A circular fountain 10 feet in diameter has a circular walk 3 feet wide paved around it. [Solved]

u discover the world of the internal circle, which is 78.54, and u subtract that from the outer cricle, which is 132.73, and u get 54.19

It is the world of a 16 ft. circle minus the world of a ten ft. circle.
122.46 sf

It is a method across the space of a circle Pi R squared
Pi = 3.14
The three toes stroll is three toes Past the fountain.
The Fountain is 10 toes in diamater, 5 ft radius. On condition that its 3 toes huge, it might be 6 toes in diameter, or 3 foot radius.
You principally compute the world of the Stroll and the fountain mixed after which subtract the world of the fountain. So:
Each = 3.14 X 8 Ft (mixed radius) Squared or 200.96 foot space
Simply Fountain = 3.14 x 5 Ft Squared or 78.5 toes.
Now subtract each from the fountain = 122.46 toes.
Subsequently, the Space of the Stroll = 122.46 Toes.
Hope that helps.

Its like in kidz approach,,
30 sq. models!!!
I feel..
l0l.
sry if it didnt assist..

determine the world of the circle outdoors of the stroll together with the fountain and all.
Then subtract the world of the fountain. pi(10)^2

First you wish to discover space coated by fountain:
Components: r^2
r=10/2
Space=25 sq. ft.
Subsequent you wish to discover the full space coated by each the stroll and the fountain. For the reason that stroll is 3 ft. huge, then the radius r (5) will increase by 3. So r=8
Complete Space=64 sq. ft.
Now you subtract the 2 to search out the world of the stroll:
Complete space=stroll space fountain space
space of stroll = (stroll space fountain space) fountain space
space of stroll = (64) 25
space of stroll = 39

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Reply 7

It is a method across the space of a circle Pi R squared
Pi = 3.14
The three toes stroll is three toes Past the fountain.
The Fountain is 10 toes in diamater, 5 ft radius. On condition that its 3 toes huge, it might be 6 toes in diameter, or 3 foot radius.
You principally compute the world of the Stroll and the fountain mixed after which subtract the world of the fountain. So:
Each = 3.14 X 8 Ft (mixed radius) Squared or 200.96 foot space
Simply Fountain = 3.14 x 5 Ft Squared or 78.5 toes.
Now subtract each from the fountain = 122.46 toes.
Subsequently, the Space of the Stroll = 122.46 Toes.
Hope that helps.

First, a pair of definitions: whereas we have 2 circles with a difficulty-loose centre, that is acknowledged as a pair of co-axial circles. the part that is not difficulty-loose to the 2 circles is acknowledged as an annulus. second, the uncomplicated maths.: If the radius of the higher one is R with part A and the smaller one is r with part a: A = pi.R; a = pi.r {the place pi = 3.142 . . . . .} So: (A a) = pi.R pi.r = pi(R r) = pi(R r).(R r) . . . . . [a million] O.okay. third; your query: (A a) is the part of the annulus; that is what you require. all of us perceive r = 5 ft utilizing truth the trail is 3 ft big, R = r 3 = 5 3 = 8 Fourth; the reply: seek advice from [a million] above; insert numbers: (A a) = pi.(8 5).(8 5)= pi x 3 x 13 = 33.pi = 122.fifty 4 sq. ft.

Reply 7

Its like in kidz approach,,
30 sq. models!!!
I feel..
l0l.
sry if it didnt assist..

This post is last updated on hrtanswers.com at Date : 1st of September – 2022