Question

• The outer edge of a circular running track is 440 m long. If it is widened by
5 m all around, find the cost of widening it at the rate of 8 per m2.​

Answer 1

DIAGRAM:-

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Given :-

The outer edge of a circular field is 440m long

To find :-

the cost of widening it after it has widened 5m all around

Solution:-

Circumference =440m

Let

radius=r

As we know that in a Circle

$\boxed{\sf Circumference =2\pi r }$

• Substitute the values

$\\\qquad\quad\sf{:}\longrightarrow 2\pi r=440$

$\\\qquad\quad\sf{:}\longrightarrow 2\times \dfrac {22}{7}r=440$

$\\\qquad\quad\sf{:}\longrightarrow \dfrac {44}{7}r=440$

$\\\qquad\quad\sf{:}\longrightarrow r=\cancel {440}\times \dfrac {7}{\cancel {44}}$

$\\\qquad\quad\sf{:}\longrightarrow r=10\times 7$

$\\\qquad\quad\sf{:}\longrightarrow r=70m$

• if it is widened 5m all around then

New radius=r+5=70+5=75m

$\\\qquad\quad\sf{:}\longrightarrow Area\: of\: inner\: circle =\pi r^2$

$\\\qquad\quad\sf{:}\longrightarrow Area_{(inner\;Circle)}=\dfrac {22}{7}(70)^2$

$\\\qquad\quad\sf{:}\longrightarrow Area_{(inner\;Circle)}=\dfrac {22}{7}\times 4900$

$\\\qquad\quad\sf{:}\longrightarrow Area_{(inner\;Circle)}=22\times 700$

$\\\qquad\quad\sf{:}\longrightarrow Area_{(inner\;Circle)}=15400m^2$

• Again

$\\\qquad\quad\sf{:}\longrightarrow Area_{(outer\;Circle)}=\pi r^2$

$\\\qquad\quad\sf{:}\longrightarrow Area_{(outer\;Circle)}=\dfrac {22}{7}(75)^2$

$\\\qquad\quad\sf{:}\longrightarrow Area_{(outer\;Circle)}=\dfrac {22}{7}\times 5625$

$\\\qquad\quad\sf{:}\longrightarrow Area_{(outer\;Circle)}=\dfrac {123750}{7}$

$\\\qquad\quad\sf{:}\longrightarrow Area_{(outer\;Circle)}=17678.5m^2$

• Now

$\\\qquad\quad\sf{:}\longrightarrow Area_{(Track)}=Area_{(outer\;circle)}-Area_{(inner\:circle)}$

$\\\qquad\quad\sf{:}\longrightarrow Area_{(Track)}=17678.5-15400$

$\\\qquad\quad\sf{:}\longrightarrow Area_{(Track)}=2278.5m^2$

Cost of widening 1m=8

$\\\qquad\quad\sf{:}\longrightarrow Cost\:of\:widening\:Track=2278.5\times 8$

$\\\qquad\quad\sf{:}\longrightarrow Cost\:of\:widening\:Track=18228units$

$\\\\\therefore\underline{\underline{\sf Cost\:of\:widening\:the\:track\:is\:18228units.}}$

Formulas to remember:-

$\\ \\ \star\sf Square=(side)^2 \\ \star\sf Rectangle=Length\times Breadth \\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2$