Question

$\huge \fbox \red{❥ Question}$


The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field​

Answer 1

$\large{\underline{ \mathfrak{ \red{Given}}}}$

• The diagonal of a rectangular field is 60 metres more than the shorter side.
• The longer side is 30 metres more than the shorter side.

$\large{ \underline{ \mathfrak{To \: find}}}$

• The sides of the field

$\large{ \purple{ \underbrace{ \mathfrak{Solution}}}}$

We have,

Long side of rectangular field is 60 meters more then the shorter side.

Let the longer side will be (y+30)mtr.

And the length of the diagonal will be (y+60) mtr.

Now, According to given question.

Using Pythagoras theorem,

(Diagonal)^2 =(shorterside)^2 +(longerside)^2

$(y+60) {}^{2} =y {}^{2} +(y+30) {}^{2} \\ ⇒y {}^{2} +3600+120y=y {}^{2} +y {}^{2} +900+60y \\ ⇒3600+120y−y {}^{2} −900−60y=0 \\ ⇒−y {}^{2} +60y+2700=0 \\ ⇒y {}^{2} −60y−2700=0 \\ ⇒y {}^{2} −(90−30)y−2700=0 \\ ⇒y2−90y+30y−2700=0 \\ ⇒y(y−90)+30(y−90)=0 \\ ⇒(y−90)(y+30)=0$

If,

y+30=0

y=−30(notpossible)

If,

y−90=0

y=90

So, the length of shorter side =y−30=90−39=60

Longer side 90+30=120

Hence,the answer is 90 meters and 120 meters.

❣Hope this helps you✌

Answer 2

$\LARGE\textbf{\underline{\underline{Solution :-}}}$

$\large\texttt{↦ Let the smaller side be ' x ' .}\\\\\large\texttt{↦ So , the diagonal will be ' 60 + x ' .}\\\\\large\texttt{↦ And , the longer side will be ' 30 + x ' .}$

______________________________________

• As we know that the small side , the long side and the diagonal of the Rectangle follow the ' Pythagoras Theorem ' .

_______________________________________

So ,

______________________________________

$\large:\: \bigstar\textsf\textcolor{orange}{\: \: \: Diagonal ² = Small side ² + Long side ² .}\\\\\\\large: \: \Longrightarrow\textsf{( x + 60 )² = x² + ( 30 + x )² .}\\\\\\\large: \: \Longrightarrow\textsf{↦ By using ( a + b )² = a² + b² + 2ab .}\\\\\\\large: \: \Longrightarrow\textsf{x² + 3600 + 120x = x² + 900 + x² + 60x .}\\\\\\\large: \: \Longrightarrow\textsf{3600 - 900 + 120x - 60x = x² + x² - x²}\\\\\\\large: \: \Longrightarrow\textsf{2700 + 60x = x²}\\\\\\\large : \: \Longrightarrow\underline{\boxed{\textsf\textcolor{purple}{- x² + 60x + 2700 = 0}}}$

______________________________________

$\large:\: \bigstar\textsf\textcolor{orange}{\: \: \: By factorisation method :-}\\\\\\\large: \: \Longrightarrow\textsf{- x² + 60x + 2700 = 0}\\\\\\\large: \: \Longrightarrow\textsf{- x² + 90x - 30x + 2700 = 0}\\\\\\\large: \: \Longrightarrow\textsf{- x ( x - 90 ) - 30 ( x - 90 ) = 0}\\\\\\\large: \: \Longrightarrow\textsf{( x - 90 ) ( - x - 30 ) = 0}$

_____________________________________

Now : -

_____________________________________

$\large: \: \Longrightarrow\textsf{x - 90 = 0 \: \: \: or \: \: \: - x - 30 = 0}\\\\\\\large: \: \Longrightarrow\textsf{x = 90 \: \: \: or \: \: \: - x = 30 }\\\\\\\large: \: \Longrightarrow\textsf{x = 90 \: \: \: or \: \: \: x = - 30}\\\\\\\large: \: \Longrightarrow\textsf{As the side can't be negative :- }\\\\\\\large: \: \Longrightarrow\textsf\textcolor{purple}{So , the value of the x = 90}$

______________________________________

$\\\\\\\large : \: \Longrightarrow\underline{\boxed{\textsf\textcolor{magenta}{Small side = 90 units }}}\\\\\\\large : \: \Longrightarrow\underline{\boxed{\textsf\textcolor{magenta}{Long side = 30 + 90 = 120 units}}}\\\\\\\large : \: \Longrightarrow\underline{\boxed{\textsf\textcolor{magenta}{Diagonal = 60 + 90 = 150 units}}}$