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The centre of gravity of a trapezium with parallel sides a and b lies at a distance of y from the base b, as shown in the below figure. The value of y is

$h(\frac { 2a + b } { a + b })$

$\frac{h} {2}(\frac { 2a + b } { a + b })$

$\frac{h} {3}(\frac { 2a + b } { a + b })$

$\frac{h} {3} (\frac { a + b } { 2a + b })$

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Answer : Option C

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