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Category: Geometry

cube-ABCD-EFGH-side-4-cm-point-P-is-center-BF-and-Q-center-AD-distance-E-to-PQG-

Question Number 126609 by abdullahquwatan last updated on 22/Dec/20 $$\mathrm{cube}\:\mathrm{ABCD}.\mathrm{EFGH}\:{side}\:\mathrm{4}\:{cm}\:{point}\:{P}\:{is}\:{center}\:{BF}\:{and}\:{Q}\:{center}\:{AD}.\:{distance}\:{E}\:{to}\:{PQG} \\ $$ Answered by liberty last updated on 22/Dec/20 $${the}\:{equation}\:{of}\:{plane}\:{passes}\:{trought} \\ $$$${P}\left(\mathrm{4},\mathrm{4},\mathrm{2}\right),{Q}\left(\mathrm{2},\mathrm{0},\mathrm{0}\right),{G}\left(\mathrm{0},\mathrm{4},\mathrm{4}\right) \\ $$$$\Rightarrow\:\begin{vmatrix}{{x}−\mathrm{2}\:\:\:\:\:\:\:{y}\:\:\:\:\:\:\:\:\:\:{z}}\\{−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\:\:\mathrm{4}}\\{\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\end{vmatrix}=\:\mathrm{0} \\…

Question-60987

Question Number 60987 by ajfour last updated on 28/May/19 Commented by ajfour last updated on 28/May/19 $${Find}\:{the}\:{illuminated}\:{area}\:{of} \\ $$$${the}\:{inner}\:{curved}\:{surface}\:{of}\: \\ $$$${shown},\:{hollow}\:{open}\:{cylinder}. \\ $$$$\:\:\:\:\:\:\left[{where}\:\:{a}>\mathrm{2}{R}\left(\frac{{H}}{{h}}−\mathrm{1}\right)\right]\:\:\:\:\: \\ $$…

Question-191962

Question Number 191962 by ajfour last updated on 04/May/23 Answered by mr W last updated on 04/May/23 $$\left[\sqrt{\left({a}+{x}\right)^{\mathrm{2}} −\left({a}−{x}\right)^{\mathrm{2}} }−\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −\left({a}−{b}\right)^{\mathrm{2}} }\right]^{\mathrm{2}} =\left({b}+{x}\right)^{\mathrm{2}} −\left(\mathrm{2}{a}−{b}−{x}\right)^{\mathrm{2}} \\…

Question-60888

Question Number 60888 by ajfour last updated on 26/May/19 Commented by ajfour last updated on 27/May/19 $${If}\:{length}\:\boldsymbol{{l}}\:{encloses}\:{maximum} \\ $$$${such}\:{area},\:{find}\:{the}\:{area}. \\ $$$$\left({a}\right)\:{if}\:{l}\:{be}\:{circular} \\ $$$$\left({b}\right)\:{if}\:{shape}\:{of}\:{l}\:{not}\:{be}\:{given} \\ $$$$\left({someone}\:{please}\:{consider}\right.…

Question-191873

Question Number 191873 by cherokeesay last updated on 02/May/23 Answered by mehdee42 last updated on 03/May/23 $${let}\:\::\:\angle{RPS}={p}_{\mathrm{1}} \:\&\angle\:{QPS}={p}_{\mathrm{2}} \\ $$$${p}_{\mathrm{1}} =\mathrm{45}−{x}\:\:\:,\:\:{p}_{\mathrm{2}} =\mathrm{135}−\mathrm{3}{x} \\ $$$$\frac{{PS}}{{QS}}=\frac{{sin}\mathrm{3}{x}}{{sinp}_{\mathrm{2}} }=\frac{{sinx}}{{sinp}_{\mathrm{1}}…

Question-191856

Question Number 191856 by TUN last updated on 02/May/23 Answered by Subhi last updated on 02/May/23 $$ \\ $$$${put}\:{AD}={y} \\ $$$$\frac{{BD}}{{sin}\left(\mathrm{30}\right)}=\frac{{y}}{{sin}\left(\mathrm{10}\right)} \\ $$$${BD}=\frac{{sin}\left(\mathrm{30}\right)}{{sin}\left(\mathrm{10}\right)}{y}\:\:\:\:\:\:\:{by}\left[{sin}\:{law}\right] \\ $$$$\frac{{CD}}{{sin}\left(\mathrm{50}\right)}=\frac{{y}}{{sin}\left({x}\right)}\:.{hence}\:{CD}=\frac{{sin}\left(\mathrm{50}\right)}{{sin}\left({x}\right)}{y}…