(i) Cost of whitewashing the dome from inside = Rs 498.96
Cost of white washing 1m2 area = Rs 2
Therefore, CSA of the inner side of dome =\(\frac{4989.60}{20} \)= 249.48 m2
(ii) Let the inner radius of the hemispherical dome be r.
CSA of inner side of dome = 249.48 m2
\(2\pi r^2\) = 249.48 m2
\(⇒\) r2 = \(\frac{249.48}{2\pi}\) m2
\(⇒\) r2 = 249.48 \(÷\) (2 ×\(\frac{ 22}{7}\))
\(⇒\) r2 = 39.69
\(⇒\) r = \(\sqrt{39.69}\)
\(⇒\) r = 6.3 m
Volume of air inside the dome = Volume of hemispherical dome
\(=\frac{ 2}{3}\pi\)r3
\(= \frac{2}{3}\) × \(\frac{22}{7}\) × 6.3 m × 6.3 m × 6.3 m
= 523.9 m3 (approximately)
Therefore, the volume of air inside the dome is 523.9 m3 .
List-I | List-II | ||
(A) | Volume of cone | (I) | \(\frac{1}{3}\pi h(r_1^2+r_2^2+r_1r_2)\) |
(B) | Volume of sphere | (II) | \(\frac{1}{3}\pi r^2h\) |
(C) | Volume of Frustum | (III) | \(\pi r^2h\) |
(D) | Volume of cylinder | (IV) | \(\frac{4}{3}\pi r^3\) |
In Fig. 9.26, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠ BAC.