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Find % increase in the area of a triangle if its each side is doubled |
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Answer» Area increases four times Let a,b,c be the sides of the original ∆ & s be its semi perimeter.S= (a+b+c)/22s= a+b+c.................(1)The sides of a new ∆ are 2a,2b,2c [ given: Side is doubled]Let s\' be the new semi perimeter.s\'= (2a+2b+2c)/2s\'= 2(a+b+c) /2s\'= a+b+cS\'= 2s. ( From eq 1)......(2)Let ∆= area of original triangle∆= √s(s-a)(s-b)(s-c).........(3)& ∆\'= area of new Triangle∆\' = √s\'(s\'-2a)(s\'-2b)(s\'-2c)∆\'= √ 2s(2s-2a)(2s-2b)(2s-2c) [From eq. 2]∆\'= √ 2s×2(s-a)×2(s-b)×2(s-c)= √16s(s-a)(s-b)(s-c)∆\'= 4 √s(s-a)(s-b)(s-c)∆\'= 4∆. (From eq (3)Increase in the area of the triangle= ∆\'- ∆= 4∆ - 1∆= 3∆%increase in area= (increase in the area of the triangle/ original area of the triangle)× 100% increase in area= (3∆/∆)×100% increase in area= 3×100=300 %Hence, the percentage increase in the area of a triangle is 300% |
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