Question Number 190564 by Best1 last updated on 06/Apr/23
Answered by a.lgnaoui last updated on 06/Apr/23
![•1a perimetere(p) [ (2,8×2)+1,6]+((1,6×π)/2) =7,1+((4π)/5)⇒ p=9,61m •2a Area =(2,8×1,6 ) +π(0,8)^2 A=6,49 m^2 1b perimetre p=(68×2)+25𝛑 cm 2b Area =68×50+π×(25)^2 =5 363,5 cm^2 =5,364m^2 (to follow for the rest...)](https://www.tinkutara.com/question/Q190571.png)
$$\bullet\mathrm{1}\boldsymbol{{a}}\:\:\:\:{perimetere}\left(\boldsymbol{{p}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\:\left(\mathrm{2},\mathrm{8}×\mathrm{2}\right)+\mathrm{1},\mathrm{6}\right]+\frac{\mathrm{1},\mathrm{6}×\pi}{\mathrm{2}} \\ $$$$\:\:\:=\mathrm{7},\mathrm{1}+\frac{\mathrm{4}\pi}{\mathrm{5}}\Rightarrow\:\:\:\:\boldsymbol{{p}}=\mathrm{9},\mathrm{61}{m} \\ $$$$\:\bullet\mathrm{2}\boldsymbol{{a}}\:\:\:\:\boldsymbol{{A}}{rea}\: \\ $$$$\:=\left(\mathrm{2},\mathrm{8}×\mathrm{1},\mathrm{6}\:\right)\:\:+\pi\left(\mathrm{0},\mathrm{8}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{A}}=\mathrm{6},\mathrm{49}\:\:{m}^{\mathrm{2}} \\ $$$$\mathrm{1}\boldsymbol{{b}}\:\:\:\boldsymbol{{p}}{erimetre} \\ $$$$\:\:\:\:\:\boldsymbol{{p}}=\left(\mathrm{68}×\mathrm{2}\right)+\mathrm{25}\boldsymbol{\pi}\:\:\:{cm} \\ $$$$\mathrm{2}\boldsymbol{{b}}\:\:\boldsymbol{{A}}{rea}\:=\mathrm{68}×\mathrm{50}+\pi×\left(\mathrm{25}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{5}\:\mathrm{363},\mathrm{5}\:{cm}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{5},\mathrm{364}{m}^{\mathrm{2}} \\ $$$$\left({to}\:{follow}\:{for}\:{the}\:{rest}…\right) \\ $$
Answered by a.lgnaoui last updated on 06/Apr/23
![e) let A_1 =Area C_1 [R_1 =15cm] A_2 : Area C_2 [R_2 =5cm ] A_3 : Area C3 [R_3 =R_2 ] A_4 : Area C_4 [R_4 =R_2 ] A=((π(R_1 ^2 −R_2 ^2 ))/2)+πR_2 ^2 =((πR_1 ^2 )/2)+((πR_2 ^2 )/2)=(π/2)(R_1 ^2 +R_2 ^2 ) =(π/2)(15^2 +5^2 ) Area=392,7](https://www.tinkutara.com/question/Q190582.png)
$$\left.{e}\right)\:\:{let}\:\:{A}_{\mathrm{1}} ={Area}\:\:{C}_{\mathrm{1}} \:\:\:\:\:\left[{R}_{\mathrm{1}} =\mathrm{15}{cm}\right] \\ $$$$\:{A}_{\mathrm{2}} :\:\:{Area}\:{C}_{\mathrm{2}} \:\:\left[{R}_{\mathrm{2}} =\mathrm{5}{cm}\:\:\right] \\ $$$${A}_{\mathrm{3}} :\:\:{Area}\:{C}\mathrm{3}\:\:\left[{R}_{\mathrm{3}} ={R}_{\mathrm{2}} \right] \\ $$$${A}_{\mathrm{4}} :\:\:\:{Area}\:{C}_{\mathrm{4}} \:\:\left[{R}_{\mathrm{4}} ={R}_{\mathrm{2}} \right] \\ $$$${A}=\frac{\pi\left({R}_{\mathrm{1}} ^{\mathrm{2}} −{R}_{\mathrm{2}} ^{\mathrm{2}} \right)}{\mathrm{2}}+\pi{R}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\:\:\:=\frac{\pi{R}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}}+\frac{\pi{R}_{\mathrm{2}} ^{\mathrm{2}} }{\mathrm{2}}=\frac{\pi}{\mathrm{2}}\left({R}_{\mathrm{1}} ^{\mathrm{2}} +{R}_{\mathrm{2}} ^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:=\frac{\pi}{\mathrm{2}}\left(\mathrm{15}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} \right)\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{Area}}=\mathrm{392},\mathrm{7} \\ $$$$\:\: \\ $$