Question Number 158893 by ajfour last updated on 10/Nov/21 Answered by ajfour last updated on 10/Nov/21 $$\frac{{r}^{\mathrm{4}} −\left({r}+{c}\right)^{\mathrm{2}} }{{c}}=\frac{{r}^{\mathrm{2}} }{\:{r}^{\mathrm{2}} −\mathrm{1}} \\ $$$${r}^{\mathrm{3}} −{r}={c}\:\:\:\:\Rightarrow\:\:{r}^{\mathrm{6}} =\left({r}+{c}\right)^{\mathrm{2}}…
Question Number 158749 by ajfour last updated on 08/Nov/21 Commented by ajfour last updated on 08/Nov/21 $${Find}\:{extreme}\:{values}\:{of}\:{R}. \\ $$ Answered by mr W last updated…
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Question Number 27627 by Rasheed.Sindhi last updated on 11/Jan/18 Commented by Rasheed.Sindhi last updated on 11/Jan/18 $$\mathrm{Given}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{r}.\:\mathrm{AB}\:\&\:\mathrm{CD} \\ $$$$\mathrm{are}\:\mathrm{two}\:\mathrm{chords}\:\mathrm{intersecting}\:\mathrm{at}\:\mathrm{E} \\ $$$$\mathrm{OX}\:\mathrm{and}\:\mathrm{OY}\:\mathrm{are}\:\mathrm{respective}\:\mathrm{distances} \\ $$$$\mathrm{of}\:\mathrm{chords}\:\mathrm{CD}\:\mathrm{and}\:\mathrm{AB}\:\:\mathrm{from}\:\mathrm{center} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{circle}.…
Question Number 158664 by mr W last updated on 07/Nov/21 Commented by mr W last updated on 07/Nov/21 $${AQ}={CP}={BR} \\ $$$${find}\:{the}\:{smallest}\:{area}\:{of}\:\Delta{PQR}\: \\ $$$${in}\:{terms}\:{of}\:{a},\:{b}. \\ $$…
Question Number 158622 by ajfour last updated on 06/Nov/21 Commented by ajfour last updated on 06/Nov/21 $${Find}\:{maximum}\:{side}\:{length} \\ $$$${of}\:{inscribed}\:{square}\:\left({s}_{{max}} \right). \\ $$ Answered by mr…
Question Number 158619 by amin96 last updated on 06/Nov/21 Commented by amin96 last updated on 06/Nov/21 $$ \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 158586 by ajfour last updated on 06/Nov/21 Commented by mr W last updated on 06/Nov/21 Commented by mr W last updated on 06/Nov/21…
Question Number 158559 by mr W last updated on 06/Nov/21 Answered by som(math1967) last updated on 06/Nov/21 Commented by som(math1967) last updated on 06/Nov/21 $$\angle{ACB}=\mathrm{90}\:\:\left[{semicircular}\:\right]…
Question Number 158543 by ajfour last updated on 06/Nov/21 Commented by ajfour last updated on 06/Nov/21 $${The}\:{quadrilateral}\:{is}\:{a}\:{square}. \\ $$ Commented by mr W last updated…