The area of a square whose side is a diameter of a circle of area 4s^2

1.	The area of a square whose side is a diameter of a circle of area 4s^2 π isa) 252b) 452c) 1682d) 6452
Answer» AnswEr : Area is 16 s² (Value isn't their in your options !!) . ⠀⠀━━━━━━━━━━━━ GivEn : Area of a square is the diameter itself of a circle with area 4s²π. SoluTion : As the area of circle is given, which is 4s²π !! We know that, Area of circle = πr² which MEAN's, $\dashrightarrow \sf \: \: 4 {s}^{2}\pi = \pi {r}^{2} \\ \\ \\ \dashrightarrow \sf \: \:4 {s}^{2} = {r}^{2} \\ \\ \\ \dashrightarrow \sf \: \:r = \sqrt{4 {s}^{2} } \\ \\ \\ \dashrightarrow \sf \: \:{ \underline{ \boxed{ \sf{ \purple{ \: r = 4s \: }}}}}\: \bigstar$ ⠀⠀━━━━━━━━━━━━ Hence, $\dag \: \: {\large{ \underline{ \underline{ \sf { \orange{Area \: of \: the \: square : - }}}}}}$ $\dashrightarrow \sf \: \: \: {l}^{2} \\ \\ \\ \dashrightarrow \sf \: \: \: {4s}^{2} \\ \\ \\ \dashrightarrow \sf \: \: \:4s \times 4s \\ \\ \\ \dashrightarrow \sf \: \: \:{ \underline{ \boxed{ \sf{ \blue{ \: 16 {s}^{2} \: }}}}} \: \bigstar$ THEREFORE, 16s² is the area of a square WHOSE side is diameter of a circle of area 4s²π. ⠀⠀━━━━━━━━━━━━

The area of a square whose side is a diameter of a circle of area 4s^2 π isa) 252b) 452c) 1682d) 6452

Answer»

AnswEr :

Area is 16 s² (Value isn't their in your options !!) .

⠀⠀━━━━━━━━━━━━

GivEn :

Area of a square is the diameter itself of a circle with area 4s²π.

SoluTion :

As the area of circle is given, which is 4s²π !!

We know that,

Area of circle = πr²

which MEAN's,

$\dashrightarrow \sf \: \: 4 {s}^{2}\pi = \pi {r}^{2} \\ \\ \\ \dashrightarrow \sf \: \:4 {s}^{2} = {r}^{2} \\ \\ \\ \dashrightarrow \sf \: \:r = \sqrt{4 {s}^{2} } \\ \\ \\ \dashrightarrow \sf \: \:{ \underline{ \boxed{ \sf{ \purple{ \: r = 4s \: }}}}}\: \bigstar$

⠀⠀━━━━━━━━━━━━

Hence,

$\dag \: \: {\large{ \underline{ \underline{ \sf { \orange{Area \: of \: the \: square : - }}}}}}$

$\dashrightarrow \sf \: \: \: {l}^{2} \\ \\ \\ \dashrightarrow \sf \: \: \: {4s}^{2} \\ \\ \\ \dashrightarrow \sf \: \: \:4s \times 4s \\ \\ \\ \dashrightarrow \sf \: \: \:{ \underline{ \boxed{ \sf{ \blue{ \: 16 {s}^{2} \: }}}}} \: \bigstar$

THEREFORE, 16s² is the area of a square WHOSE side is diameter of a circle of area 4s²π.