Question

A right circular cone with a radius to height ratio is 12: 5 is cut parallel to its base to get a smaller cone and a frustum. If the height of the smaller cone to the height of the frustum is in the ratio 3:1, by what percentage is the combined total surface area of the smaller cone and frustum more with respect to the original cone.

(a) 22% 

(b) 32% 

(c) 46% 

(d) None of these

Answer 1

Answer: (d) = none of these

Let the radius and height of the circular cone be 12 k and 5 k respectively.

Given:

\(\frac{Height (h_1) \,of\, smaller\, cone}{Height (h_2 ) \,of \,frustum} =\frac{3}{1}\)

∴  h₁ = \(\frac{3}{4}\)× 5k and h2 = \(\frac{1}{4}\)× 5k 

△AFE ~ △AGC, so

\(\frac{FE}{GC} = \frac{AF}{AG}\) 

⇒ \(\frac{r_1}{r_2} = \frac{h_1}{AG}\) 

⇒ \(\frac{r_1}{12k} =\frac{\frac{15k}{4}}{5k}\) 

⇒ \(r_1 = \frac{3}{4}\times 12k =9k\)  

Total surface area of original cone = πr₂l + πr₂² 

= \(\pi \times 12k \times \sqrt{(12k)^2+(5k)^2}+\pi \times(12k)^2\)  

= π × 12k × 13k + 144 πk² 

= 156 πk² + 144 πk² 

= 300 πk² 

Total surface area of smaller cone

\(\pi \times 9k\times \sqrt{\big(\frac{15k}{4}\big)^2 +(9k)^2} +\pi (9k)^2\) 

\(=\pi \times 9k \times \frac{39}{4} k+81 \pi k^2 = \frac{351\pi k^2}{4} +81 \pi k^2 = \frac{675\pi k^2}{4}\)   

Total surface area of frustum

= \(\pi \times (9k+12k) \times \sqrt{\big(\frac{5k}{4}\big)^2+3k^2} +\pi [(12k)^2 +(9k)^2]\) 

= \(\pi \times 21k\times \frac{13}{4}k+ \pi(144k^2 +81k^2)\) 

= \(\frac{273\pi k^2}{4} +225 \pi k^2 = \frac{1173}{4} \pi k^2\)   

∴  Reqd. difference = Total surface area of (smaller cone + frustum) – Total surface area of original cone

= \(\big(\frac{1173}{4} \pi k^2 +\frac{675}{4} \pi k^2\big) -300\pi k^2\)   

= \(\frac{1848}{4}\pi k^2 -300 \pi k^2 = 462 \pi k^2 -300\pi k^2 =162\pi k^2\)  

∴  Reqd. percent = \(\big(\frac{162\pi k^2}{300 \pi k^2}\times 100 \big)\) % 

= 54%