{{D}^{2}} \\
 & ED=\sqrt{O{{E}^{2}}-O{{D}^{2}}}=\sqrt{{{\left( 6\sqrt{5} \right)}^{2}}-{{\left( 4\sqrt{5} \right)}^{2}}}=\sqrt{180-80}=\sqrt{100}=\sqrt{10}cm \\
\end{align}\]
Hence we got the required length, ED = 10 cm.
\[\therefore \]Option D is the correct answer.
Note:
By considering the figure, you might mistakenly take that AB and EC will have the same length and as\[OA=4\sqrt{5}\], you might take\[ED=4\sqrt{5}\]. But this method is wrong. It is only given that\[AE||BC\]. It’s not given that \[AE||EC\]and their length may vary.