Question

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A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of ₹498.96. If the cost of white-washing is ₹2.00 per square metre, find the
1.inside surface area of dome
2. radius of dome
3.volume of dome
​

Answer 1

Answer:

Given That... A dome of a building is in the form of hemisphere and we know that the volume of hemisphere = 2/3 πr³

And Total cost = 498.96 Rs.

(1) We Know that... Total cost = Area whitewash×cost of white wash

=> Area whitewashed = Total cost/cost of white wash

=> Area whitewashed = 498.96/2

=> Area whitewashed = 249.48 Rs.

And according to the question, the area whitewashed is Curved surface area of hemisphere.

Therefore, inner surface area of dome=298.48m²

(2) Let, the radius of the base of the dome =

Surface area of the dome = 249.48m²

According to the question....

2πr² = 249.48 m²

=> r² = 249.48/2π

=> r² = 249.48×7/2×22

=> r² = 39.69

=> r = √39.68 = 6.3 m

we know that... volume of inside the dome = volume of hemisphere

=> volume of inside the dome = 2/3πr³

=> 2/3×22/6×(6.3)³

=> 2/3×22/7×6.3×6.3×6.3

=> 523.908 m³ (approx.)

Answer 2

Cost of white washing dome = Rs.498.96

Cost of white washing 1m² area = Rs.2

$\sf \: CSA= \frac{498.96}{2} = 249.48 \: {m}^{2}$

Let, radius = r ,CSA =249.48 m²

$\sf \: CSA= 2\pi {r}^{2}$

$\sf \:249.48= 2 \times 3.14 \times {r}^{2}$

$\sf {r}^{2} = 37.9 \: {m}^{2}$

$\sf \: r = 6.3 \: m$

$\sf\: volume\: of \: dome = \frac{2}{3} \pi {r}^{3}$

$\sf \: vol. \: of \: dome = 523.9 \: {m}^{3}$