Problem Solving

rfshocker2255 wrote:A 40m long wire is cut into 2pcs. One forms a circle with radius r, and the other piece forms a square. What is the total area of the circle and square?

Here's the original question in its entirety:
A thin piece of wire 40 meters long is cut into two pieces. One piece is used to form a circle with radius r, and the other is used to form a square. No wire is left over. Which of the following represents the total area, in square meters, of the circular and the square regions in terms of r?

a) (pi)r^2
b) (pi)r^2 + 10
c) (pi)r^2 + (1/4)(pi)^2(r)^2
d) (pi)r^2 + [40 - 2(pi)r]^2
e) (pi)r^2 + [1O - (1/2)(pi)r]^2

Here's the algebraic approach:

Since r is the radius of the circle, the area of the circle will be (pi)r^2.

If r is the radius of the circle, the length of wire used for this circle will equal its circumference which is 2(pi)r

So, the length of wire to be used for the square must equal 40 - 2(pi)r

In other words, the perimeter of the square will be 40 - 2(pi)r

Since squares have 4 equal sides, the length of each side of the square will be [40 - 2(pi)r]/4, which simplifies to be 10 - (pi)r/2

If each side of the square has length 10 - (pi)r/2, the area of the square will be [10 - (pi)r/2]^2

So, the total area will equal (pi)r^2 + [10 - (pi)r/2]^2, which is the same as E

Cheers,
Brent

rfshocker2255 wrote: For all positive integers, [m]=m/2 for all even integers and [m]=3m for all odd integers. What does [9] x [6] equal?
I keep coming up with 81, and am having a hard time understanding what I am missing.

You are correct in that [9] x [6] = 81

9 is odd, so [9] = (3)(9) = 27
6 is even, so [6] = 6/2 = 3
So, [9] x [6] = 27 x 3 = 81

However, the answer choices are as follows:

a) [81]
b) [54]
c) [37]
d) [27]
e) [18]

So, which of these equals 81?

Since 27 is odd, [27] = (3)(27) = 81

So, the correct answer is D

Cheers,
Brent

Refer to the posts as follows...

For problem 1 >> https://www.beatthegmat.com/geometry-t131113.html#516326
For problem 2 >> https://www.beatthegmat.com/function-pro ... tml#327643