If l x w = A, then we will call the new area B. B results when l has been decreased by 20% and w has been increased by 20%: in other words, 80% of l and 120% of w.

This is expressed as (0.8) l * (1.2) w = B

The simplest way to determine the result is to create rectangles with actual integers and see what happens. Let's say our first rectangle is a square 10 units by 10 units.

10 * 10 = 100 for our A.

Plugging these numbers into our new equation gives us: 

(0.8)10 * (1.2)10 = 96.

So the final step is to compare the areas.  Because we used a 10 x 10 rectangle as our example, this step will be easy, since 100% = 100.

96/100 = 0.96

0.96 * 100 = 96%

Which reflects a 4% reduction in the size of the rectangle.

Any other length and width used for this rectangle will result in a 4% reduction in size when the above parameters are applied.

Because Amy is building her patio in a rectangle, and the bricks are rectangular, the simplest way to find this solution is to determine how many bricks long and how many bricks wide her patio will be.

To do this, begin by converting the patio's dimensions into inches, since the bricks Amy is using are measured in inches.

Length:

12 feet * 12 inches in a foot = 144 inches long.

Width:

8 feet * 12 inches in a foot = 96 inches wide.

Now, all that's left to do is figure out how many bricks will fit in these spaces.  Let's be intuitive and assume the long sides of the bricks will align with the long sides of the patio.

144 inches long / bricks that are 6 inches long = 24 bricks long

Amy will need columns of 24 bricks to cover the length of her patio.  Now, for the width.

96 inches wide / bricks that are 4 inches wide = 24 bricks wide

Amy will need rows of 24 bricks to cover the width of her patio.

To determine how many bricks Amy will need in total, the last step is to multiply how many rows of bricks she will need by how many columns.

24 bricks to a row * 24 bricks to a column = 576

Amy needs 576 bricks to build her patio.

The area of the two shapes depends on their dimensions, which can vary greatly among different combinations. A rectangle with a perimeter of 40 could have dimensions of \(\displaystyle 19*1\) (smaller area) or \(\displaystyle 10*10\) (larger area), for instance. Similarly, a triangle could have sides of 10--11--19 (smaller area) or 13--13--14 (larger area). Thus, the relationship cannot be determined without more specific details.

Recall that perimeter is defined as:

\(\displaystyle P=2W+2L\)

For our data, this is:

\(\displaystyle 130=2*15+2L\)

Solve for \(\displaystyle L\):

\(\displaystyle 130=30+2L\)

\(\displaystyle 100=2L\)

Therefore, \(\displaystyle L=50 \:in\)

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

\(\displaystyle A=WL=50*15=750\:in^2\)

Area of rectangle \(\displaystyle \dpi{100} \small A\) is \(\displaystyle \dpi{100} \small 20\times 3=60\).

Area of rectangle \(\displaystyle \dpi{100} \small B\) is \(\displaystyle \dpi{100} \small 9\times 10=90\)

\(\displaystyle \dpi{100} \small 60x=90\)

\(\displaystyle \dpi{100} \small x=1.5\)

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:

\(\displaystyle L=2H\)

The second sentence means:

\(\displaystyle P=20\:in\), which is the same as:

\(\displaystyle 2L+2H=20\)

Now, we can replace \(\displaystyle L\) with \(\displaystyle 2H\) in the second equation:

\(\displaystyle 2(2H)+2H=20\)

\(\displaystyle 4H+2H=20\)

\(\displaystyle 6H=20\)

\(\displaystyle 3H=10\)

Therefore, \(\displaystyle H=\frac{10}{3}\:in\)

Now, this means that:

\(\displaystyle L=2*\frac{10}{3}=\frac{20}{3}\:in\)

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

\(\displaystyle A=LH=\frac{10}{3}*\frac{20}{3}=\frac{200}{9}\:in^2\)

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

\(\displaystyle \frac{1}{4}L=W\).

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

\(\displaystyle A=W\cdot L\)

\(\displaystyle A=\left(\frac{1}{4}L\right)\cdot L\)

\(\displaystyle A=\frac{1}{4}L^2\)

Solve for length:

\(\displaystyle 64=\frac{1}{4}L^2\)

\(\displaystyle 256=L^2\)

\(\displaystyle L=16\)

Using the length, solve for Width:

\(\displaystyle W=\frac{1}{4}(16)\)

\(\displaystyle W=4\)

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

\(\displaystyle P=2W+2L\)

\(\displaystyle P=2(4)+2(16)\)

\(\displaystyle P=8+32\)

\(\displaystyle P=40\)

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:

\(\displaystyle Perimeter = 2L+2W=14\)

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

\(\displaystyle L^{2}+W^{2}=5^{2}=25\)

This provides two equations and two unknowns. Redefining the first equation to isolate \(\displaystyle L\) gives:

\(\displaystyle L=7-W\)

Plugging this into the second equation in turn gives:

\(\displaystyle 49-14W+W^{2}+W^{2}=25\)

Which can be reduced to:

\(\displaystyle 2W^{2}-14W+24 = 0\)

or

\(\displaystyle 2(W-4)(W-3)=0\)

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

\(\displaystyle L=7-3=4\)

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

\(\displaystyle Area =L\cdot W=3\cdot 4=12\)

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:

\(\displaystyle 5x + 12.5\) and \(\displaystyle 10x -25\)

For this, you have:

\(\displaystyle \frac{5x+12.5}{10x-25}\)

Now, if you factor out \(\displaystyle 2.5\), you have:

\(\displaystyle \frac{2.5(2x+5)}{2.5(4-10)}\)

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