Question

Prove that the area of a circular path of uniform width h surrounding a circular region of radius r is π h( 2r+ h)

Answer 1

Answer:

πh(2r + h)

Step-by-step explanation:

Please see the attached diagram for the problem description.  

We are measuring the circular path colored in the diagram.  

Inner Circle radius is r.  

Since the road width is h, the outer circle radius is (r + h)

Area of outer circle = π(r + h)^2

Area of inner circle = πr^2

Area of circular path = area of outer circle – area of inner circle

= π(r + h)^2 – πr^2

= π[(r + h)^2 – r^2]

= π[(r + h + r)(r + h – r)]       (Since a^2 – b^2 = (a + b)(a - b)]

= π(2r + h)(h)

= πh(2r + h)

Answer 2

Answer:

Area of Path = $\pi h (2r+h)$

QED

Step-by-step explanation:

we have to Prove that the area of a circular path of uniform width h surrounding a circular region of radius r is π h( 2r+ h)

There would be two circles

one inner circle with Radius = r

and one outer circle with Radius = r + h

as we know area of circle = $\pi\times R^{2}$ Where R = Radius

Area of Inner circle = $\pi \times r^{2}$

Area of Outer circle = $\pi \times (r+h)^{2}$

Area of Path = area of outer circle - area of Inner circle

=> Area of Path = $\pi \times (r+h)^{2} - \pi \times r^{2}$

=> Area of Path = $\pi \times ((r+h)^{2} - r^{2})$

now applying formula

a² - b² = (a+b)(a-b)  here a = r+h & b = r

=> Area of Path = $\pi \times (r+h +r) \times (r+h- r)$

=> Area of Path = $\pi \times (2r+h) \times (h)$

=> Area of Path = $\pi h (2r+h)$

QED ( hence Proved)