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Question Number 88936 by naka3546 last updated on 14/Apr/20

Answered by ajfour last updated on 14/Apr/20

let   a=(√(196−c^2 ))    from 1^(st)   b^2 = 2c(√(196−c^2 ))−27   from 2^(nd)   and    b^2 =(c+(√(225−c^2 )) )^2    from 3^(rd)   ⇒     2c(√(196−c^2 ))−27  = (c+(√(225−c^2 )) )^2     ......

$${let}\:\:\:{a}=\sqrt{\mathrm{196}−{c}^{\mathrm{2}} }\:\:\:\:{from}\:\mathrm{1}^{{st}} \\ $$$${b}^{\mathrm{2}} =\:\mathrm{2}{c}\sqrt{\mathrm{196}−{c}^{\mathrm{2}} }−\mathrm{27}\:\:\:{from}\:\mathrm{2}^{{nd}} \\ $$$${and}\:\:\:\:{b}^{\mathrm{2}} =\left({c}+\sqrt{\mathrm{225}−{c}^{\mathrm{2}} }\:\right)^{\mathrm{2}} \:\:\:{from}\:\mathrm{3}^{{rd}} \\ $$$$\Rightarrow\:\: \\ $$$$\:\mathrm{2}{c}\sqrt{\mathrm{196}−{c}^{\mathrm{2}} }−\mathrm{27}\:\:=\:\left({c}+\sqrt{\mathrm{225}−{c}^{\mathrm{2}} }\:\right)^{\mathrm{2}} \\ $$$$\:\:...... \\ $$

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