Question Number 183962 by Acem last updated on 01/Jan/23 Commented by Acem last updated on 01/Jan/23 $$\:{TB}\:{is}\:{a}\:{tangent}\:{to}\:{the}\:{circle} \\ $$ Answered by mr W last updated…
Question Number 52881 by ajfour last updated on 14/Jan/19 Commented by ajfour last updated on 14/Jan/19 $${To}\:{find}\:\boldsymbol{{r}}\:{in}\:{terms}\:{of}\:{a},{b},{c}. \\ $$ Commented by mr W last updated…
Question Number 183945 by AROUNAMoussa last updated on 31/Dec/22 Answered by HeferH last updated on 31/Dec/22 Commented by HeferH last updated on 31/Dec/22 $$\mathrm{16}\:=\:\mathrm{3}\left(\mathrm{2}{r}\:−\:\mathrm{3}\right) \\…
Question Number 183951 by Acem last updated on 01/Jan/23 Commented by Acem last updated on 01/Jan/23 $$\:\bullet{c}\:\:{Are}\:{the}\:{points}\:{A},\:{F},\:{N},\:{M}\:\in\:{a}\:{circle}? \\ $$$$\:\:\:\:\:\:\:\:\:{if}\:{they}\:{are}\:{then}\:{find}\:{it}\:{center}\:{position}\:{and}\:{it}\:{radius} \\ $$$$\:\ast\:{AN}\bot{BC},\:{BA}\bot{CA} \\ $$$$\: \\ $$…
Question Number 183886 by Acem last updated on 31/Dec/22 Answered by HeferH last updated on 31/Dec/22 $${A}\:=\:\left(\frac{{BC}\:+\:{AD}}{\mathrm{2}}\right)\mathrm{15} \\ $$$$\:{AE}\:=\:\sqrt{\mathrm{17}^{\mathrm{2}} −\mathrm{15}^{\mathrm{2}} }\:=\:\sqrt{\mathrm{2}\centerdot\mathrm{32}}\:=\:\mathrm{8} \\ $$$$\:{B}'{D}\:=\:\sqrt{\mathrm{39}^{\mathrm{2}} −\mathrm{15}^{\mathrm{2}} }\:=\:\mathrm{36}…
Question Number 52806 by mr W last updated on 13/Jan/19 Commented by mr W last updated on 13/Jan/19 $${I}\:{found}\:{this}\:{unsolved}\:{question}\:{in} \\ $$$${Q}\mathrm{35115}\:\left({from}\:{ajfour}\:{sir}\right): \\ $$$${Find}\:{r}\:{in}\:{terms}\:{of}\:{a},{b},{c}. \\ $$…
Question Number 52780 by ajfour last updated on 12/Jan/19 Commented by ajfour last updated on 13/Jan/19 $${If}\:{the}\:{two}\:{small}\:{circles}\:{have}\:{the} \\ $$$${same}\:{radii},\:{find}\:{length}\:{AB}\:{in} \\ $$$${terms}\:{of}\:{semicircle}\:{radius}\:{R}, \\ $$$${and}\:\angle\:{between}\:{their}\:{tangent}\: \\ $$$$\left({red}\:{line}\right)\:{and}\:{diameter}\:{of}\:{the}…
Question Number 183815 by AROUNAMoussa last updated on 30/Dec/22 Answered by mnjuly1970 last updated on 30/Dec/22 $$\:\:\:\:{r}_{{l}} \:−{r}_{{s}} =\:\mathrm{5} \\ $$$$\:\:\:\:\:{r}_{{l}} \:+{r}_{{s}} =\mathrm{13} \\ $$$$\:\:\:\:\:{r}_{\:{l}}…
Question Number 183795 by AROUNAMoussa last updated on 30/Dec/22 Answered by HeferH last updated on 30/Dec/22 $$\:\left({a}+{b}\right)^{\mathrm{2}} −\left({a}−{b}\right)^{\mathrm{2}} =\:\mathrm{4}{ab}\:\Rightarrow \\ $$$$\:\sqrt{\left(\mathrm{325}\right)^{\mathrm{2}} −\left(\mathrm{125}\right)^{\mathrm{2}} }\:=\:\sqrt{\mathrm{400}{r}}\:+\:\sqrt{\mathrm{4}\centerdot\mathrm{225}\centerdot{r}} \\ $$$$\:\sqrt{\mathrm{450}\centerdot\mathrm{200}}=\:\mathrm{20}\sqrt{{r}}\:+\:\mathrm{30}\sqrt{{r}}…
Question Number 118251 by oustmuchiya@gmail.com last updated on 16/Oct/20 Answered by bobhans last updated on 16/Oct/20 $$\left({a}\right){perimeter}\:=\:\mathrm{24}\:{cm} \\ $$$$\left({b}\right)\:{area}\:=\:\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{4}×{h}\:=\:\mathrm{8}\sqrt{\mathrm{6}}\:{cm}^{\mathrm{2}} \\ $$$$\:{where}\:{h}\:=\:\sqrt{\mathrm{100}−\mathrm{4}}\:=\:\sqrt{\mathrm{96}}\:=\:\mathrm{4}\sqrt{\mathrm{6}}\: \\ $$ Terms of…