Answer to Question #149419 in Geometry for Favour

Question #149419
A lampshade in form of a frustum of cone has height of 12cm upper and lower diameter of 10cm and 20cm respectively what area of material is required to cover the curved surface area
1
Expert's answer
2020-12-08T17:59:27-0500

Explanations & Calculations




  • Lateral surface area of a cone is given by A=πrL\small A = \pi rL
  • Therefore, by careful inspection, the needed area is

A1=πR(L+x)πrx=π[RL+x(Rr)](1)\qquad\qquad \begin{aligned} \small A_1 &=\small\pi R(L+x)-\pi rx\\ \small &= \small \pi \big[RL+x(R-r)\big]\cdots(1) \end{aligned} R=Base face radiusr=upper face radius of the fusrtum\qquad\qquad \begin{aligned} \small R &=\small \text{Base face radius}\\ \small r &= \small\text{upper face radius of the fusrtum} \end{aligned}

  • Considering the equiangular triangles ABE & ACD,

510=h12+hh=12cm\qquad\qquad \begin{aligned} \small \frac{5}{10}&= \small \frac{h}{12+h}\\ \small h&= \small 12cm \end{aligned}

  • By Pythagoras theorem to ABE triangle,

x2=122+52x=13cm\qquad\qquad \begin{aligned} \small x^2 &=\small 12^2 +5^2\\ \small x &= \small 13cm \end{aligned}

  • Considering the equiangular triangles ACD & ABE,

105=L+xxL=13cm\qquad\qquad \begin{aligned} \small \frac{10}{5} &= \small \frac{L+x}{x}\\ \small L &= \small 13cm \end{aligned}

  • Therefore, by (1),

A1=π[10×13+13×(105)]=612.61cm2\qquad\qquad \begin{aligned} \small A_1 &= \small \pi\big[10\times 13+13\times (10-5)\big]\\ &= \small \bold{612.61cm^2} \end{aligned}


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