A cone has a height of #7 cm# and its base has a radius of #5 cm#. If the cone is horizontally cut into two segments #1 cm# from the base, what would the surface area of the bottom segment be?

Question

A cone has a height of #7  cm# and its base has a radius of #5  cm#.  If the cone is horizontally cut into two segments #1 cm# from the base, what would the surface area of the bottom segment be?

Wyatt Anderson · Accepted Answer

Total surface area of bottom segment is #172.09(2dp)#sq.cm

The cone is cut at 1 cm from base, So upper radius of the frustum of cone is #r_2=(7-1)/7*5=30/7~~4.286#cm ; slant ht #l~~sqrt(1^2+(5-4.29)^2)~~1.23# Top surface area #A_t=pi*4.286^2~~57.70# sq.cm Bottom surface area #A_b=pi*5^2~~78.54# sq.cm Slant Area #A_s=pi*l*(r_1+r_2)=pi*1.23*(5+4.285)=35.85# sq.cm Total surface area of bottom segment #=A_t+A_b+A_s~~57.70+78.54+35.85 ~~172.09(2dp)#sq.cm[Ans]

Wyatt Anderson · Answer

undefined To find the surface area of the bottom segment of the cone, we need to calculate the area of the circular base and the lateral surface area of the remaining frustum (the portion of the cone above the cut).

First, let's calculate the area of the circular base using the formula for the area of a circle: $ \pi \times radius^2 $. Given that the radius $ r $ of the circular base is 5 cm, the area of the base is:

$$ A_{\text{base}} = \pi \times 5^2 = 25\pi \text{ cm}^2 $$

Next, we need to find the slant height of the frustum, which can be calculated using the Pythagorean theorem. The slant height ($ l $) and the height ($ h $) of the frustum form a right triangle, where $ l $ is the hypotenuse and $ h $ is one of the legs. We know that the height of the frustum is 6 cm (7 cm - 1 cm = 6 cm), and the radius of the circular base is 5 cm. Thus, the slant height can be found as follows:

$$ l = \sqrt{r^2 + h^2} = \sqrt{5^2 + 6^2} = \sqrt{25 + 36} = \sqrt{61} \text{ cm} $$

Now, let's calculate the lateral surface area of the frustum using the formula $ A_{\text{lateral}} = \pi \times (r_1 + r_2) \times l $, where $ r_1 $ is the radius of the base and $ r_2 $ is the radius of the top surface of the frustum (which is 1 cm less than the radius of the base).

$$ A_{\text{lateral}} = \pi \times (5 + 4) \times \sqrt{61} = 9\pi \times \sqrt{61} \text{ cm}^2 $$

Therefore, the surface area of the bottom segment (the circular base) is $ 25\pi $ square centimeters.