space = pi(1.5)^2 = about 7.068583471 m^2

we could eind the radius of the outside of water. by related triangle- correct of cone/correct of water=radius of cone/radius of water flooring 10/5=3/r r=one million.5 meter flooring portion of correct of water=(22/7)r^2 = (22/7)x1.5^2 = 3.142.25 =7.0.5 m^2

we could eind the radius of the outside of water. by related triangle- correct of cone/correct of water=radius of cone/radius of water flooring 10/5=3/r r=one million.5 meter flooring portion of correct of water=(22/7)r^2 = (22/7)x1.5^2 = 3.142.25 =7.0.5 m^2

10/5 = 6/X

X = 3

So the bottom of the triangle is 3 m, and the radius of the cone at this level is 3/2 m.

A circle (floor of the water) with radius 3/2m has an space of

piR^2

or

pi* 9/4 sq. meters

or approx. 7.068 sq. meters.

A circle (floor of the water) with radius 3/2m has an space of

piR^2

or

pi* 9/4 sq. meters

or approx. 7.068 sq. meters.

we could eind the radius of the outside of water. by related triangle- correct of cone/correct of water=radius of cone/radius of water flooring 10/5=3/r r=one million.5 meter flooring portion of correct of water=(22/7)r^2 = (22/7)x1.5^2 = 3.142.25 =7.0.5 m^2

the radius is 1.5 cm

Think about the cone as a triangle, 10m excessive with a (32)=6m base.

space = pi(1.5)^2 = about 7.068583471 m^2

the radius is 1.5 cm

Think about the half crammed with water as an identical triangle 5m excessive. The bottom could be half the bottom of the opposite triangle or (1.52)=3m.

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