Question Number 69643 by ahmadshahhimat775@gmail.com last updated on 26/Sep/19 Answered by MJS last updated on 26/Sep/19 $${l}+{w}=\mathrm{8} \\ $$$${l}^{\mathrm{2}} +{w}^{\mathrm{2}} =\mathrm{6}^{\mathrm{2}} \\ $$$$\Rightarrow\:{l}=\mathrm{4}\pm\sqrt{\mathrm{2}}\wedge{w}=\mathrm{4}\mp\sqrt{\mathrm{2}} \\ $$$$\mathrm{perimeter}\:{P}=\mathrm{10}+\mathrm{3}\pi…
Question Number 4080 by Rasheed Soomro last updated on 27/Dec/15 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{inside}\:\mathrm{a}\:\mathrm{square},\:\mathrm{a}\:\mathrm{semi}-\mathrm{circle}, \\ $$$$\mathrm{which}\:\mathrm{touches}\:\mathrm{all}\:\mathrm{the}\:\mathrm{four}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{square},\:\mathrm{is}\:\mathrm{possible}\:\mathrm{with}\:\boldsymbol{\mathrm{ruler}}\:\mathrm{and}\:\boldsymbol{\mathrm{compass}}. \\ $$ Commented by Rasheed Soomro last updated on 29/Dec/15…
Question Number 4077 by Rasheed Soomro last updated on 28/Dec/15 $$\mathrm{For}\:\mathrm{two}\:\boldsymbol{\mathrm{co}}-\boldsymbol{\mathrm{planer}}\:\mathrm{circles}\:\mathrm{to}\:\mathrm{be}\:\mathrm{tangent} \\ $$$$\:\mathrm{necessary}\:\mathrm{and}\:\mathrm{sufficient}\:\mathrm{condition}\:\mathrm{is}, \\ $$$$\mathrm{I}\:\mathrm{think} \\ $$$$\:\:\:\:\:\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{the}\:\mathrm{centers}\:\mathrm{of}\: \\ $$$$\:\:\:\:\mathrm{circles}\:\:\mathrm{must}\:\mathrm{be}\:\mathrm{equal}\:\mathrm{to}\:\boldsymbol{\mathrm{r}}_{\mathrm{1}} +\boldsymbol{\mathrm{r}}_{\mathrm{2}} \:\mathrm{or}\:\mid\boldsymbol{\mathrm{r}}_{\mathrm{1}} −\boldsymbol{\mathrm{r}}_{\mathrm{2}} \mid\:, \\ $$$$\:\:\:\:\:\mathrm{where}\:\:\boldsymbol{\mathrm{r}}_{\mathrm{1}}…
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Question Number 69609 by TawaTawa last updated on 25/Sep/19 Commented by Prithwish sen last updated on 25/Sep/19 $$\mathrm{I}\:\mathrm{think} \\ $$$$\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{inner}\:\mathrm{section}\:=\:\pi\left(\mathrm{75}\right)^{\mathrm{2}} \mathrm{sq}.\:\mathrm{cm} \\ $$$$\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{outer}\:\mathrm{section}\:=\:\pi\left(\mathrm{100}−\mathrm{75}\right)\left(\mathrm{100}+\mathrm{75}\right) \\ $$$$=\:\pi×\mathrm{25}×\mathrm{175}\:\mathrm{sq}.\:\mathrm{cm}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…
Question Number 69608 by TawaTawa last updated on 25/Sep/19 Commented by Prithwish sen last updated on 27/Sep/19 $$\left.\boldsymbol{\mathrm{a}}\right)\:\mid\boldsymbol{\mathrm{AB}}\mid=\mathrm{2}\sqrt{\mathrm{2}}=\:\mathrm{2}.\mathrm{8m} \\ $$$$\left.\boldsymbol{\mathrm{b}}\right)\:\boldsymbol{\mathrm{perimeter}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{ALB}}\:=\frac{\mathrm{2}×\pi×\mathrm{2}×\mathrm{270}}{\mathrm{360}}\mathrm{m}=\:\mathrm{9}.\mathrm{4m} \\ $$ Commented by TawaTawa…
Question Number 135128 by bramlexs22 last updated on 10/Mar/21 Commented by mr W last updated on 10/Mar/21 $${is}\:{ABCD}\:{a}\:{square}? \\ $$ Commented by mr W last…
Question Number 4033 by Rasheed Soomro last updated on 27/Dec/15 $$\:\:\:\:\mathrm{One}\:\mathrm{circle}\:\mathrm{in}\:\mathrm{a}\:\mathrm{plane}\:\mathrm{can}\:\mathrm{produce}\:\mathrm{one}\:\:\mathrm{closed} \\ $$$$\mathrm{region}\:\mathrm{at}\:\mathrm{most}\left(\mathrm{It}\:\mathrm{produces}\:\mathrm{one}\:\mathrm{closed}\:\mathrm{region}\:\right. \\ $$$$\left.\mathrm{at}\:\mathrm{least}\right).\mathrm{Two}\:\:\mathrm{circles}\:\mathrm{in}\:\mathrm{a}\:\mathrm{plane}\:\:\mathrm{can}\:\mathrm{produce} \\ $$$$\mathrm{at}\:\mathrm{most}\:\mathrm{three}\:\:\mathrm{regions}\left(\mathrm{They}\:\mathrm{produce}\:\mathrm{at}\:\mathrm{least}\right. \\ $$$$\left.\mathrm{two}\:\:\mathrm{regions}\right).\mathrm{Three}\:\mathrm{circles}\:\mathrm{can}\:\mathrm{produce}\:\mathrm{seven} \\ $$$$\mathrm{closed}\:\mathrm{regions}\:\mathrm{at}\:\mathrm{most}\left(\mathrm{They}\:\mathrm{produce}\:\mathrm{three}\right. \\ $$$$\left.\mathrm{closed}\:\mathrm{regions}\:\mathrm{at}\:\mathrm{least}\right). \\ $$$$…
Question Number 69545 by TawaTawa last updated on 24/Sep/19 Commented by mr W last updated on 24/Sep/19 $$\mathrm{30}×\mathrm{20}−\mathrm{28}×\mathrm{18}=\mathrm{96}\:{m}^{\mathrm{2}} \\ $$$$\mathrm{96}×\mathrm{0}.\mathrm{08}=\mathrm{7}.\mathrm{68}\:{m}^{\mathrm{3}} \\ $$$$\frac{\mathrm{7}.\mathrm{68}}{\mathrm{1}}=\mathrm{7}.\mathrm{68}\:\Rightarrow\:\mathrm{8}\:\:{bags} \\ $$$$\mathrm{8}×\mathrm{600}=\mathrm{4800}\:{N} \\…
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