Math Is Fun Forum

mathxyz

Member

From: Brooklyn, NY

Registered: 2024-02-24

Posts: 1,053

Rectangle Inscribed In Circle

A rectangle is Inscribed in a circle of radius 2.
Let P = (x, y) be a point in quadrant 1 that is a vertex of the rectangle and is on the circle.

A. Express the area A of the rectangle as a function of x.

B. Express the perimeter p of the rectangle as a function of x.

Let me see.

Part A

We want A(x).
Area of square = (side)^2.
Perimeter of square = 4(side), where side is x.

A(x) = (perimeter of square of length x)^2(something)

A(x) = (4x)(something)

I need help finding "something" for A(x).

Part B

In quadrant 1 we have a square of length x.

We want P(x).

Length of side of square in quadrant 1 = x.

Perimeter of square = 4(side).

P(x) = 4x + 4(something)

I need finding "something" for P(x).

Last edited by mathxyz (2024-04-27 07:53:56)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,637

Re: Rectangle Inscribed In Circle

It's a rectangle not a square.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

mathxyz

Member

From: Brooklyn, NY

Registered: 2024-02-24

Posts: 1,053

Re: Rectangle Inscribed In Circle

It's a rectangle not a square.

Bob

The entire image is a rectangle but we are in quadrant 1. The image in quadrant 1 is a little square.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,637

Re: Rectangle Inscribed In Circle

Sorry but it isn't.  The corner is (x,y) so unless x=y the length and width are different.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

mathxyz

Member

From: Brooklyn, NY

Registered: 2024-02-24

Posts: 1,053

Re: Rectangle Inscribed In Circle

Sorry but it isn't.  The corner is (x,y) so unless x=y the length and width are different.

Bob

Thank you. I will return to the beginning of this section because I obviously did not learn a thing.