Semi-perimeter of the triangle, s=2a+b+c
Area of triangle =s(s−a)(s−b)(s−c).
When each side is doubled, the new sides are 2a,2b,2c.
Hence, new s′=22a+2b+2c=2(2a+b+c)=2s
New area A′=2s(2s−2a)(2s−2b)(2s−2c)
=2×2×2×2×s(s−a)(s−b)(s−c)
=4s(s−a)(s−b)(s−c).
∴ % change in Area =AA′−A×100
=s(s−a)(s−b)(s−c)4s(s−a)(s−b)(s−c)−s(s−a)(s−b)(s−c)
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