Question

Diagonal AC of a rectangle ABCD is produced to the point E such that AC:CE \[ = {\text{ }}2:1,\] AB \[ = {\text{ }}8{\text{ }}cm\] and BC \[ = {\text{ }}6{\text{ }}cm.\] Find the length of DE (in cm)
A. $\sqrt {17} $
B. $3\sqrt {17} $
C. $2\sqrt {17} $
D. $4\sqrt {17} $

Answer 1

Hint: To solve this question, we will first draw a diagram using the giving information. Assuming CE as x, and AC as 2x, we will proceed further and then using Pythagoras theorem we will evaluate the value of x. Then again applying Pythagoras theorem, we will find the value of DE and hence get our required answer.

Complete step-by-step answer:
We have been given that a diagonal AC of a rectangle ABCD is produced to the point E such that AC:CE \[ = {\text{ }}2:1,\]
AB \[ = {\text{ }}8{\text{ }}cm\] and BC \[ = {\text{ }}6{\text{ }}cm.\] We need to find the length of DE.
Let, CE \[ = {\text{ }}x,\] then AC \[ = {\text{ }}2x.\]
So, using the given information let us draw a figure, to understand better.

We can see in the figure that the \[\Delta ABC\] is a right-angled triangle.
So, to find the diagonal of a rectangle we will use the Pythagoras theorem.
\[\Rightarrow A{C^2} = A{B^2} + B{C^2}\]
On putting the values in the above formula, we get
\[
   {{{\left( {2x} \right)}^2} = {8^2} + {6^2}} \\
   {\;4{x^2} = 64 + 36} \\
      {{x^2} = 25} \\
  { \Rightarrow x = 5cm}
\]
So, we get, CE \[ = {\text{ }}5cm,\] AC \[ = {\text{ }}10cm\] and also, AE \[ = {\text{ }}15cm.\]
Now, we can see in the figure that $\Delta DEA$ is a right-angled triangle.
So, again we will apply the Pythagoras theorem.
\[\Rightarrow A{E^2} = D{E^2} + A{D^2}\]
\[
\Rightarrow D{E^2} = A{E^2} - A{D^2} \\
\Rightarrow D{E^2} = {15^2} - {6^2} \\
\Rightarrow D{E^2} = 225 - 36 \\
\Rightarrow D{E^2} = 189 \\
\Rightarrow DE = 3\sqrt {17} cm \\
\]
So, the length of DE is $3\sqrt {17} cm.$

So, the correct answer is “Option C”.

Note: In the solutions, we have applied Pythagora's theorem. It is a well-known geometric theorem; it is applied in a right-angled triangle where the square of hypotenuse or the longest side is equal to sum of square of base and square of perpendicular.