Giventhat
\(x + y = \frac{4a}{5}\) ,\(y + z = \frac{7a}{5}\) and\(z + x = \frac{9a}{5}\)
Adding all 3, we get
\(2 * (x + y + z) = \frac{4a}{5} + \frac{7a}{5} + \frac{9a}{5}\)
\(2 * (x + y + z) = \frac{(4a+7a+9a)}{5}\)\(x + y + z = \frac{20a}{5}*\frac{1}{2}\)
\(x + y + z = 2a\) (Option C)
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\(x + y = \frac{4a}{5}\) ,\(y + z = \frac{7a}{5}\) and\(z + x = \frac{9a}{5}\)
Adding all 3, we get
\(2 * (x + y + z) = \frac{4a}{5} + \frac{7a}{5} + \frac{9a}{5}\)
\(2 * (x + y + z) = \frac{(4a+7a+9a)}{5}\)\(x + y + z = \frac{20a}{5}*\frac{1}{2}\)
\(x + y + z = 2a\) (Option C)
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