# Thread: Nice Qu but i need help pleeeease

1. ## Nice Qu but i need help pleeeease

hi guys ,

please , i need ur supporting ,,,

how i can solve this qu

about A i solved as the below :
supposing that L1 = perimeter of square , L2 = circumference of circle
L = L1 + L2
L - 4s = L2
s = (L-L2) / 4
area of square = (L - L2)/4)^2
area of circle = pi ( L2 / 2pi )^2

( (L - L2)/4)^2 + pi ( L2 / 2pi )^2 ,

Is This right ???

2. Hello, Learner guy!

I'll get you started . . .

A circle and a square are to be constructed
from a piece of wire of length $L.$

(a) Give an expression for the total area of the circle and the square.
I think you have too many variables.
I would do it like this . . .

Let $x$ = length of wire for the circle.
Then $L-x$ = length of wire for the square.

The circumference of a circle is: . $2\pi r \:=\:C$
. . So we have: . $2\pi r \:=\:x\quad\Rightarrow\quad r \:=\:\frac{x}{2\pi}$

The area of a circle is: . $A \:=\:\pi r^2$
. . So we have: . $A_c \:=\:\pi\left(\frac{x}{2\pi}\right)^2 \:=\:\frac{x^2}{4\pi}$

The perimeter of a square is: . $4s \:=\:P$
. . So we have: . $4s \:=\:L-x \quad\Rightarrow\quad s \:=\:\frac{L-x}{4}$
The area of a square is: . $A \:=\:s^2$

. . So we have: . $A_s \;=\;\left(\frac{L-x}{4}\right)^2$

Therefore, the total area is: . $A \;=\;\frac{1}{4\pi}x^2 + \frac{1}{16}(L-x)^2$