If 4r = h + \(\sqrt {{r^2} + {h^2}} \) then r2 : h2 is? (r ≠ 0)
If 4r = h + \(\sqrt {{r^2} + {h^2}} \) then r2 : h2 is? (r ≠ 0)
1). 289 : 64
2). 64 : 289
3). 64 : 225
4). 225 : 64
1 answers
$(\begin{array}{l} 4r = h + \sqrt {{r^2} + {h^2}} \\ \Rightarrow 4{\rm{r}} - {\rm{h}} = \sqrt {{r^2} + {h^2}} \end{array})$
Squaring both sides, we get
16r2 + h2 – 8rh = r2 + h2
⇒ 15r2 – 8rh = 0
⇒ r(15r – 8h) = 0
? r ≠ 0
∴ 15r – 8h = 0
$(\Rightarrow \frac{{\rm{r}}}{{\rm{h}}} = \frac{8}{{15}})$
∴ r2 : h2 = 64 : 225Other Questions
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