Question-92684

Question Number 92684 by I want to learn more last updated on 08/May/20

Commented by I want to learn more last updated on 08/May/20

Diagonal = 10(√5) (10(√5) + r)^2 = 2r^2 r = 10((√5) + (√(10))) ?????

$$\mathrm{Diagonal}\:\:=\:\:\mathrm{10}\sqrt{\mathrm{5}} \\ $$$$\:\:\:\:\left(\mathrm{10}\sqrt{\mathrm{5}}\:\:+\:\:\mathrm{r}\right)^{\mathrm{2}} \:\:=\:\:\mathrm{2r}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\mathrm{r}\:\:\:=\:\:\mathrm{10}\left(\sqrt{\mathrm{5}}\:\:+\:\:\sqrt{\mathrm{10}}\right)\:\:????? \\ $$

Commented by Prithwish Sen 1 last updated on 08/May/20

r^2 =(r−10)^2 +(r−20)^2 r^2 −50r+500=0 r=((50±(√(2500−2000)))/2) = ((50±10(√5))/2) = 25±5(√5) =36.18 and 13.81 ∵13.81 <20 it cannot be the radius. ∴ the radius is 36.18 please check

$$\mathrm{r}^{\mathrm{2}} =\left(\mathrm{r}−\mathrm{10}\right)^{\mathrm{2}} +\left(\mathrm{r}−\mathrm{20}\right)^{\mathrm{2}} \\ $$$$\mathrm{r}^{\mathrm{2}} −\mathrm{50r}+\mathrm{500}=\mathrm{0} \\ $$$$\mathrm{r}=\frac{\mathrm{50}\pm\sqrt{\mathrm{2500}−\mathrm{2000}}}{\mathrm{2}}\:=\:\frac{\mathrm{50}\pm\mathrm{10}\sqrt{\mathrm{5}}}{\mathrm{2}}\:=\:\mathrm{25}\pm\mathrm{5}\sqrt{\mathrm{5}} \\ $$$$=\mathrm{36}.\mathrm{18}\:\mathrm{and}\:\mathrm{13}.\mathrm{81} \\ $$$$\because\mathrm{13}.\mathrm{81}\:<\mathrm{20}\:\:\mathrm{it}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{the}\:\mathrm{radius}. \\ $$$$\therefore\:\mathrm{the}\:\mathrm{radius}\:\mathrm{is}\:\mathrm{36}.\mathrm{18}\:\:\:\mathrm{please}\:\mathrm{check} \\ $$

Commented by Prithwish Sen 1 last updated on 08/May/20

sorry i was in a hurry , and made a mistake. please foegive me.

$$\mathrm{sorry}\:\mathrm{i}\:\mathrm{was}\:\mathrm{in}\:\mathrm{a}\:\mathrm{hurry}\:,\:\mathrm{and}\:\mathrm{made}\:\mathrm{a}\:\mathrm{mistake}.\:\mathrm{please} \\ $$$$\mathrm{foegive}\:\mathrm{me}. \\ $$

Commented by I want to learn more last updated on 08/May/20