Question Number 151973 by ajfour last updated on 24/Aug/21 Commented by ajfour last updated on 24/Aug/21 $${OB}={OC}=\sqrt{{OA}}\:\:;\:{OM}=\mathrm{1} \\ $$$${yellow}\:{area}=\mathrm{1}/\mathrm{4}\:,\:{then}\:{find} \\ $$$${OB}={OC}\:={x}\:\:\:{or}\:\:{OA}={x}^{\mathrm{2}} . \\ $$ Answered…
Question Number 151961 by Tawa11 last updated on 24/Aug/21 Commented by Tawa11 last updated on 24/Aug/21 $$\mathrm{The}\:\mathrm{pencils}\:\mathrm{has}\:\mathrm{diameter}\:\:\:\mathrm{7mm}. \\ $$$$ \\ $$$$\mathrm{Question} \\ $$Seven pencils are…
Question Number 86379 by jagoll last updated on 28/Mar/20 Commented by jagoll last updated on 28/Mar/20 $$\mathrm{to}\:\mathrm{miss}\:\mathrm{tawa} \\ $$ Commented by TawaTawa1 last updated on…
Question Number 86282 by ajfour last updated on 27/Mar/20 Commented by ajfour last updated on 27/Mar/20 $${Find}\:{s}/{r}\:.\:\:\left({s}\:{being}\:{side}\:{of}\:{square}\right). \\ $$ Commented by Prithwish Sen 1 last…
Question Number 86256 by TawaTawa1 last updated on 27/Mar/20 Answered by ajfour last updated on 27/Mar/20 $$\pi{r}^{\mathrm{2}} =\pi{a}^{\mathrm{2}} \left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$ Answered by ajfour last…
Question Number 86244 by TawaTawa1 last updated on 27/Mar/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 20599 by Tinkutara last updated on 28/Aug/17 $$\mathrm{In}\:\mathrm{a}\:\mathrm{rectangle}\:{ABCD},\:{E}\:\mathrm{is}\:\mathrm{the}\:\mathrm{midpoint} \\ $$$$\mathrm{of}\:{AB};\:{F}\:\mathrm{is}\:\mathrm{a}\:\mathrm{point}\:\mathrm{on}\:{AC}\:\mathrm{such}\:\mathrm{that}\:{BF} \\ $$$$\mathrm{is}\:\mathrm{perpendicular}\:\mathrm{to}\:{AC};\:\mathrm{and}\:{FE} \\ $$$$\mathrm{perpendicular}\:\mathrm{to}\:{BD}.\:\mathrm{Suppose}\:{BC}\:=\:\mathrm{8}\sqrt{\mathrm{3}}. \\ $$$$\mathrm{Find}\:{AB}. \\ $$ Answered by ajfour last updated…
Question Number 20545 by Tinkutara last updated on 23/Sep/17 $$\mathrm{Let}\:\mathrm{ABC}\:\mathrm{be}\:\mathrm{an}\:\mathrm{acute}-\mathrm{angled}\:\mathrm{triangle} \\ $$$$\mathrm{with}\:\mathrm{AC}\:\neq\:\mathrm{BC}\:\mathrm{and}\:\mathrm{let}\:\mathrm{O}\:\mathrm{be}\:\mathrm{the} \\ $$$$\mathrm{circumcenter}\:\mathrm{and}\:\mathrm{F}\:\mathrm{be}\:\mathrm{the}\:\mathrm{foot}\:\mathrm{of} \\ $$$$\mathrm{altitude}\:\mathrm{through}\:\mathrm{C}.\:\mathrm{Further},\:\mathrm{let}\:\mathrm{X}\:\mathrm{and} \\ $$$$\mathrm{Y}\:\mathrm{be}\:\mathrm{the}\:\mathrm{feet}\:\mathrm{of}\:\mathrm{perpendiculars}\:\mathrm{dropped} \\ $$$$\mathrm{from}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{respectively}\:\mathrm{to}\:\left(\mathrm{the}\right. \\ $$$$\left.\mathrm{extension}\:\mathrm{of}\right)\:\mathrm{CO}.\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{FY}\:\bot\:\mathrm{CA} \\ $$$$\mathrm{using}\:\mathrm{that}\:\angle\mathrm{CFY}\:=\:\angle\mathrm{CBY}\:=\:\angle\mathrm{CAF}. \\…
Question Number 86040 by TawaTawa1 last updated on 26/Mar/20 Answered by sakeefhasan05@gmail.com last updated on 26/Mar/20 Commented by Serlea last updated on 26/Mar/20 $$ \\…
Question Number 86008 by TawaTawa1 last updated on 26/Mar/20 Commented by jagoll last updated on 26/Mar/20 $$\mathrm{10}^{\mathrm{o}} \\ $$ Commented by TawaTawa1 last updated on…