Ex.4.3 Q6 Quadratic Equations Solutions - NCERT Maths Class 10

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Question

The diagonal of a rectangular field is \(60\) meters more than the shorter side. If the longer side is \(30\) meters more than the shorter side, find the sides of the fields.

 Video Solution
Quadratic Equations
Ex 4.3 | Question 6

Text Solution

What is known?

i) The diagonal of the rectangular field is \(60\) meters more than the shorter side.

ii) The longer side is \(30\) meters more than the shorter side.

What is Unknown?

Sides of rectangular field.

Reasoning:

Let the shorter side be \(x\) meter. Then the length of diagonal of field will be \(x+60\) and length of longer side will be \(x+30.\) Using Pythagoras theorem, value of \(x\) can be found.

By applying Pythagoras theorem:

\[\begin{align}\text{Hypotenuse }^2&= \text{ Side 1}^{2} + \text{Side 2 }\!\!{}^\text{2}\\(60 + x)^2 &= {x^2} + (30 + x)^2\end{align}\]

Steps:

\[\begin{align}{{(60 + x)}^2} &= {x^2} + {{(30 + x)}^2} \\
60 + 2(60)x + {x^2} &= {x^2} + {{30}^2} + 2(30)x + {x^2}\\&\qquad\because {{\left( {a + b} \right)}^2} = {a^2} + 2ab + {b^2} \\3600 + 120x + {x^2} &= {x^2} + 900 + 60x + {x^2} \\3600 + 120x + {x^2} - {x^2} - 900 - 60x - {x^2}&= 0 \\2700 + 60x - {x^2} &= 0\end{align}\]

Multiplying both sides by \(-1:\)

\[{x^2} - 60x - 2700 = 0\]

Solving by quadratic formula:

Comparing with \(ax^{2}+bx+c=0\)

\[a =1,\; b=  - 60,\; c = -2700\]

\[\begin{align}{b^2} - 4ac&= {{( - 60)}^2} - 4(1)( - 2700)\\& = 3600 + 10800\\{b^2} - 4ac& = 14400 > 0\end{align}\]

\(\therefore\;\)Roots exist.

\[\begin{align}{{x}} &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\
 &= \frac{{ - ( - 60) \pm \sqrt {14400} }}{2}\\ &= \frac{{60 \pm \sqrt {14400} }}{2}\\x &= \frac{{60 \pm 120}}{2}\\x &= \frac{{60 + 120}}{2} \quad x = \frac{{60 - 120}}{2}\\x &= \frac{{180}}{2}\qquad \;\;x = \frac{{ - 60}}{2}\\x &= 90 \qquad \;\;x = - 30\end{align}\]

Length can’t be a negative value. 

\(\therefore x = 90\)

Length of shorter side \(x = 90 \,\rm{m}\)

Length of longer side \(=\text{ }30+x= 30 + 90 = 120\,\rm{ m.}\)