If the error in measuring the side l of an eq | vedclass

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(A) Given, the error in side $l$ is $dl = 0.01$. The area $A$ of an equilateral triangle is given by $A = \frac{\sqrt{3}}{4} l^2$. Differentiating $A$

Vedclass · Accepted Answer

$\frac{2}{l}$

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(A) Given, the error in side $l$ is $dl = 0.01$. The area $A$ of an equilateral triangle is given by $A = \frac{\sqrt{3}}{4} l^2$. Differentiating $A$ with respect to $l$, we get $\frac{dA}{dl} = \frac{\sqrt{3}}{2} l$. Thus, the approximate change in area is $dA = \frac{\sqrt{3}}{2} l \cdot dl$. The percentage error in area is given by $\frac{dA}{A} 	imes 100$. Substituting the values: $\frac{dA}{A} 	imes 100 = \frac{\frac{\sqrt{3}}{2} l \cdot dl}{\frac{\sqrt{3}}{4} l^2} 	imes 100$. Simplifying this, we get $\frac{dA}{A} 	imes 100 = \frac{2 \cdot dl}{l} 	imes 100$. Given $dl = 0.01$, the percentage error is $\frac{2 	imes 0.01}{l} 	imes 100 = \frac{2}{l}$.

If the error in measuring the side $l$ of an equilateral triangle is $0.01$, then the percentage error in the area of the triangle, in terms of its side $l$ is:

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