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The length and width of a rectangular room are measured to be $3.95 \pm 0.05 \mathrm{~m}$ and $3.05 \pm 0.05 \mathrm{~m}$, respectively, the area of the floor is
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The correct answer is:
$12.05 \pm 0.34 \mathrm{~m}^{2}$
$\mathrm{A}=\ell \mathrm{B}$
$\mathrm{d} \mathrm{A}=\ell \mathrm{dB}+\mathrm{B} \mathrm{d} \ell$
$\frac{\mathrm{d} \mathrm{A}}{\mathrm{A}}=\frac{\ell \mathrm{dB}}{\ell \mathrm{B}}+\frac{\mathrm{Bd} \ell}{\ell \mathrm{B}}$
$\frac{\mathrm{d} \mathrm{A}}{\mathrm{A}}=\frac{\mathrm{dB}}{\mathrm{B}}+\frac{\mathrm{d} \ell}{\ell}$
$\quad=\frac{0.05}{3.05}+\frac{0.05}{3.95}$
$\quad=0.016+0.012$
$\quad=0.028 \times 12.05$
$\mathrm{dA}=0.33$
$12.05 \pm 0.34$ Ans. (C)
$\mathrm{d} \mathrm{A}=\ell \mathrm{dB}+\mathrm{B} \mathrm{d} \ell$
$\frac{\mathrm{d} \mathrm{A}}{\mathrm{A}}=\frac{\ell \mathrm{dB}}{\ell \mathrm{B}}+\frac{\mathrm{Bd} \ell}{\ell \mathrm{B}}$
$\frac{\mathrm{d} \mathrm{A}}{\mathrm{A}}=\frac{\mathrm{dB}}{\mathrm{B}}+\frac{\mathrm{d} \ell}{\ell}$
$\quad=\frac{0.05}{3.05}+\frac{0.05}{3.95}$
$\quad=0.016+0.012$
$\quad=0.028 \times 12.05$
$\mathrm{dA}=0.33$
$12.05 \pm 0.34$ Ans. (C)
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