Question Number 147232 by Tawa11 last updated on 19/Jul/21 Commented by Rasheed.Sindhi last updated on 19/Jul/21 $${Square}−{quarter}\:{circle}\:{of}\:{radius}\:\mathrm{9} \\ $$$$\mathrm{9}^{\mathrm{2}} −\frac{\pi.\mathrm{9}^{\mathrm{2}} }{\mathrm{4}}=\frac{\mathrm{324}−\mathrm{81}\pi}{\mathrm{4}} \\ $$ Commented by…
Question Number 147229 by Tawa11 last updated on 19/Jul/21 Answered by Olaf_Thorendsen last updated on 19/Jul/21 $$\mathrm{2}×\left(\mathrm{42}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}×\pi×\mathrm{42}^{\mathrm{2}} \right)\:=\:\mathrm{882}\left(\mathrm{4}−\pi\right) \\ $$ Commented by Tawa11 last…
Question Number 16140 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 18/Jun/17 Answered by ajfour last updated on 18/Jun/17 $${Area}_{\Delta{ABC}} \:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right)\: \\ $$ Commented by b.e.h.i.8.3.4.1.7@gmail.com…
Question Number 81649 by Power last updated on 14/Feb/20 Commented by Power last updated on 14/Feb/20 $$\mathrm{BDM}\:\:\mathrm{triangular}\:\mathrm{face}−? \\ $$ Commented by Power last updated on…
Question Number 16110 by vpawarksp@gmail.com last updated on 18/Jun/17 $$\mathrm{let}\:\mathrm{a}_{\mathrm{1}} >\mathrm{a}_{\mathrm{2}} >\mathrm{0}\:\mathrm{and}\:\mathrm{a}_{\mathrm{n}+\mathrm{1}} =\sqrt{\mathrm{a}_{\mathrm{n}} \mathrm{a}_{\mathrm{n}−\mathrm{1}\:\:\:} } \\ $$$$\mathrm{where}\:\mathrm{n}\:\mathrm{is}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{2}\: \\ $$$$\mathrm{Then} \\ $$$$\mathrm{The}\:\mathrm{sequence}\:\left\{\mathrm{a}_{\mathrm{2n}} \right\}\:\mathrm{is}\: \\ $$$$\left(\mathrm{1}\right)\:\mathrm{monotonic}\:\mathrm{increasing} \\…
Question Number 16108 by Joel577 last updated on 18/Jun/17 Commented by Joel577 last updated on 18/Jun/17 $$\mathrm{How}\:\mathrm{to}\:\mathrm{find}\:\angle{PAD}\:? \\ $$ Answered by ajfour last updated on…
Question Number 81610 by TawaTawa last updated on 14/Feb/20 Answered by MJS last updated on 14/Feb/20 $$\mathrm{2}\leqslant{x}\leqslant\mathrm{4} \\ $$$$\mathrm{let}\:{x}={t}^{\mathrm{2}} +\mathrm{2}\:\Rightarrow\:\mathrm{0}\leqslant{t}^{\mathrm{2}} \leqslant\mathrm{2} \\ $$$$\mathrm{3}{t}−\mathrm{12}\sqrt{\mathrm{2}−{t}^{\mathrm{2}} }+\mathrm{21}{t}^{\mathrm{2}} −\mathrm{40}+{t}\sqrt{\mathrm{2}−{t}^{\mathrm{2}}…
Question Number 16077 by Tinkutara last updated on 17/Jun/17 $$\mathrm{Let}\:{ABCDE}\:\mathrm{be}\:\mathrm{an}\:\mathrm{equiangular} \\ $$$$\mathrm{pentagon}\:\mathrm{whose}\:\mathrm{side}\:\mathrm{lengths}\:\mathrm{are} \\ $$$$\mathrm{rational}\:\mathrm{numbers}.\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{pentagon}\:\mathrm{is}\:\mathrm{regular}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 16074 by Tinkutara last updated on 17/Jun/17 $$\mathrm{Let}\:{K},\:{L},\:{M}\:\mathrm{and}\:{N}\:\mathrm{be}\:\mathrm{the}\:\mathrm{midpoints}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{sides}\:{AB},\:{BC},\:{CD}\:\mathrm{and}\:{DA}, \\ $$$$\mathrm{respectively},\:\mathrm{of}\:\mathrm{a}\:\mathrm{cyclic}\:\mathrm{quadrilateral} \\ $$$${ABCD}.\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{orthocenters} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{triangles}\:{AKN},\:{BKL},\:{CLM}\:\mathrm{and} \\ $$$${DMN}\:\mathrm{are}\:\mathrm{the}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{a} \\ $$$$\mathrm{parallelogram}. \\ $$ Terms…
Question Number 147144 by gsk2684 last updated on 18/Jul/21 $${prove}\:{that}\: \\ $$$${equation}\:{of}\:{a}\:{circle}\:{passing}\:{through} \\ $$$${the}\:{points}\:{of}\:{intersection}\:{of}\:{a}\:{circle} \\ $$$${S}=\mathrm{0}\:{and}\:{a}\:{line}\:{L}=\mathrm{0}\:{can}\:{be}\:{taken}\:{as} \\ $$$${S}+\lambda{L}=\mathrm{0}\:{where}\:\lambda\:{is}\:{a}\:{parameter} \\ $$ Answered by mr W last…