Question Number 191714 by Mingma last updated on 29/Apr/23 Commented by AST last updated on 29/Apr/23 Answered by AST last updated on 29/Apr/23 $${Let}\:{AC}\:{intersect}\:{the}\:{circle}\:{from}\:{A} \\…
Question Number 191715 by Shlock last updated on 29/Apr/23 Answered by mr W last updated on 29/Apr/23 Commented by Shlock last updated on 29/Apr/23 Sir,canyon use the similarity of the right triangles? Can you give more explanation in details?…
Question Number 191676 by ajfour last updated on 28/Apr/23 Commented by ajfour last updated on 28/Apr/23 $${If}\:{first}\:{time}\:{dipped}\:{area}\:{of} \\ $$$${biscuit}\:{is}\:{A}_{\mathrm{1}} . \\ $$$${Find}\:\:{max}.\:{of}\:\frac{{A}_{\mathrm{2}} }{{A}_{\mathrm{1}} }.\:\: \\…
Question Number 191457 by Mingma last updated on 24/Apr/23 Commented by Tinku Tara last updated on 24/Apr/23 $$\mathrm{The}\:\mathrm{user}\:\mathrm{is}\:\mathrm{blocked}\:\mathrm{from}\:\mathrm{making}\:\mathrm{further} \\ $$$$\mathrm{posts}. \\ $$ Commented by mr…
Question Number 191458 by Mingma last updated on 24/Apr/23 Answered by a.lgnaoui last updated on 26/Apr/23 $$\bigtriangleup\mathrm{ACD}\:\:\mathrm{et}\:\bigtriangleup\mathrm{BCD}\:\:\mathrm{semblablables} \\ $$$$\mathrm{AB}\:\mathrm{tangdnte}\:\mathrm{au}\:\mathrm{quart}\:\mathrm{cercle}\:\mathrm{Rouge}\:\mathrm{en}\:\mathrm{D} \\ $$$$\mathrm{CD}=\mathrm{R2}\:\:\:\mathrm{CD}\bot\mathrm{AB}\:\:;\mathrm{BC}=\mathrm{R1}+\mathrm{R2} \\ $$$$\left(\mathrm{R1}\:\mathrm{rayon}\:\mathrm{du}\:\:\mathrm{quart}\:\:\mathrm{cercle}\:\mathrm{vert}.\right. \\ $$$$\mathrm{ABC}\:\mathrm{triangle}\:\:\mathrm{recrangle}\:\mathrm{en}\:\mathrm{C}…
Question Number 191335 by Mingma last updated on 23/Apr/23 Answered by mr W last updated on 23/Apr/23 $${radius}\:{of}\:{big}\:{quatercircle}\:={a} \\ $$$${radius}\:{of}\:{big}\:{semicircle}\:={b} \\ $$$${radius}\:{of}\:{small}\:{quatercircle}\:={c} \\ $$$${b}^{\mathrm{2}} +\left(\mathrm{2}{b}\right)^{\mathrm{2}}…
Question Number 191325 by Mingma last updated on 23/Apr/23 Answered by HeferH last updated on 23/Apr/23 $$\mathrm{16}\:=\:\mathrm{2}×\mathrm{a}\left(\mathrm{a}+\mathrm{b}\right) \\ $$$$\mathrm{Yellow}\:=\:\mathrm{8}\: \\ $$ Commented by Mingma last…
Question Number 191324 by Mingma last updated on 23/Apr/23 Answered by HeferH last updated on 23/Apr/23 $$\mathrm{r}\:=\:\mathrm{20} \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{15}×\mathrm{20}}{\mathrm{2}}\:+\:\frac{\mathrm{7}×\mathrm{24}}{\mathrm{2}}\:=\:\mathrm{150}+\mathrm{84}\:=\:\mathrm{234}\:\mathrm{u}^{\mathrm{2}} \: \\ $$ Commented by Mingma…
Question Number 191302 by Mingma last updated on 22/Apr/23 Answered by a.lgnaoui last updated on 22/Apr/23 $$\boldsymbol{\mathrm{S}}_{\mathrm{1}} =\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{OAB}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{AB}×\mathrm{h}_{\mathrm{1}} \:\:\:\:=\frac{\mathrm{CD}×\mathrm{h}_{\mathrm{1}} }{\mathrm{4}}\:\:\:\left(\mathrm{h}_{\mathrm{1}} =\mathrm{OH}_{\mathrm{1}} \right) \\ $$$$\:\:\:\frac{\mathrm{CD}×\mathrm{h}_{\mathrm{1}} }{\mathrm{4}}=\mathrm{8}\:\:\:\Rightarrow\boldsymbol{\mathrm{CD}}.\boldsymbol{\mathrm{h}}_{\mathrm{1}}…
Question Number 191276 by Mingma last updated on 22/Apr/23 Answered by mr W last updated on 22/Apr/23 $$\frac{{c}}{{a}}=\frac{{a}+{b}}{{b}−{a}} \\ $$$$\Rightarrow{c}=\frac{{a}\left({a}+{b}\right)}{{b}−{a}} \\ $$ Commented by Mingma…