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Two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3cm/sec. How fast is the area decreasing when the two equal sides are equal to the base?

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1 Answer

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  • Let two equal sides be $ x $
  • $ \perp$ from vertex of the isoceles $ \Delta$ will divide the base into equal ratio.
  • Area $ A = \large\frac{1}{2}\: b \sqrt{x^2-\large\frac{b^2}{4}}$
Step 1
Given $ \large\frac{dx}{dt}=3$
$ \large\frac{dA}{dt}=?$ when $ x = b$
Step 2
$ A = \large\frac{1}{4} b \sqrt{4x^2-b^2}$
$ \Large\frac{dA}{dt}=\Large\frac{bx}{\sqrt{4x^2-b^2}}.\large\frac{dx}{dt}$
Ans $ \large\frac{dA}{dt} = \sqrt 3b$ when $ x = b$
answered Apr 12, 2013 by thanvigandhi_1

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