Question

Please guys, I have functions problem... Here it is, A piece of wire length 23 centimeters is to be cut into two pieces. One of the pieces will be bent into a square; the other will be bent into a circle. If the length of the piece used for the square is x, express the total area of the square and circular regions formed in terms of x. Help me please. Thanks!

Answer 1

a be side of a square. circumference of circle is (23-x)
then 4a =x which implies , a=X/4 radius = (23-x)/2*3.14

Answer 2

@mtalampas21 The side length of the square is x/4. What is the area of the square?

Answer 3

A = a^2 then A = (x/4)^2 ?

Answer 4

Correct! The radius of the circle is given by:
\[r=\frac{23-x}{2 \pi}\]
Can you state the area of the circle?

Answer 5

\[\frac{ x ^{2} }{ 4^{2} } + {(23-8)^2}/4\pi \] this is the req. area when you will solve u will get it.

Answer 6

how did you get that?

Answer 7

adding area of square and circle.

Answer 8

I'm still confused with the area of the circle

Answer 9

ok look, u have a wire of 23 cm, now what u did was cut it in two pieces.
let length of one wire is 'x' . then length of other wire is (23-x).

Answer 10

now, bent wire of length 'x' to make a square. and wire of length (23-x) to make circle.
therefore, length of wire =perimeter of square.
and length of other wire = circumference of circle.

Answer 11

length of wire= 'x' = 4a i.e. perimeter of square
therefore, a=x/4.
similarly length of other wire is (23-x) = circumference of circle
therefore, 2πr =(23-x)
r= (23-x)/2π

Answer 12

this is my best i have tried.

Answer 13

Finally, I got it. Thanks.

Answer 14

wait, was that really 8? (23-8)^2 ?