Step-by-step explanation:
well, the question is in essence, how much larger the surface of the "mantle" of the large cone is than the surface of the small cone.
that means we need to calculate the surface areas (only the mantle without the circle base area of the mathematical cone) of both cones and determine the difference.
and that means we need to find out of the given information all the data necessary to make these calculations.
so, let's start.
the surface just of the outside mantle area of a cone is
pi×r×lateral height
the lateral height is the straight distance from the circular rim of the cone to the tip of the cone.
remember Pythagoras
c² = a² + b²
c = sqrt(a² + b²)
with c being the Hypotenuse (the baseline opposite of the 90° angle) of the right-angled triangle.
so, we get the lateral height as being the Hypotenuse of the right-angled triangle with the radius and the inner height as legs.
therefore, the lateral height = sqrt(r² + height²).
and so, the surface of the cone mantle is
pi×r×sqrt(r² + height²)
remember, the radius is half of the diameter.
so, the radius of the small cone is 50/2 = 25 mm
the radius of the large cone is 50×2/2 = 50 mm
because the diameter of the large cone is twice the diameter of the small cone.
the inner height of the small cone is 60 mm
the inner height of the large cone is 60 × 3/2 = 90 mm
we get this out of "the height of the small cone is 2/3 of the height of the large cone". so, the height of the large cone is 3/2 the height of the small cone.
and now we calculate
the large cone has
pi×50×sqrt(50² + 90²) = pi×50×sqrt(2500 + 8100) =
= pi×50×sqrt(10600) = 16,172.33801... mm²
the small cone has
pi×25×sqrt(25² + 60²) = pi×25×sqrt(625 + 3600) =
= pi×25×sqrt(4225) = pi×25×65 = 5,105.088062... mm²
therefore, we need
16,172.33801... - 5,105.088062... =
= 11,067.24995... mm² ≈ 11,067.25 mm²
more waffle to make a large waffle cone than to make the small waffle cone.
