Question

a circular field of radius 105m has a circular path of uniform width of 5m along and insideits boundary . find the area of the path

Answer 1

Area of the path is $\bold{3221.428\ m^{2}}$ in circular field radius is 105 m which has uniform circular form of width 5 m along and inside it is boundary.

Solution:

Radius of outer circle = 105 m  

Radius of inner circle = 100 m

Width of path (outer circle radius minus inner circle radius) = 105 – 100 = 5 m.

Outer circle area is $A_{o} =\pi \times 105 \times 105=(105)^({2}\pi)$

Inner circle area is $A_{i}=\pi \times 100 \times 100=(100)^{2} \pi$

Remaining area is area of circular path: $\bold{\pi\left(105^{2}-100^{2}\right)=3221.428 \mathrm{m}^{2}}$

Answer 2

Answer:

$Area \: of \: the \: path \\=1025\pi \:m^{2} = 3221.43 \:m^{2}$

Step-by-step explanation:

$Given \: Radius \: of \: a\\</p><p>circular \: field (R)= 105\:m$

$width \: of \: uniform \: path (w) = 5\:m$

$Now, \\radius \: of \: inner\: circle (r)= R-w\\r=105-5\\=100\:m$

$Area \: of \: the \: path \\=\pi R^{2}-\pi r^{2}\\=\pi (R^{2}-r^{2})\\=\pi (R+r)(R-r)\\=\pi (105+100)(105-100)\\=\pi \times 205\times 5\\=1025\pi \:m^{2}$

Or

$=1025 \times \frac{22}{7}\\= 3221.43\: m^{2}$

Therefore,

$Area \: of \: the \: path \\=1025\pi \:m^{2} = 3221.43 \:m^{2}$

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