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**Hint**: Here in this question, we have to find the length of the side AD in the given rectangular diagram AD, this can be solve by applying a Pythagoras theorem i.e., \[hy{p^2} = op{p^2} + ad{j^2}\] in triangles \[\Delta DEC\], \[\Delta EAD\], and \[\Delta EBC\] then further simplify by using a basic arithmetic operation we get the required length of the side AD.

**:**

__Complete step-by-step answer__Consider the given diagram \[ABCD\] is a rectangular and point E lies on the side AB, and the given line DE and EC which have a length 3 and 4 respectively and the point E makes angle \[{90^ \circ }\].

In rectangle \[ABCD\], we found three triangles i.e., \[\Delta DEC\], \[\Delta EAD\] and \[\Delta EBC\]. E be the common point in all three triangles.

Now, consider the triangle \[\Delta DEC\] is a right-angled triangle.

In \[\Delta DEC\], \[\left| \!{\underline {

E }} \right. = {90^ \circ }\]

Then by the Pythagoras theorem

\[ \Rightarrow {\left( {DC} \right)^2} = {\left( {DE} \right)^2} + {\left( {EC} \right)^2}\]

Given, \[DE = 3\] and \[EC = 4\], on substituting we have

\[ \Rightarrow {\left( {DC} \right)^2} = {\left( 3 \right)^2} + {\left( 4 \right)^2}\]

\[ \Rightarrow {\left( {DC} \right)^2} = 9 + 16\]

\[ \Rightarrow {\left( {DC} \right)^2} = 25\]

Taking square root on both side, then

\[ \Rightarrow DC = \pm \sqrt {25} \]

\[ \Rightarrow DC = \pm 5\]

When, measuring the length of any shape we only consider the positive value

\[\therefore DC = 5\]

Now, consider triangle \[\Delta EAD\] is a right-angle triangle

In, \[\Delta EAD\], \[\left| \!{\underline {

A }} \right. = {90^ \circ }\]and let take \[AE = x\]

Then by the Pythagoras theorem

\[ \Rightarrow {\left( {DE} \right)^2} = {\left( {AD} \right)^2} + {\left( {AE} \right)^2}\]

\[ \Rightarrow {\left( 3 \right)^2} = {\left( {AD} \right)^2} + {\left( x \right)^2}\]

\[ \Rightarrow 9 = {\left( {AD} \right)^2} + {x^2}\]

On rearranging, we have

\[ \Rightarrow {\left( {AD} \right)^2} = 9 - {x^2}\]-------(1)

Now, consider triangle \[\Delta EBC\] is a right-angle triangle

In, \[\Delta EBC\], \[\left| \!{\underline {

B }} \right. = {90^ \circ }\] and

Let, \[EB = AB - AE\]

But \[AB = DC = 5\], then

\[ \Rightarrow EB = 5 - x\]

Then by the Pythagoras theorem

\[ \Rightarrow {\left( {EC} \right)^2} = {\left( {BC} \right)^2} + {\left( {EB} \right)^2}\]

But \[BC = AD\], then

\[ \Rightarrow {\left( 4 \right)^2} = {\left( {AD} \right)^2} + {\left( {5 - x} \right)^2}\]

\[ \Rightarrow 16 = {\left( {AD} \right)^2} + {\left( {5 - x} \right)^2}\]

On applying a algebraic identity: \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\], then

\[ \Rightarrow 16 = {\left( {AD} \right)^2} + {5^2} + {x^2} - 10x\]

\[ \Rightarrow 16 = {\left( {AD} \right)^2} + 25 + {x^2} - 10x\]

On rearranging, we have

\[ \Rightarrow {\left( {AD} \right)^2} = 16 - 25 - {x^2} + 10x\]

\[ \Rightarrow {\left( {AD} \right)^2} = 10x - 9 - {x^2}\] ------(2)

From equation (1) and (2), we have

\[ \Rightarrow 9 - {x^2} = 10x - 9 - {x^2}\]

Take variable \[x\] terms in to the LHS, then

\[ \Rightarrow {x^2} - {x^2} - 10x = - 9 - 9\]

On simplification, we have

\[ \Rightarrow - 10x = - 18\]

Divide both side by -10, then we get

\[ \Rightarrow x = \dfrac{{ - 18}}{{ - 10}}\]

\[ \Rightarrow x = 1.8\]

Substitute \[x\] value in equation (1), then

\[ \Rightarrow {\left( {AD} \right)^2} = 9 - {\left( {1.8} \right)^2}\]

\[ \Rightarrow {\left( {AD} \right)^2} = 9 - 3.24\]

\[ \Rightarrow {\left( {AD} \right)^2} = 5.76\]

Take square root on both side, we have

\[ \Rightarrow AD = \pm \sqrt {5.76} \]

\[ \Rightarrow AD = \pm 2.4\]

\[\therefore AD = 2.4\]

Hence, the length of the side \[AD = 2.4\].

Therefore, Option (A) is correct.

**So, the correct answer is “Option A”.**

**Note**: To solve rectangular based problems, remember the properties of triangle i.e., the parallel sides are equal and when triangle is right angle it‘s obey the Pythagoras theorem which stated as “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides” i.e., \[hy{p^2} = op{p^2} + ad{j^2}\], and main thing is measurement of length of any shape should be in positive.

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