Question Number 187581 by HeferH last updated on 19/Feb/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 187557 by Rupesh123 last updated on 18/Feb/23 Answered by HeferH last updated on 19/Feb/23 Commented by HeferH last updated on 19/Feb/23 $${C}\:{is}\:{centroid}. \\…
Question Number 187549 by Rupesh123 last updated on 18/Feb/23 Answered by HeferH last updated on 18/Feb/23 $${The}\:{height}\:{of}\:\bigtriangleup{BIC}\:{is}\:{the}\:{inradius}. \\ $$$${Area}\:=\:{s}\centerdot{I} \\ $$$$\:\frac{\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)}}{{s}}\:=\:{I} \\ $$$$\:{s}\:=\:\frac{\mathrm{7}+\mathrm{9}+\mathrm{12}}{\mathrm{2}}\:=\:\mathrm{14} \\ $$$$\:{I}\:=\:\sqrt{\frac{\left(\mathrm{14}−\mathrm{9}\right)\left(\mathrm{14}−\mathrm{7}\right)\left(\mathrm{14}−\mathrm{12}\right)}{\mathrm{14}}}\:=\:\sqrt{\mathrm{5}}…
Question Number 187535 by ajfour last updated on 18/Feb/23 Commented by ajfour last updated on 18/Feb/23 $${Q}.\mathrm{187482} \\ $$ Answered by ajfour last updated on…
Question Number 187524 by sonukgindia last updated on 18/Feb/23 Answered by HeferH last updated on 18/Feb/23 Commented by HeferH last updated on 18/Feb/23 $$\:{Area}\:{of}\:{triangle}\:\left(\sqrt{\mathrm{20}},\sqrt{\mathrm{20}},\sqrt{\mathrm{8}}\right)=\:{A} \\…
Question Number 121985 by Dwaipayan Shikari last updated on 13/Nov/20 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n}\begin{pmatrix}{\mathrm{4n}}\\{\mathrm{2n}}\end{pmatrix}} \\ $$ Commented by Dwaipayan Shikari last updated on 13/Nov/20 $$\underset{{n}=\mathrm{1}} {\overset{\infty}…
Question Number 187482 by ajfour last updated on 17/Feb/23 Commented by ajfour last updated on 17/Feb/23 $${If}\:\:\:\:{cone}'{s}\:\:\:\:\:\frac{{R}}{{h}}={m},\:\:{green}\:{ball} \\ $$$${radius}={a}.\:\:{Find}\:{b}\:{the}\:{radii}\:{of} \\ $$$${upper}\:{balls}\:\left({equal}\right),\:{all}\:{in}\:{contact} \\ $$$${the}\:{way}\:{shown}. \\ $$…
Question Number 187473 by Rupesh123 last updated on 17/Feb/23 Answered by HeferH last updated on 17/Feb/23 Commented by HeferH last updated on 17/Feb/23 $$\:\sqrt{\left(\mathrm{5}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{1}}\:=\:\mathrm{7}\:…
Question Number 187452 by Rupesh123 last updated on 17/Feb/23 Answered by mr W last updated on 17/Feb/23 $${side}\:{length}\:{of}\:{octagon}\:=\:{a} \\ $$$${area}\:{of}\:{octagon}\:{A}_{\mathrm{8}} =\left({a}+\mathrm{2}×\frac{{a}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} =\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){a}^{\mathrm{2}} \\ $$$${black}\:{area}\:{A}_{{black}}…
Question Number 187451 by Rupesh123 last updated on 17/Feb/23 Answered by HeferH last updated on 17/Feb/23 Commented by HeferH last updated on 17/Feb/23 $$\mathrm{8}{x}\:=\:\mathrm{60}° \\…