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Question Number 155056 by mathdanisur last updated on 24/Sep/21

Answered by mr W last updated on 24/Sep/21

BF^2 =AB^2 +((√2)AC)^2 +2×AB×(√2)AC×cos (A+(π/4))   ..(i)  CD^2 =AC^2 +((√2)AB)^2 +2×AC×(√2)AB×cos (A+(π/4))   ..(ii)  (i)−(ii):  BF^2 −CD^2 =AC^2 −AB^2   ⇒AB^2 +BF^2 =AC^2 +CD^2

$${BF}^{\mathrm{2}} ={AB}^{\mathrm{2}} +\left(\sqrt{\mathrm{2}}{AC}\right)^{\mathrm{2}} +\mathrm{2}×{AB}×\sqrt{\mathrm{2}}{AC}×\mathrm{cos}\:\left({A}+\frac{\pi}{\mathrm{4}}\right)\:\:\:..\left({i}\right) \\ $$$${CD}^{\mathrm{2}} ={AC}^{\mathrm{2}} +\left(\sqrt{\mathrm{2}}{AB}\right)^{\mathrm{2}} +\mathrm{2}×{AC}×\sqrt{\mathrm{2}}{AB}×\mathrm{cos}\:\left({A}+\frac{\pi}{\mathrm{4}}\right)\:\:\:..\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$${BF}^{\mathrm{2}} −{CD}^{\mathrm{2}} ={AC}^{\mathrm{2}} −{AB}^{\mathrm{2}} \\ $$$$\Rightarrow{AB}^{\mathrm{2}} +{BF}^{\mathrm{2}} ={AC}^{\mathrm{2}} +{CD}^{\mathrm{2}} \\ $$

Commented by mathdanisur last updated on 24/Sep/21

very nice Ser thank you

$$\mathrm{very}\:\mathrm{nice}\:\mathrm{Ser}\:\mathrm{thank}\:\mathrm{you} \\ $$

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