Question

Class 9 ch: heron's formula.....plz solve it gyse fast....xD​

Answer 1

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$Let \: the \: side \: be \: "a" \: units \\ area = \sqrt{3} \frac{ {(3a)}^{2} }{4} \\ now \: side \: is \: tripled \\ new \: side =b = 3a \\ new \: area = \sqrt{3} \frac{ {(3a)}^{2} }{4} \\ = 9 \times \sqrt{3} \frac{ {(3a)}^{2} }{4} \\ = 9 \times initial \: area$

$increase \: in \: area \: \\ = 9 \times initial \: area \: - initial \: area \\ = 8 \times initial \: area \: \\ \: percent \: increase \\ = ( \frac{8 \times initial \: area}{initial \: area} ) \times 100 \\ = 800 \: percent$