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#### Need solution for RD Sharma maths class 12 chapter 12 Derivative as a Rate Measurer exercise multiple choise question 23

$\frac{dx}{dt}=V\! s=6\: ft/sec$

Hint:

Here, we use basic concept of volume algebra

Given:

$hm=6\, f\! t \quad V\! m=9\, f\! t/sec \quad h_{1}=15\, f\! t$

Solution:

\begin{aligned} &\rightarrow \text { Consider } \Delta A E C \text { and } \Delta B E D \\ &\angle A E D=\angle B E D=\theta \\ &\angle E A C=\angle E B D=90^{\circ} \end{aligned}

\begin{aligned} &\text { Therefore, } \triangle A E C \approx \Delta B E D \text { by AA criteria }\\ &\frac{A C}{B D}=\frac{A E}{B E} \end{aligned}

\begin{aligned} &\rightarrow \text { Substituting values, we get }\\ &\frac{15}{6}=\frac{9 t+x}{x} \end{aligned}

\begin{aligned} &\rightarrow \text { Simplifying the equation }\\ &\frac{15}{6}=\frac{9 t}{x+x}\\ &x=6 t\\ &V \! s=\frac{d x}{d t}=6 \mathrm{ft} / \mathrm{sec} \end{aligned}

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