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.
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.
Solution:
In rectangle ABCD, let the shorter side BC = x metres.
Then AB = (x + 30) metres and diagonal
AC = (x + 60) metres.
By Pythagoras Theorem we have
$\mathrm{AB}^{2}+\mathrm{BC}^{2}=\mathrm{AC}^{2}$
$\Rightarrow(x+30)^{2}+x^{2}$
$=(x+60)^{2}$
$\Rightarrow x^{2}+60 x+900+x^{2}$
$=x^{2}+120 x+3600$
$\Rightarrow x^{2}-60 x-2700=0$
$a=1, b=-60, c=-2700$
$D=(-60)^{2}-4 \times 1 \times(-2700)$
$=3600+10800=14400$
$\Rightarrow \sqrt{10}=\sqrt{14400}=120$
Then $x=\frac{-\mathbf{b} \pm \sqrt{\mathbf{D}}}{\mathbf{2 a}}=\frac{\mathbf{6 0} \pm \mathbf{1 2 0}}{\mathbf{2}}=90,-30$
We reject $x=-30 \quad(\because x+0)$
$\Rightarrow x=90$
$\Rightarrow \mathrm{BC}=90 \mathrm{~m}$ and $\mathrm{AB}=(90+30) \mathrm{m}=120 \mathrm{~m}$
Hence, the sides of the rectangular field are $90 \mathrm{~m}$ and $120 \mathrm{~m}$.
In rectangle ABCD, let the shorter side BC = x metres.
Then AB = (x + 30) metres and diagonal
AC = (x + 60) metres.
By Pythagoras Theorem we have
$\mathrm{AB}^{2}+\mathrm{BC}^{2}=\mathrm{AC}^{2}$
$\Rightarrow(x+30)^{2}+x^{2}$
$=(x+60)^{2}$
$\Rightarrow x^{2}+60 x+900+x^{2}$
$=x^{2}+120 x+3600$
$\Rightarrow x^{2}-60 x-2700=0$
$a=1, b=-60, c=-2700$
$D=(-60)^{2}-4 \times 1 \times(-2700)$
$=3600+10800=14400$
$\Rightarrow \sqrt{10}=\sqrt{14400}=120$
Then $x=\frac{-\mathbf{b} \pm \sqrt{\mathbf{D}}}{\mathbf{2 a}}=\frac{\mathbf{6 0} \pm \mathbf{1 2 0}}{\mathbf{2}}=90,-30$
We reject $x=-30 \quad(\because x+0)$
$\Rightarrow x=90$
$\Rightarrow \mathrm{BC}=90 \mathrm{~m}$ and $\mathrm{AB}=(90+30) \mathrm{m}=120 \mathrm{~m}$
Hence, the sides of the rectangular field are $90 \mathrm{~m}$ and $120 \mathrm{~m}$.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.