# A cone has a height of 6 cm and its base has a radius of 4 cm. If the cone is horizontally cut into two segments 5 cm from the base, what would the surface area of the bottom segment be?

Feb 23, 2017

Surface area of bottom segment$= 137 c {m}^{2}$

#### Explanation:

In triangle ABG: $\frac{A G}{B G} = \frac{t}{a} = \tan \angle B \Rightarrow \frac{6}{4} = \tan \angle B = 1.5$

angleB=56°18’36’’

In triangle ADF:$\frac{A F}{D F} = \frac{t}{a} = \tan \angle D$----DE parallel to BC

angleD=56°18’36’’

:.a/t=cot56°18’36’’ xx 1

multiply both sides by$\frac{t}{1}$

$a = 0.666664964 c m \Rightarrow$ radius top cone

Surface area of a cone$= \pi r l$

Area of circle$= \pi {r}^{2}$

Side of cone=l :.s/t=cosec56°18’36’’

multiply both sides by$\frac{t}{1}$

s=6 xx cosec56°18’36’’

$s = 7.211 = l$ side of cone

Side of top cone=s/t=cosec56°18’36’’

multiply both sides by$\frac{t}{1}$

s=1 xx cosec56°18’36’’

$s = 1.201850646 c m = l$ slope side of top cone

Total SA$= \pi {r}^{2} + \pi r l$

SA$= \pi \times {4}^{2} + \pi \times 4 \times 7.211$

Total SA$= 140.882 c {m}^{2}$

SA ot top cone: $= \pi {r}^{2} + \pi r l$

$= \pi \times {.666664964}^{2} + \pi \times 0.666664964 \times 1.202$

Top SA$= 3.913 c {m}^{2}$

SA of bottom part$= 140.882 - 3.913 = 136.968 = 137 c {m}^{2}$