The centre of gravity of the trapezium as shown in below figure from the side is at a distance of

A. $$\frac{{\text{h}}}{3} \times \frac{{{\text{b}} + {\text{2a}}}}{{{\text{b}} + {\text{a}}}}$$
B. $$\frac{{\text{h}}}{3} \times \frac{{2{\text{b}} + {\text{a}}}}{{{\text{b}} + {\text{a}}}}$$
C. $$\frac{{\text{h}}}{2} \times \frac{{{\text{b}} + {\text{2a}}}}{{{\text{b}} + {\text{a}}}}$$
D. $$\frac{{\text{h}}}{2} \times \frac{{2{\text{b}} + {\text{a}}}}{{{\text{b}} + {\text{a}}}}$$
Answer: Option A

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