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Plzz solve this it is of maths |
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Answer» 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%Step-by-step explanation:please mark my answer as brainliest |
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