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 \sqrt{5}

{(10 \sqrt{5} + r)}^{2} = {2 r}^{2}

r = 10 (\sqrt{5} + \sqrt{10}) ? ? ? ? ?

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

r^{2} = {(r - 10)}^{2} + {(r - 20)}^{2}

r^{2} - 50 r + 500 = 0

r = \frac{50 \pm \sqrt{2500 - 2000}}{2} = \frac{50 \pm 10 \sqrt{5}}{2} = 25 \pm 5 \sqrt{5}

= 36.18 and 13.81

∵ 13.81 < 20 it cannot be the radius .

∴ the radius is 36.18 please 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 .

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

Thanks sir

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

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

I appreciate sir