1. A string of length 24 cm is bent first into a square and then into a right-angled triangle by keeping one side of the square fixed as its base. Then the area of triangle equals to:
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By: anil on 05 May 2019 02.28 am
String of length 24 cm is bent into square, => Perimeter of square = 24 cm Let side of square = $$a$$ cm => $$4a=24$$ => $$a=frac{24}{4}=6$$ cm Let the other side of triangle be $$b$$ and hypotenuse be $$c$$ cm => Perimeter of triangle = $$a+b+c=24$$ => $$b+c=24-6=18$$ => $$c=18-b$$ ------------(i) Also, using Pythagoras Theorem => $$6^2+b^2=c^2$$ => $$c^2-b^2=36$$ -----------(ii) Solving equations (i) and (ii), we get : $$b=8$$ cm and $$c=10$$ cm $$ herefore$$ Area of triangle = $$frac{1}{2} ab$$ = $$frac{1}{2} imes6 imes8=24$$ $$cm^2$$ => Ans - (A)
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