A pathway with a width of $1 m$ is built around a circular pond of radius $6 m$ . What is the area of pathway?

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Solution

Radius of inner circle $(r) = 6 + 1 = 7 m$

Radius of outer circle $(R) = 6 + 1 = 7 m$

Area of pathway $=$ Area of outer circle $-$ Area of inner circle

$= π R^{2} - π r^{2}$

$= π [R^{2} - r^{2}]$

$= π (R + r) (r - r)$

$= π (7 + 6) (7 - 6)$

$= 13 π$

$= 40.84 m^{2}$ .

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A pathway with a width of 1 m is built around a circular pond of radius 6 m. What is the area of pathway?

Radius of inner circle(r)=6+1=7mRadius of outer circle(R)=6+1=7mArea of pathway = Area of outer circle − Area of inner circle=πR2−πr2=π[R2−r2]=π(R+r)(r−r)=π(7+6)(7−6)=13π=40.84m2.

A pathway with a width of $1 m$ is built around a circular pond of radius $6 m$ . What is the area of pathway?

Radius of inner circle $(r) = 6 + 1 = 7 m$

Radius of outer circle $(R) = 6 + 1 = 7 m$

Area of pathway $=$ Area of outer circle $-$ Area of inner circle

$= π R^{2} - π r^{2}$

$= π [R^{2} - r^{2}]$

$= π (R + r) (r - r)$

$= π (7 + 6) (7 - 6)$

$= 13 π$

$= 40.84 m^{2}$ .