Question Number 175172 by cherokeesay last updated on 21/Aug/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 44103 by Raj Singh last updated on 21/Sep/18 Commented by Raj Singh last updated on 21/Sep/18 $${in}\:{this}\:{figure}\:{PQ}\mid\mid{BA}\:{and}\:{PR}\mid\mid{CA} \\ $$$${if}\:{PD}=\mathrm{12}\:{find}\:{BD}×{CD} \\ $$ Answered by…
Question Number 175166 by cherokeesay last updated on 21/Aug/22 Answered by ajfour last updated on 22/Aug/22 $${let}\:{centre}\:{of}\:{required}\:{circle} \\ $$$${be}\:\:\left({h},{R}\right) \\ $$$${h}=−\frac{{r}}{\mathrm{2}\sqrt{\mathrm{2}}}=−\mathrm{4} \\ $$$$\Rightarrow\:\:{r}=\mathrm{8}\sqrt{\mathrm{2}} \\ $$$$\left({h}−\frac{{r}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}}…
Question Number 175149 by cherokeesay last updated on 20/Aug/22 Answered by som(math1967) last updated on 21/Aug/22 Commented by som(math1967) last updated on 21/Aug/22 $${AB}=\mathrm{2}{R}\:\:,{BC}={R} \\…
Question Number 109591 by 1549442205PVT last updated on 25/Aug/20 $$\mathrm{An}\:\mathrm{isosceles}\:\mathrm{AEF}\:\mathrm{is}\:\mathrm{inscribed}\:\mathrm{into}\:\mathrm{a} \\ $$$$\mathrm{square}\:\mathrm{ABCD}\:\mathrm{such}\:\mathrm{that}\:\mathrm{pointE}\:\mathrm{is}\:\mathrm{on} \\ $$$$\mathrm{side}\:\mathrm{BC},\mathrm{point}\:\mathrm{F}\:\mathrm{is}\:\mathrm{on}\:\mathrm{side}\:\mathrm{CDand}\:\mathrm{AE}=\mathrm{EF}. \\ $$$$\mathrm{Knownthat}\:\mathrm{tan}\widehat {\mathrm{AEF}}=\mathrm{2}.\mathrm{Find}\:\mathrm{tan}\widehat {\mathrm{FEC}} \\ $$ Commented by 1549442205PVT last updated…
Question Number 44017 by ajfour last updated on 20/Sep/18 Commented by ajfour last updated on 20/Sep/18 $${A}\:{cone}\:{is}\:{cut}\:{by}\:{a}\:{plane}\:\:{at}\:{an} \\ $$$${angle}\:\theta\:{to}\:{its}\:{base}\:{and}\:{passes} \\ $$$${through}\:{a}\:{point}\:{at}\:{the}\:{base}\:{of} \\ $$$${cone}.\:{Find}\:{ratio}\:{of}\:\boldsymbol{{upper}}\:\boldsymbol{{cone}} \\ $$$$\boldsymbol{{volume}}\:\boldsymbol{{to}}\:\boldsymbol{{lower}}\:\boldsymbol{{cone}}\:\boldsymbol{{volume}}.…
Question Number 43980 by Tawa1 last updated on 19/Sep/18 Answered by MJS last updated on 19/Sep/18 $${a}=\mathrm{2}+\mathrm{5}=\mathrm{7} \\ $$$$\alpha=\mathrm{60}° \\ $$$$ \\ $$$${b}=\mathrm{2}+{p} \\ $$$${c}=\mathrm{5}+{p}…
Question Number 109508 by Study last updated on 24/Aug/20 Commented by som(math1967) last updated on 24/Aug/20 $$\:\frac{\mathrm{1}}{\mathrm{4}}\pi×\mathrm{2}^{\mathrm{2}} −\left\{\frac{\mathrm{1}}{\mathrm{2}}\pi×\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1}×\mathrm{1}\right\} \\ $$$$\left(\pi−\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{cm}^{\mathrm{2}} \\ $$ Answered by…
Question Number 175045 by 2kdw last updated on 17/Aug/22 $${In}\:{the}\:{figure}\:{ABCD}\:{is}\:{an}\:{square}\:{and} \\ $$$${BM}={MC}.\:{If}\:{the}\:{area}\:{of}\:{PCD}=\mathrm{14}{u}.\: \\ $$$${What}\:{is}\:{the}\:{value}\:{of}: \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:{Area}\:{of}\:{ABC}; \\ $$$$\left(\mathrm{2}\right)\:{Area}\:{of}\:\:{ABMP}; \\ $$$$\left(\mathrm{3}\right)\:{The}\:{area}\:{of}\:{ABCD} \\ $$$$ \\…
Question Number 43944 by Tawa1 last updated on 18/Sep/18 $$\mathrm{If}\:\mathrm{the}\:\mathrm{points}\:\left(−\mathrm{3},\:\mathrm{5}\right)\:,\:\left(\mathrm{4},\:−\mathrm{2}\right)\:\mathrm{and}\:\left(\mathrm{6},\:\mathrm{2}\right)\:\mathrm{are}\:\mathrm{the}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}. \\ $$$$\left(\mathrm{i}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{perpendicular}\:\mathrm{bisector}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sides} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{coordinate}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circumcenter}.\:\left(\mathrm{The}\:\mathrm{circumcenter}\:\mathrm{of}\:\mathrm{a}\right. \\ $$$$\mathrm{triangle}\:\mathrm{is}\:\mathrm{the}\:\mathrm{point}\:\mathrm{of}\:\mathrm{intersection}\:\mathrm{of}\:\mathrm{the}\:\mathrm{perpendicular}\:\mathrm{bisector}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{side} \\ $$$$\left(\mathrm{iii}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circumcircle}. \\ $$ Answered by tanmay.chaudhury50@gmail.com…