Rectangles - GRE Quantitative Reasoning

Suppose ABCD has sides a and b.

The length of one of ABCD's diagonals is given by a2+ b2 = c2, where c is one of the diagonals.

Note that both diagonals are of the same length.

Quantity A: The length of diagonal AC times the length of diagonal BD

This is c * c = c2.

Quantity A = c2 = a2+ b2

Now for Quantity B, remember that the perimeter of a rectangle with sides a and b is Perimeter = 2(a + b).

Half of Perimeter = (a + b)

Square Half of Perimeter = (a + b)2

Use FOIL: (a + b)2 = a2+ 2ab + b2

Quantity B = (a + b)2 = a2+ 2ab + b2

The question is asking us to compare a2+ b2 with a2+ 2ab + b2.

Note that as long as a and b are positive numbers (in this case a and b are dimensions of a rectangle, so they must be positive), the second quantity will be greater.

Lets call our longer side L and our shorter side W.

If the perimeter is equal to 68, then

.

We also have that

.

If we then plug this into our equation for perimeter, we get .

Therefore, and . Using the Pythagorean Theorem, we have so .

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.

Area of rectangle is .

Area of rectangle is

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 (smaller area) or (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:

For our data, this is:

Solve for :

Therefore,

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

We can set up our data into the following two equations:

(Area) LH = 48

(Perimeter) 2L + 2H = 28

Solve the area equation for one of the two variables (here, length): L = 48 / H

Place that value for L into ever place you find L in the perimeter equation: 2(48 / H) + 2H = 28; then simplify:

96/H + 2H = 28

Multiply through by H: 96 + 2H2 = 28H

Get everything on the same side of the equals sign: 2H2 - 28H + 96 = 0

Divide out the common 2: H2 - 14H + 48 = 0

Factor: (H - 6) (H - 8) = 0

Either of these multiples can be 0, therefore, consider each one separately:

H - 6 = 0; H = 6

H - 8 = 0; H = 8

Because this is a rectangle, these two dimensions are the height and width. If you choose 6 for the "height" the other perpendicular dimension would be 8 and vice-versa. Therefore, the dimensions are 6 x 8.

For any rectangle, , where , , and .

In this problem, we are given that (length is three times the width), so replace in the perimeter equation with :

Plug in our value for the perimeter, :

Simplify:

We know that the following two equations hold for rectangles. For area:

For perimeter:

Now, for our data, we know:

Now, solve the first equation for one of the variables:

Now, substitute this value into the second equation:

Solve for :

Multiply both sides by :

Solve as a quadratic. Divide through by :

Now, get the equation into standard form:

Factor this:

This means that (or ) would equal either or . Therefore, your answer is .

Suppose ABCD has sides a and b.

The length of one of ABCD's diagonals is given by a2+ b2 = c2, where c is one of the diagonals.

Note that both diagonals are of the same length.

Quantity A: The length of diagonal AC times the length of diagonal BD

This is c * c = c2.

Quantity A = c2 = a2+ b2

Now for Quantity B, remember that the perimeter of a rectangle with sides a and b is Perimeter = 2(a + b).

Half of Perimeter = (a + b)

Square Half of Perimeter = (a + b)2

Use FOIL: (a + b)2 = a2+ 2ab + b2

Quantity B = (a + b)2 = a2+ 2ab + b2

The question is asking us to compare a2+ b2 with a2+ 2ab + b2.

Note that as long as a and b are positive numbers (in this case a and b are dimensions of a rectangle, so they must be positive), the second quantity will be greater.

Lets call our longer side L and our shorter side W.

If the perimeter is equal to 68, then

.

We also have that

.

If we then plug this into our equation for perimeter, we get .

Therefore, and . Using the Pythagorean Theorem, we have so .

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.

Area of rectangle is .

Area of rectangle is

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 (smaller area) or (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:

For our data, this is:

Solve for :

Therefore,

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

Suppose ABCD has sides a and b.

The length of one of ABCD's diagonals is given by a2+ b2 = c2, where c is one of the diagonals.

Note that both diagonals are of the same length.

Quantity A: The length of diagonal AC times the length of diagonal BD

This is c * c = c2.

Quantity A = c2 = a2+ b2

Now for Quantity B, remember that the perimeter of a rectangle with sides a and b is Perimeter = 2(a + b).

Half of Perimeter = (a + b)

Square Half of Perimeter = (a + b)2

Use FOIL: (a + b)2 = a2+ 2ab + b2

Quantity B = (a + b)2 = a2+ 2ab + b2

The question is asking us to compare a2+ b2 with a2+ 2ab + b2.

Note that as long as a and b are positive numbers (in this case a and b are dimensions of a rectangle, so they must be positive), the second quantity will be greater.

Lets call our longer side L and our shorter side W.

If the perimeter is equal to 68, then

.

We also have that

.

If we then plug this into our equation for perimeter, we get .

Therefore, and . Using the Pythagorean Theorem, we have so .

Suppose ABCD has sides a and b.

The length of one of ABCD's diagonals is given by a2+ b2 = c2, where c is one of the diagonals.

Note that both diagonals are of the same length.

Quantity A: The length of diagonal AC times the length of diagonal BD

This is c * c = c2.

Quantity A = c2 = a2+ b2

Now for Quantity B, remember that the perimeter of a rectangle with sides a and b is Perimeter = 2(a + b).

Half of Perimeter = (a + b)

Square Half of Perimeter = (a + b)2

Use FOIL: (a + b)2 = a2+ 2ab + b2

Quantity B = (a + b)2 = a2+ 2ab + b2

The question is asking us to compare a2+ b2 with a2+ 2ab + b2.

Note that as long as a and b are positive numbers (in this case a and b are dimensions of a rectangle, so they must be positive), the second quantity will be greater.