Question Number 47407 by ajfour last updated on 09/Nov/18 Answered by ajfour last updated on 10/Nov/18 $${let}\:{BD}\:=\:{x}\:,\:\angle{A}\:=\:\phi,\:\angle{C}\:=\:\theta \\ $$$${S}_{{ABCD}} \:=\:\frac{{ab}\mathrm{sin}\:\theta}{\mathrm{2}}+\frac{{cd}\mathrm{sin}\:\phi}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} =\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\mathrm{cos}\:\theta…
Question Number 47382 by ajfour last updated on 09/Nov/18 Commented by ajfour last updated on 09/Nov/18 $${The}\:{sides}\:{of}\:{inclined}\:{triangle} \\ $$$${are}\:{p},\:{q},\:{and}\:{r}. \\ $$$${Find}\:{P}\:{G}=\:{h}\:. \\ $$$${The}\:{sides}\:{of}\:{the}\:{prism}\:{box}\:{are} \\ $$$${squares},\:{and}\:{base}\:{equilateral}.…
Question Number 178450 by mr W last updated on 16/Oct/22 Commented by mr W last updated on 16/Oct/22 $${find}\:{the}\:{area}\:{of}\:{shaded}\:{rectangle}. \\ $$ Commented by cortano1 last…
Question Number 47365 by Raj Singh last updated on 09/Nov/18 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 47350 by behi83417@gmail.com last updated on 08/Nov/18 Commented by behi83417@gmail.com last updated on 08/Nov/18 $${ABDC},{is}\:{a}:\:\mathrm{2}×\mathrm{3}\:{rectangle}.{circle} \\ $$$${with}\:{A},{as}\:{center}\:{and}\:{AD},{as}\:{radius}, \\ $$$${meets}\:{extention}\:{of}\:{BC}\:{at}\::{E}\:{and}\:{F}. \\ $$$$….\boldsymbol{\mathrm{EF}}=? \\ $$…
Question Number 47349 by behi83417@gmail.com last updated on 08/Nov/18 Answered by behi83417@gmail.com last updated on 09/Nov/18 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 47282 by behi83417@gmail.com last updated on 07/Nov/18 Answered by MrW3 last updated on 07/Nov/18 Commented by MrW3 last updated on 08/Nov/18 $${AB}={c} \\…
Question Number 47250 by Raj Singh last updated on 07/Nov/18 Answered by MJS last updated on 07/Nov/18 $$\mathrm{surface}\:\mathrm{of}\:\mathrm{cuboid} \\ $$$$\mathrm{2}×\left(\mathrm{15}×\mathrm{10}+\mathrm{15}×\mathrm{5}+\mathrm{10}×\mathrm{5}\right)=\mathrm{550} \\ $$$$\mathrm{surface}\:\mathrm{of}\:\mathrm{cylinder}\:\left(\mathrm{without}\:\mathrm{bottom\&top}\right) \\ $$$$\mathrm{2}×\mathrm{3}.\mathrm{5}\pi×\mathrm{5}=\mathrm{35}\pi \\…
Question Number 178323 by mr W last updated on 15/Oct/22 Commented by mr W last updated on 15/Oct/22 $${find}\:{shaded}\:{area} \\ $$ Commented by cortano1 last…
Question Number 47243 by MrW3 last updated on 06/Nov/18 Answered by MJS last updated on 07/Nov/18 $$\mathrm{left}\:\mathrm{picture} \\ $$$$\frac{\mathrm{8}}{{r}}=\frac{{H}}{{R}}\:\Rightarrow\:{r}=\frac{\mathrm{8}{R}}{{H}} \\ $$$${v}_{{blue}} =\frac{\pi{R}^{\mathrm{2}} \left({H}^{\mathrm{2}} +\mathrm{8}{H}+\mathrm{64}\right)\left({H}−\mathrm{8}\right)}{\mathrm{3}{H}^{\mathrm{2}} }…