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Category: Mensuration

Question-177044

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-177047

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-177046

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) \\…