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Category: Geometry

Question-22661

Question Number 22661 by ajfour last updated on 21/Oct/17 Commented by ajfour last updated on 21/Oct/17 $${Q}.\mathrm{22657}\:\left({solution}\:;\:{here}\:\:{becoz}\right. \\ $$$$\:\:\:\:\:\left({images}\:{from}\:{my}\:{cellphone}\right. \\ $$$$\:\:\:\:\:{gets}\:{uploaded}\:{only}\:{as}\:{new}\: \\ $$$$\:\:\:\:\:\:{question},\:{and}\:{rarely}\:{as}\:{ans} \\ $$$$\left.\:\:\:\:\:\:{or}\:{comment}\right).…

Question-88188

Question Number 88188 by mr W last updated on 08/Apr/20 Commented by mr W last updated on 08/Apr/20 $${Given}:\:{triangle}\:{ABC}\:{with}\:{side}\:{lengthes} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a},\:{b},\:{c}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{P}\:{is}\:{midpoint}\:{of}\:{BC}. \\ $$$${Find}:\:\:\:{perimeter}\:{of}\:{inscribed}\:{triangle}…

x-1-2-x-1-2-x-1-3-dx-

Question Number 22545 by vajpaithegrate@gmail.com last updated on 20/Oct/17 $$\int\frac{\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{2}}} −\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{3}}} }\mathrm{dx}= \\ $$$$ \\ $$ Answered by $@ty@m last updated on 20/Oct/17 $${Its}\:{similar}\:{to}\:{Q}.\:{No}.\:\mathrm{20540}…

Question-22525

Question Number 22525 by ajfour last updated on 19/Oct/17 Commented by ajfour last updated on 19/Oct/17 $${A}\:{hemisphere}\:{is}\:{inscribed}\:{in}\:{a} \\ $$$${regular}\:{tetrahedron}\:,{of}\:{edge}\:\boldsymbol{{a}}\:, \\ $$$${such}\:{that}\:{three}\:{faces}\:{of}\:{tetrahedron} \\ $$$${are}\:{tangent}\:{to}\:{its}\:{spherical} \\ $$$${surface},\:{and}\:{the}\:{fourth}\:{serves}…

In-a-quadrilateral-ABCD-it-is-given-that-AB-is-parallel-to-CD-and-the-diagonals-AC-and-BD-are-perpendicular-to-each-other-Show-that-a-AD-BC-AB-CD-b-AD-BC-AB-CD-

Question Number 22515 by Tinkutara last updated on 19/Oct/17 $$\mathrm{In}\:\mathrm{a}\:\mathrm{quadrilateral}\:{ABCD},\:\mathrm{it}\:\mathrm{is}\:\mathrm{given} \\ $$$$\mathrm{that}\:{AB}\:\mathrm{is}\:\mathrm{parallel}\:\mathrm{to}\:{CD}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{diagonals}\:{AC}\:\mathrm{and}\:{BD}\:\mathrm{are}\:\mathrm{perpendicular} \\ $$$$\mathrm{to}\:\mathrm{each}\:\mathrm{other}. \\ $$$$\mathrm{Show}\:\mathrm{that} \\ $$$$\left(\mathrm{a}\right)\:{AD}.{BC}\:\geqslant\:{AB}.{CD}; \\ $$$$\left(\mathrm{b}\right)\:{AD}\:+\:{BC}\:\geqslant\:{AB}\:+\:{CD}. \\ $$ Terms…