Answer:
Here given that perimeter of circular running track =330m
Let radius=r
As we know that in a circle
[tex]{\boxed{\sf Perimeter=2\pi r}}[/tex]
[tex]\qquad \quad{:}\longmapsto\tt 2\times\dfrac {22}{7}×r=330 [/tex]
[tex]\qquad \quad{:}\longmapsto\tt \dfrac {44}{7}r=330 [/tex]
[tex]\qquad \quad{:}\longmapsto\tt r=\dfrac {330×7 }{44}[/tex]
[tex]\qquad \quad{:}\longmapsto\tt r=\dfrac{2310}{44}[/tex]
[tex]\qquad \quad{:}\longmapsto\tt r=52.5 [/tex]
[tex]\therefore\sf r=53m (Approx)[/tex]
Again
[tex]{\boxed{\sf Area=\pi r^2 }}[/tex]
[tex]\qquad \quad{:}\longmapsto\tt Area=\dfrac{22}{7}×(53)^2 [/tex]
[tex]\qquad \quad{:}\longmapsto\tt Area=8828.2m^2 [/tex]
[tex]\therefore\sf Area=8828m^2 (Approx)[/tex]
Now
the radius is increased by 7m
Hence
New radius=(x+7)m=53+7=60m
New Area=
[tex]\qquad \quad{:}\longmapsto\tt \pi. r^2 [/tex]
[tex]\qquad \quad{:}\longmapsto\tt \dfrac {22}{7}(60)^2 [/tex]
[tex]\qquad \quad{:}\longmapsto\tt \dfrac {22}{7}×3600 [/tex]
[tex]\qquad \quad{:}\longmapsto\tt 11314.28m^2 [/tex]
[tex]\therefore\sf AREA=11314m^2 (Approx).[/tex]
now
Area of widened Area=New area-Old Area
[tex]\qquad \quad{:}\longmapsto\tt 11314-8828 [/tex]
[tex]\qquad \quad{:}\longmapsto\tt 2486m^2 [/tex]
- Cost of widening per square=8
Total cost=
[tex]\qquad \quad{:}\longmapsto\tt 2486×8 [/tex]
[tex]\qquad \quad{:}\longmapsto\tt 19888[/tex]
[tex]\therefore\sf Total\:cost\:is\: 19888.[/tex]
Can't understand the attachment?
Here is a latex diagram for your question.
[tex]\Huge\downarrow [/tex]
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