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Quadratic Equations — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsQuadratic Equations5 Marks
Question
A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m. Find the length and breadth of the park.
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Answer


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Let the breadth of rectangular park is ' $x$ ' m.
$\therefore$ Length $=( x +3) m$
Now, Area of rectangular park
$=x(x+3) m^2$
From question, base of isosceles triangle
$=x m \text { and height }=12 m$
$\therefore$ Area of isosceles triangle
$=\left(\frac{1}{2} \times x \times 12\right) m^2=6 x m^2$
According to question,
$\begin{array}{l}x(x+3)=6 x+4 \\\Rightarrow x^2+3 x=6 x+4 \\\Rightarrow x^2+3 x-6 x-4=0 \\\text { Now } x=\frac{-b \pm \sqrt{b^2-4 ac}}{2 a} \\=\frac{-(-3) \pm \sqrt{(-3)^2-4 \cdot 1 \cdot(-4)}}{2} \\=\frac{3 \pm \sqrt{25}}{2}\end{array}$
$\therefore x =4$ or $x =-1$ (not possible)
So, Breadth $=4 m$
and Length $=(4+3) m =7 m$

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