Inner radius of hemispherical bowl = 5 cm
Thickness of the bowl = 0.25 cm
Outer radius (r) of hemispherical bowl = (5 + 0.25) cm = 5.25 cm
Outer CSA of hemispherical bowl =\( 2\pi r^2\)
= 2 × \(\frac{22}{7} \) × 5.25 cm × 5.25 cm
= 173.25 cm2
Therefore, the outer curved surface area of the bowl is 173.25 cm2 .
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.