Recall that perimeter is defined as:

For our data, this is:

Solve for :

Therefore, 

This means that the area of the rectangle is equal to:

Area of rectangle  is .

Area of rectangle  is

The diagonal of a square creates a 45-45-90 triangle; therefore, considering x as the value for the sides of the square in A, set up the ratio: 1/√2 = x/12 → x = 12/√2

Simplify the square root out of the denominator: x = (12^√2)/2 = 6^√2. The perimeter is equal to 4x; therefore, quantity A is 24^√2.

Use the Pythagorean Theorem for B. We know that one side of the triangle is x and the other must be 3x. Furthermore, we know the hypotenuse is 10; therefore:

x² + (3x)² = 10² → x² + 9x² = 100 → 10x² = 100 → x² = 10 → x = √10. 

Transform the √10 into √2  * √5 ; therefore, x = √2  * √5 . The perimeter of the rectangle will equal 3x + 3x + x + x = 8x = 8√2 * √5 .

Compare these two values. Quantity A has a coefficient of 24 in for its √2 . Quantity B has a coefficient of 8√5, which must be smaller than 24, because 8 * 3 = 24, but √5 is less than 3 (which would be √9 ); therefore, A is larger.

Based on our first sentence, we know:

The second sentence means:

, which is the same as:

Now, we can replace  with  in the second equation:

Therefore, 

Now, this means that:

If these are our values, then the area of the rectangle is:

Begin by setting up a ratio of width to length for the rectangle. We know that the width is one-fourth the length, therefore 

.

Now, we can substitute W in the rectangle area formula with our new variable to solve for the length.

Solve for length:

Using the length, solve for Width:

Now, plug the length and width into the formula for the perimeter of a rectangle to solve:

The perimeter of the rectangle is 40.

One potentially helpful first step is to draw the rectangle described in the problem statement:

After that, it's a matter of using the other information given. The perimeter is given as 14, and can be written in terms of the length and width of the rectangle:

Furthermore, notice that the diagonal forms the hypotenuse of a right triangle. The Pythagorean Theorem may be applied:

This provides two equations and two unknowns. Redefining the first equation to isolate  gives:

Plugging this into the second equation in turn gives:

Which can be reduced to:

or

Note that there are two possibile values for ; 3 or 4. The one chosen is irrelevant. Choosing a value 3, it is possible to then find a value for :

This in turn allows for the definition of the rectangle's area:

So Quantity B is 12, which is less than Quantity A.

For this problem, you need to find the pair of sides that would reduce to the same ratio as the original set of sides. This is a little tricky at first, but consider the set:

 and 

For this, you have:

Now, if you factor out , you have:

Thus, the proportions are the same, meaning that the two rectangles would be similar.

Based on the information given, we know that  could be either the longer or the shorter side of the similar rectangle. Similar rectangles have proportional sides. We might need to test both, but let us begin with the easier proportion, namely:

 as 

For this proportion, you really do not even need fractions. You know that  must be .

This means that the figure would have a perimeter of 

Luckily, this is one of the answers!

36

A square has four equal sides and its area = side². Therefore, you can find the side length by taking the square root of the area √81 = 9. Then, find the perimeter by multiplying the side length by 4:

4 * 9 = 36 

The center of an inscribed square lies on the center of the circle. Thus, the line joining the center to a vertex of the square is also the radius. If we join the center with two adjacent vertices we can create a 45-45-90 right isosceles triangle, where the diagonal of the square is the hypotenuse. Since the radius is 2, the hypotenuse (a side of the square) must be  .

Finally, the perimeter of a square is .