The two equal sides of an isosceles triangle | vedclass

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Question

(B) Let $	riangle ABC$ be an isosceles triangle where $BC$ is the base of fixed length $b$. Let the length of the two equal sides be $a$.Draw $AD \pe

Vedclass · Accepted Answer

$\sqrt{3}b \ cm^2/s$

Vedclass · Answer

(B) Let $	riangle ABC$ be an isosceles triangle where $BC$ is the base of fixed length $b$. Let the length of the two equal sides be $a$.Draw $AD \perp BC$. In $	riangle ADC$,by the Pythagoras theorem,the height $h = AD = \sqrt{a^2 - (b/2)^2} = \sqrt{a^2 - b^2/4}$.The area $A$ of the triangle is given by $A = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes b 	imes \sqrt{a^2 - b^2/4} = \frac{b}{2} \sqrt{a^2 - b^2/4}$.Differentiating with respect to time $t$,we get $\frac{dA}{dt} = \frac{b}{2} 	imes \frac{1}{2\sqrt{a^2 - b^2/4}} 	imes 2a 	imes \frac{da}{dt} = \frac{ab}{2\sqrt{a^2 - b^2/4}} 	imes \frac{da}{dt}$.Given $\frac{da}{dt} = -3 \ cm/s$. When $a = b$,the height $h = \sqrt{b^2 - b^2/4} = \sqrt{3b^2/4} = \frac{\sqrt{3}b}{2}$.Substituting $a = b$ and $\frac{da}{dt} = -3$ into the derivative:$\frac{dA}{dt} = \frac{b \cdot b}{2(\sqrt{3}b/2)} 	imes (-3) = \frac{b^2}{\sqrt{3}b} 	imes (-3) = -\frac{3b}{\sqrt{3}} = -\sqrt{3}b$.The area is decreasing at the rate of $\sqrt{3}b \ cm^2/s$.

The two equal sides of an isosceles triangle with a fixed base $b$ are decreasing at the rate of $3 \ cm/s$. How fast is the area decreasing when the two equal sides are equal to the base?

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