surface arena of a cone:
A=pi*r(r+sqrt(h^2+r^2))
where r=radius
Given:
A= 258 cm^2
r=4 cm
find h
258=(3.14*4)(4+sqrt(h^2+4^2))
258=12.56(4+sqrt(h^2+16))
258/12.56 = 4+sqrt(h^2+16)
16.54 = sqrt(h^2+16)
square both sides
274.233 = h^2 + 16
258.233 = h^2
h=16.0696
A right pyramid has a surface area of 258 cm2.
A right cone has a base radius of 4 cm. The
cone and pyramid have equal surface areas.
What is the height of the cone to the nearest
tenth of a centimetre?
1 answer