Question Number 56289 by mr W last updated on 13/Mar/19 Commented by mr W last updated on 13/Mar/19 $${See}\:{Q}\mathrm{56264} \\ $$$${Find}\:{the}\:{max}.\:{and}\:{min}.\:{circumscribing} \\ $$$${equilateral}\:{triangle}\:{of}\:{a}\:{given}\:{triangle} \\ $$$${with}\:{side}\:{lengthes}\:{a},\:{b}\:{and}\:{c}.…
Question Number 121802 by prakash jain last updated on 11/Nov/20 Commented by prakash jain last updated on 11/Nov/20 $$\mathrm{ajfour}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{cannot}\:\mathrm{find}\:\mathrm{the}\:\mathrm{original} \\ $$$$\mathrm{question}\:\mathrm{but}\:\mathrm{i}\:\mathrm{got}\:{x}\notin\mathbb{R}. \\ $$$$\mathrm{Please}\:\mathrm{check}.\:\mathrm{I}\:\mathrm{will}\:\mathrm{also}\:\mathrm{recheck}. \\ $$$$\mathrm{Thanks}.…
Question Number 187312 by Ari last updated on 15/Feb/23 Commented by Frix last updated on 15/Feb/23 $$\alpha=\mathrm{0}\:\Rightarrow\:{AB}=\mathrm{2} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 187274 by normans last updated on 15/Feb/23 Commented by normans last updated on 15/Feb/23 $${suppose}\:{point}\:\boldsymbol{{P}}\:{are}\:{in}\:\bigtriangleup\boldsymbol{{ABC}} \\ $$$${then}\:{the}\:{bisector}\:\angle\boldsymbol{{BPC}},\angle\boldsymbol{{CPA}},\:\angle\boldsymbol{{APB}} \\ $$$${cut}\:{the}\:{sides}\:\boldsymbol{{BC}},\boldsymbol{{CA}}\:{and}\:\boldsymbol{{AB}}\:\:{each}\:{on}\:{point}\:\boldsymbol{{X}},\boldsymbol{{Y}}\:{and}\:\boldsymbol{{Z}}. \\ $$$${show}\:{that}\:\boldsymbol{{AX}},\boldsymbol{{BY}}\:\:{and}\:\boldsymbol{{CY}}\:{is}\:{a}\:{kongruen}. \\ $$$$…
Question Number 187257 by ajfour last updated on 15/Feb/23 Commented by ajfour last updated on 15/Feb/23 $$\left({x}^{\mathrm{2}} −\mathrm{1}\right){x}={c}\:\:\:\:\:\:{Find}\:{x}. \\ $$ Commented by ajfour last updated…
Question Number 187205 by mr W last updated on 14/Feb/23 $${if}\:{the}\:{sides}\:{of}\:{a}\:{triangle}\:{are} \\ $$$$\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} },\:\sqrt{{a}^{\mathrm{2}} +\mathrm{4}{b}^{\mathrm{2}} },\:\sqrt{\mathrm{4}{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:{respectively}, \\ $$$${what}\:{is}\:{its}\:{area}? \\ $$ Answered by…
Question Number 187168 by Rupesh123 last updated on 14/Feb/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 187167 by Rupesh123 last updated on 14/Feb/23 Answered by mr W last updated on 14/Feb/23 Commented by mr W last updated on 14/Feb/23…
Question Number 121615 by ajfour last updated on 10/Nov/20 Commented by ajfour last updated on 10/Nov/20 $${The}\:{blue}\:{triangle}\:{is}\:{right}\:{angled}\:{and} \\ $$$${isosceles}\:{with}\:\:{AF}={BF}=\mathrm{3}. \\ $$$${Radius}\:{of}\:{circle}\:{r}=\mathrm{2}. \\ $$$${Find}\:{maximum}\:{side}\:{length}\:{of} \\ $$$${equilateral}\:\bigtriangleup{DEF}.…
Question Number 187146 by Rupesh123 last updated on 14/Feb/23 Answered by a.lgnaoui last updated on 14/Feb/23 $$\bigtriangleup{SMI}\:\:{et}\:{RNI}\:{semblables}\:\:\: \\ $$$${I}\:{centre}\:{de}\:{SRMN}\:\:{SK}=\frac{{PQ}}{\mathrm{2}}=\mathrm{3} \\ $$$$\bigtriangleup{BRQ}\:\:\:{RIO}\:\:{Semblables} \\ $$$$\frac{{OR}}{{OI}}=\frac{\frac{{RN}}{\mathrm{2}}}{{HQ}}=\frac{{RQ}}{{BQ}}\Rightarrow\:\:\:\frac{{RN}}{\mathrm{2}{HQ}}=\frac{{RQ}}{{BQ}} \\ $$$${HQ}=\mathrm{3}\:\:{RQ}=\mathrm{6}\:\:\:{BQ}={BP}+\mathrm{6}…