An altitude of a triangle is 5/3 the length of its corresponding base.If the altitude be increased by 4 cm and the base decreased by 2 cm.The area of the triangle remains same.Find the base and altitude of triangle
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ATQ,
1/2bh=1/2b'h'(b=x,h=5x/3,b'=x-2,h'=5x/3+4)
1/2*5x/3*x=1/2*(5x/3+4)*(x-2)
5x^2/6=(5x+12)(x-2)/6
5x^2/6=(5x^2-10x+12x-24)/6
0=2x-24(5x^2 and 6 cancelled)
24=2x
x=24/2
x=12cm
hence x i.e. base=12cm and 5x/3
i.e.altitude =20cm.
1/2bh=1/2b'h'(b=x,h=5x/3,b'=x-2,h'=5x/3+4)
1/2*5x/3*x=1/2*(5x/3+4)*(x-2)
5x^2/6=(5x+12)(x-2)/6
5x^2/6=(5x^2-10x+12x-24)/6
0=2x-24(5x^2 and 6 cancelled)
24=2x
x=24/2
x=12cm
hence x i.e. base=12cm and 5x/3
i.e.altitude =20cm.
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