how value of d is calculated in this equation

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February 24th 2011, 07:19 AM
moonnightingale

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how value of d is calculated in this equation

kindly help me to solve this equation
how he calculated vlue of d to be 16.1
February 24th 2011, 08:17 AM
harish21

$RHS=0.022\times10^{-6}\times 3\times10^8=6.6$

Therefore, you have:

$2\sqrt{\bigg(\dfrac{d}{2}\bigg)^2+8^2}-d= 6.6$

$2\sqrt{\bigg(\dfrac{d^2}{4}\bigg)+64}=6.6+d$

squaring both sides gives you:

$\bigg[2\sqrt{\bigg(\dfrac{d^2}{4}\bigg)+64}\bigg]^2=(6.6+d)^2$

$4 \cdot \bigg(\dfrac{d^2}{4}+64\bigg) = 6.6^2+2d(6.6)+ d^2$

$d^2+256=43.56+13.2d+d^2$

the $d^2$ terms cancel each other out and simplification gives you the value of d.

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