# 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
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:

${x^2} - 60x - 2700 = 0$

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 $$=\text{ }30+x= 30 + 90 = 120\,\rm{ m.}$$

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