Question and Answers Forum

All Questions      Topic List

Mensuration Questions

Previous in All Question      Next in All Question      

Previous in Mensuration      Next in Mensuration      

Question Number 81188 by jagoll last updated on 10/Feb/20

Commented by jagoll last updated on 10/Feb/20

dear mr W

$${dear}\:{mr}\:{W} \\ $$

Answered by mr W last updated on 10/Feb/20

Commented by mr W last updated on 10/Feb/20

WAY 1 R_A =2+1=3 AB=3−2=1 AC=3−1=2 BD=2+R_D AD=3−R_D CD=1+R_D cos α=((2^2 +(3−R_D )^2 −(1+R_D )^2 )/(2×2(3−R_D ))) cos β=((1^2 +(3−R_D )^2 −(2+R_D )^2 )/(2(3−R_D ))) α+β=180° cos α=−cos β ((2^2 +(3−R_D )^2 −(1+R_D )^2 )/(2×2(3−R_D )))=−((1^2 +(3−R_D )^2 −(2+R_D )^2 )/(2(3−R_D ))) ((4+4(2−2R_D ))/2)=−((1+5(1−2R_D ))/1) 6=7R_D ⇒R_D =(6/7)

$${WAY}\:\:\mathrm{1} \\ $$$${R}_{{A}} =\mathrm{2}+\mathrm{1}=\mathrm{3} \\ $$$${AB}=\mathrm{3}−\mathrm{2}=\mathrm{1} \\ $$$${AC}=\mathrm{3}−\mathrm{1}=\mathrm{2} \\ $$$${BD}=\mathrm{2}+{R}_{{D}} \\ $$$${AD}=\mathrm{3}−{R}_{{D}} \\ $$$${CD}=\mathrm{1}+{R}_{{D}} \\ $$$$\mathrm{cos}\:\alpha=\frac{\mathrm{2}^{\mathrm{2}} +\left(\mathrm{3}−{R}_{{D}} \right)^{\mathrm{2}} −\left(\mathrm{1}+{R}_{{D}} \right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{2}\left(\mathrm{3}−{R}_{{D}} \right)} \\ $$$$\mathrm{cos}\:\beta=\frac{\mathrm{1}^{\mathrm{2}} +\left(\mathrm{3}−{R}_{{D}} \right)^{\mathrm{2}} −\left(\mathrm{2}+{R}_{{D}} \right)^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{3}−{R}_{{D}} \right)} \\ $$$$\alpha+\beta=\mathrm{180}° \\ $$$$\mathrm{cos}\:\alpha=−\mathrm{cos}\:\beta \\ $$$$\frac{\mathrm{2}^{\mathrm{2}} +\left(\mathrm{3}−{R}_{{D}} \right)^{\mathrm{2}} −\left(\mathrm{1}+{R}_{{D}} \right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{2}\left(\mathrm{3}−{R}_{{D}} \right)}=−\frac{\mathrm{1}^{\mathrm{2}} +\left(\mathrm{3}−{R}_{{D}} \right)^{\mathrm{2}} −\left(\mathrm{2}+{R}_{{D}} \right)^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{3}−{R}_{{D}} \right)} \\ $$$$\frac{\mathrm{4}+\mathrm{4}\left(\mathrm{2}−\mathrm{2}{R}_{{D}} \right)}{\mathrm{2}}=−\frac{\mathrm{1}+\mathrm{5}\left(\mathrm{1}−\mathrm{2}{R}_{{D}} \right)}{\mathrm{1}} \\ $$$$\mathrm{6}=\mathrm{7}{R}_{{D}} \\ $$$$\Rightarrow{R}_{{D}} =\frac{\mathrm{6}}{\mathrm{7}} \\ $$

Commented by mr W last updated on 10/Feb/20

WAY 2 (−(1/3)+(1/2)+(1/1)+(1/R_D ))^2 =2((1/3^2 )+(1/2^2 )+(1/1^2 )+(1/R_D ^2 )) (1/R_D ^2 )−(7/(3R_D ))+((49)/(36))=0 ((1/R_D )−(7/6))^2 =0 ⇒R_D =(6/7)

$${WAY}\:\:\mathrm{2} \\ $$$$\left(−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{{R}_{{D}} }\right)^{\mathrm{2}} =\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{{R}_{{D}} ^{\mathrm{2}} }\right) \\ $$$$\frac{\mathrm{1}}{{R}_{{D}} ^{\mathrm{2}} }−\frac{\mathrm{7}}{\mathrm{3}{R}_{{D}} }+\frac{\mathrm{49}}{\mathrm{36}}=\mathrm{0} \\ $$$$\left(\frac{\mathrm{1}}{{R}_{{D}} }−\frac{\mathrm{7}}{\mathrm{6}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{R}_{{D}} =\frac{\mathrm{6}}{\mathrm{7}} \\ $$

Commented by jagoll last updated on 10/Feb/20

thank you mister. i like way 1

$${thank}\:{you}\:{mister}.\:{i}\:{like}\:{way}\:\mathrm{1} \\ $$

Answered by ajfour last updated on 10/Feb/20

(x−h)^2 +(y−k)^2 =r^2 h^2 +k^2 =(3−r)^2 (h+1)^2 +k^2 =(2+r)^2 (2−h)^2 +k^2 =(1+r)^2 ⇒ (3−r)^2 +2h+1=(2+r)^2 & 4−4h+(3−r)^2 =(1+r)^2 ⇒ 6+3(3−r)^2 =2(2+r)^2 +(1+r)^2 ⇒ 28r=24 ⇒ r=(6/7) .

$$\left({x}−{h}\right)^{\mathrm{2}} +\left({y}−{k}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${h}^{\mathrm{2}} +{k}^{\mathrm{2}} =\left(\mathrm{3}−{r}\right)^{\mathrm{2}} \\ $$$$\left({h}+\mathrm{1}\right)^{\mathrm{2}} +{k}^{\mathrm{2}} =\left(\mathrm{2}+{r}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{2}−{h}\right)^{\mathrm{2}} +{k}^{\mathrm{2}} =\left(\mathrm{1}+{r}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\left(\mathrm{3}−{r}\right)^{\mathrm{2}} +\mathrm{2}{h}+\mathrm{1}=\left(\mathrm{2}+{r}\right)^{\mathrm{2}} \\ $$$$\&\:\:\:\mathrm{4}−\mathrm{4}{h}+\left(\mathrm{3}−{r}\right)^{\mathrm{2}} =\left(\mathrm{1}+{r}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{6}+\mathrm{3}\left(\mathrm{3}−{r}\right)^{\mathrm{2}} =\mathrm{2}\left(\mathrm{2}+{r}\right)^{\mathrm{2}} +\left(\mathrm{1}+{r}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{28}{r}=\mathrm{24}\:\:\:\Rightarrow\:\:\boldsymbol{{r}}=\frac{\mathrm{6}}{\mathrm{7}}\:. \\ $$

Commented by jagoll last updated on 10/Feb/20

thank you mister . great..

$${thank}\:{you}\:{mister}\:.\:{great}.. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com