Menu Close

Category: Geometry

Question-128652

Question Number 128652 by ajfour last updated on 09/Jan/21 Commented by ajfour last updated on 09/Jan/21 $${Find}\:{the}\:{radius}\:{of}\:{outer}\:{circle} \\ $$$${in}\:{terms}\:{of}\:{inradii}\:{R}\:{and}\:{r}\:{of} \\ $$$${two}\:{isosceles}\:{right}\:{angled}\: \\ $$$${triangles}\:\left({as}\:{shown},\:{see}\:{fig}\right). \\ $$…

Question-63108

Question Number 63108 by ajfour last updated on 29/Jun/19 Commented by ajfour last updated on 29/Jun/19 $${Find}\:{x}_{{A}} \:{in}\:{terms}\:{of}\:{a},{b},{s}\:. \\ $$$$\:\:\:\:\:{s}_{{min}} \leqslant{s}\leqslant{s}_{{max}} \:. \\ $$$${a},{b}\:{are}\:{parameters}\:{of}\:{ellipse} \\…

Question-128586

Question Number 128586 by ajfour last updated on 08/Jan/21 Commented by ajfour last updated on 08/Jan/21 $${Cubic}\:{curve}\:{eq}.\:\:{y}={x}^{\mathrm{3}} −{x}−{c} \\ $$$${Circle}\:{eq}.\:\:\:{x}^{\mathrm{2}} +\left({y}+{k}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$ Terms…

Question-62914

Question Number 62914 by Tawa1 last updated on 26/Jun/19 Answered by MJS last updated on 27/Jun/19 $$\mathrm{this}\:\mathrm{is}\:\mathrm{very}\:\mathrm{easy} \\ $$$${R}=\mathrm{15} \\ $$$${r}=\frac{\mathrm{15}}{\mathrm{3}}=\mathrm{5} \\ $$$$\mathrm{blue}={R}^{\mathrm{2}} \pi−\mathrm{7}{r}^{\mathrm{2}} \pi=\mathrm{50}\pi…

Question-62889

Question Number 62889 by Tawa1 last updated on 26/Jun/19 Answered by Kunal12588 last updated on 26/Jun/19 $$\left({a}\right){s}={t}^{\mathrm{2}} −\mathrm{5}{t}+\mathrm{6} \\ $$$${at}\:{initial}\:{pos}.\:{t}=\mathrm{0} \\ $$$${s}\left(\mathrm{0}\right)=\mathrm{6} \\ $$$$\therefore{the}\:{initial}\:{dist}.\:{of}\:{particle}\:{from}\:{O}\:{is}\:\mathrm{6}{m} \\…

Question-62850

Question Number 62850 by ajfour last updated on 25/Jun/19 Commented by ajfour last updated on 25/Jun/19 $${Find}\:{possible}\:{values}\:{of}\:{side}\:\boldsymbol{{s}}\:{of} \\ $$$${equilateral}\:\bigtriangleup{PQR}\:{in}\:{terms}\:{of} \\ $$$${a}\:{and}\:{b}.\left({for}\:{example}\:{let}\:{a}=\mathrm{4},\:{b}=\mathrm{3}\right). \\ $$ Answered by…

Question-62839

Question Number 62839 by ajfour last updated on 25/Jun/19 Commented by ajfour last updated on 25/Jun/19 $${let}\:\theta\:{be}\:{angle}\:{between}\:{b}\:{and}\: \\ $$$${extended}\:{c}\:. \\ $$$$\left({c}+{b}\mathrm{cos}\:\theta\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta={r}^{\mathrm{2}} \\…

Question-62821

Question Number 62821 by ajfour last updated on 25/Jun/19 Commented by ajfour last updated on 25/Jun/19 $${If}\:{the}\:{right}\:{circular}\:{cone}\:{contains} \\ $$$${half}\:{the}\:{spherical}\:{volume}\:{within} \\ $$$${its}\:{lateral}\:{surface},\:{find}\:\boldsymbol{\alpha}. \\ $$ Answered by…