Question 1199008
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The figure below shows a vertical cross - section of a hemispherical bowl centre O. 
The radius OA of the bowl is 15 cm. The bowl contains water to a depth of 6 cm. 
Calculate the diameter of the water surface in the bowl.
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Make a sketch.

Draw the radius of 15 cm vertically down.

Draw the water level horizontally.


Draw another radius to the edge of the water level.


In the sketch, find a right-angled triangle.


    Its hypotenuse is the slanted radius of 15 cm length.

    Its one leg is 15-6 = 9 cm long.

    Hence, the other leg is  {{{sqrt(15^2-9^2)}}} = 12 cm.


Thus the radius of the water surface in the bowl is 12 cm.


Hence, the area of the water surface is  {{{pi*12^2}}} = 383.08 cm^2.    <U>ANSWER</U>
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Solved.