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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}} \\ $$$${again}\:\:{cosD}=\frac{\mathrm{7}^{\mathrm{2}} +{DB}^{\mathrm{2}}…
Question Number 177046 by Ar Brandon last updated on 30/Sep/22 Answered by som(math1967) last updated on 30/Sep/22 $${Area}\left(\bigtriangleup{ADE}+\bigtriangleup{BEC}\right)=\frac{\mathrm{1}}{\mathrm{2}}{AreaABCD} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{x}}{\mathrm{2}}\:\left({let}\right) \\ $$$${Area}\bigtriangleup{AEB}=\frac{\mathrm{1}}{\mathrm{2}}{ABCD}=\frac{{x}}{\mathrm{2}} \\ $$$${Area}\left({ADEP}+\bigtriangleup{ADE}\right) \\…
Question Number 177043 by Ar Brandon last updated on 30/Sep/22 Answered by a.lgnaoui last updated on 30/Sep/22 $$\left.\mathrm{A}\right){only}\:\mathrm{I} \\ $$ Commented by Rasheed.Sindhi last updated…
Question Number 176973 by Ar Brandon last updated on 28/Sep/22 Commented by Ar Brandon last updated on 28/Sep/22 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{perimeter}\:\mathrm{of}\:\mathrm{triangle}. \\ $$ Terms of Service Privacy…
Question Number 176946 by Ar Brandon last updated on 28/Sep/22 Answered by mr W last updated on 29/Sep/22 Commented by mr W last updated on…
Question Number 176933 by Ar Brandon last updated on 28/Sep/22 Answered by mr W last updated on 28/Sep/22 $${AC}+{BC}>{AB}=\mathrm{7} \\ $$$$\left({AC}+{BC}\right)_{{min}} =\mathrm{8} \\ $$$${p}_{{min}} =\left({AC}+{BC}+{AB}\right)_{{min}}…
Question Number 176936 by Ar Brandon last updated on 28/Sep/22 Answered by mr W last updated on 28/Sep/22 $$\frac{{x}}{\mathrm{sin}\:\left(\mathrm{15}+\mathrm{30}\right)}=\frac{\mathrm{4}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}{\mathrm{sin}\:\mathrm{15}} \\ $$$${x}=\frac{\mathrm{4}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)×\mathrm{4}}{\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right)}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}=\mathrm{8}\:{cm}\:\checkmark \\ $$ Commented by…