Question Number 112467 by Aina Samuel Temidayo last updated on 08/Sep/20 Answered by 1549442205PVT last updated on 08/Sep/20 $$\mathrm{Side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}\:\mathrm{equal}\:\mathrm{to}\:\frac{\mathrm{a}}{\:\sqrt{\mathrm{2}}} \\ $$$$\mathrm{Width}\:\mathrm{of}\:\mathrm{rectangle}\:\mathrm{equal}\:\mathrm{to}\:\frac{\mathrm{2a}}{\:\sqrt{\mathrm{2}}}=\mathrm{a}\sqrt{\mathrm{2}} \\ $$$$\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{rectangle}\:\mathrm{is}\:\mathrm{S}_{\mathrm{1}} =\mathrm{2a}.\mathrm{a}\sqrt{\mathrm{2}}=\mathrm{2a}^{\mathrm{2}} \sqrt{\mathrm{2}}\:\left(\mathrm{1}\right)…
Question Number 177578 by BaliramKumar last updated on 07/Oct/22 Commented by BaliramKumar last updated on 07/Oct/22 area of green colour Commented by som(math1967) last updated on 07/Oct/22 $$\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}}\:−\mathrm{1}\:?…
Question Number 111799 by ajfour last updated on 05/Sep/20 Commented by mr W last updated on 05/Sep/20 $${sir},\:{not}\:{possible}:\:{A}={B}={C} \\ $$ Commented by ajfour last updated…
Question Number 111734 by Aina Samuel Temidayo last updated on 04/Sep/20 $$\mathrm{A}\:\mathrm{chord}\:\mathrm{which}\:\mathrm{is}\:\mathrm{a}\:\mathrm{perpendicular} \\ $$$$\mathrm{bisector}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{length}\:\mathrm{18cm}\:\mathrm{in}\:\mathrm{a} \\ $$$$\mathrm{circle},\:\mathrm{has}\:\mathrm{length}. \\ $$$$ \\ $$ Answered by Rasheed.Sindhi last updated…
Question Number 177253 by BaliramKumar last updated on 02/Oct/22 $$ \\ $$The base of a right pyramid is a hexagon of side 16 cm, and…
Question Number 46085 by peter frank last updated on 20/Oct/18 Commented by maxmathsup by imad last updated on 21/Oct/18 $${let}\:{S}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:{cos}^{{n}} \theta\:{cos}\left({n}\theta\right)\:={Re}\left(\:\sum_{{n}=\mathrm{1}} ^{\infty} \:{e}^{{in}\theta}…
Question Number 177092 by BaliramKumar last updated on 30/Sep/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 177045 by Ar Brandon last updated on 30/Sep/22 Commented by a.lgnaoui last updated on 30/Sep/22 $${DC}\:;\:\:{DM};\:\:{MK};{and}\:{AB} \\ $$ Commented by a.lgnaoui last updated…
Question Number 177044 by Ar Brandon last updated on 30/Sep/22 Answered by a.lgnaoui last updated on 30/Sep/22 $$\Sigma\:{angles}\:{interieures}=\Sigma{angles}\:{exterieures} \\ $$$$\frac{\measuredangle{B}}{\mathrm{2}}+\measuredangle{KPD}+\frac{\measuredangle{D}}{\mathrm{2}}=\mathrm{90} \\ $$$$\measuredangle{EDD}=\frac{\measuredangle{D}}{\mathrm{2}}\:\:\:\left({PK}\mid\mid\:{ED}\right) \\ $$$$\:\frac{\measuredangle{B}}{\mathrm{2}}+\measuredangle{D}=\mathrm{90} \\…
Question Number 177047 by Ar Brandon last updated on 30/Sep/22 Answered by som(math1967) last updated on 30/Sep/22 $${cosD}=\frac{\mathrm{7}^{\mathrm{2}} +{DC}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} }{\mathrm{2}×\mathrm{7}{DC}}=\frac{\mathrm{24}+{DC}^{\mathrm{2}} }{\mathrm{14}{DC}} \\ $$$${again}\:\:{cosD}=\frac{\mathrm{7}^{\mathrm{2}} +{DB}^{\mathrm{2}}…