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Category: Mensuration

Question-191756

Question Number 191756 by Shlock last updated on 30/Apr/23 Answered by aleks041103 last updated on 30/Apr/23 $${side}\:{of}\:{whole}\:{square}\:\rightarrow\:{a} \\ $$$${side}\:{blue}/{red}/{yellow}\:{square}\:\rightarrow\:{b}/{r}/{y} \\ $$$${b}^{\mathrm{2}} ={r}^{\mathrm{2}} +\left({a}−\mathrm{2}{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} +{y}^{\mathrm{2}}…

Question-191458

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-191335

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-191324

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…