While lesser-used than most area formulas, the area of a trapezoid is fair game for the GMAT to test so you should be sure to know it. A trapezoid has two parallel sides, which for the purposes of calculating area will serve as your bases. Here those sides are AB and EC. The formula, then, is 

, whereas always the height must be perpendicular to the bases.

Here you're given the top base as , and need to derive the full bottom base. Since AB must equal ED, you can add ED, which is , to DC, which is , to find the second base as .

To find the height, recognize that triangle BCD is a right triangle with a hypotenuse of 5 and a shorter side of 3. This then means that you're looking at a  Pythagorean triplet, and line segment BD must then measure .

Now you can plug into the trapezoid area formula: 

Therefore  is the correct answer choice.

The area of a parallelogram is calculated as Base × Height, where the height must be perpendicular to the base (much like in a triangle).

Here you can use your knowledge of Pythagorean Theorem (and common Pythagorean triplets) to find the height. Since the right triangle on the left-hand side of the shape has a side of  and a hypotenuse of , it fits the 3-4-5 side ratio, meaning that the height must be 8. (or you could use Pythagorean Theorem and call the height , so , meaning that  so again the height is ).

You know that the area, Base × Height, is , so you can now solve for the base: , so .

You now need the perimeter, which is the sum of all four sides. That will be .

Therefore the correct answer is .

The perimeter of a rectangle can be calculated as  (where  = length and  = width). Here you're told that . This allows you to solve for the ratio of .

First, subtract  from both sides to get .

Next, divide both sides by  to get .

Finally, divide both sides by  to express the ratio as a fraction: , meaning that .

The correct answer is .

The area of any square can be calculated as . Here, you're given the area and need to work backward to find the length of a side. That means that , so the length of a side is .

What is the greatest distance between any two points on a square? It's the square's diagonal, which has a length of . That's a rule you should memorize, but of course, it derives from the hypotenuse of an isosceles right triangle.

So the calculation you're looking for is  times  (the diagonal ratio multiplied by the length of a side). That becomes .

The key to solving this problem is in dividing the given figure into two: a rectangle on the left and a right triangle on the right:

If you do so, you should recognize something familiar with the right triangle: the hypotenuse has a length of  and one side has a length of 30, meaning that this triangle will fit the  side ratio. You then know that the bottom side of the triangle must measure .

With that, you can fill in figures for the bottom of the quadrilateral. The bottom of the rectangle will measure , symmetrical to the top, and the bottom of the triangle will also measure , meaning that the entire bottom side of the quadrilateral measures . Therefore the perimeter is .

As you are given the perimeter of each square as , you can then divide  by  to determine that the side of each square equals .

From there, recognize how the dimensions of the squares fit to the dimensions of the rectangle. The rectangle's height is exactly the same as the diagonal of the square. You can calculate that diagonal using the rule for isosceles right triangles: the diagonal forms the hypotenuse of an isosceles right triangle, where the hypotenuse measures the same as one side times . So the diagonal equals .

That's the height of your rectangle, and the width is equal to  times that amount, since the width spans exactly three squares laid diagonal-to-diagonal. So you can calculate the area of the triangle as height times width:

As you are given that the total area is , and you know that that area is the sum of the areas of five identical squares, you can solve for the side length of a square using , where  represents the side length. That allows you to divide both sides by  to get . And then take the square root to get .

So you now know that the side length of a square must be 9, and the question becomes how many of those sides you add up. Recognize that the middle square is not at all part of the perimeter, and that each of the four outer squares shares one side with that middle square. So the perimeter will be composed of the other three sides of each of the outer four squares. That gives you  squares times  sides times the side length of :

The most concrete piece of given information on this problem is that the area of a rectangle is . You know that Area = Length × Width, so you can say here that . You're also told how the length and width relate to each other. The longer side (which is the one they ask you to solve) is  inches longer than the shorter side, so you can call those  (for the longer) and  for the shorter. This then means that:

You can then expand the multiplication:

Which becomes a quadratic if you subtract  from each side:

And now you have a choice: you could solve this algebraically by factoring the quadratic, but of course  may not be the easiest number to quickly factor. Instead, you could test the answer choices to see which potential  multiplies with  to yield a product of .

If you start in the middle with , you'll see that  would be . Before you calculate, first check to see whether you will indeed get a units digit of  (otherwise why do the math?). You will, but when you do do that math you'll see that  is , which is too small since you need a larger number in .

So then assess the remaining larger answers. If the answer were E, the sides would be 20×16 which will not end in a , so that's out. If it were D, then your sides would be , which does end in a , and which does yield .

Had you wished to factor the quadratic, you would find that  factors to , again yielding  as the answer for the longer side.

Since the monitor in question is square, its diagonal creates a  isosceles right triangle. Recall that the sides of such a triangle are in the ratio . In this case, the hypotenuse has length , so  and . Thus, the sides of the square monitor have length , and the area of the square is .

We are told that the screen area is three times the frame area. Now, the temptation may be to divide  by , but in fact  represents the area of the total monitor – screen plus frame – and the ratio of screen to total area is  (we can compute the ratio to the total by adding up the component ratios  and ; screen:frame:total = ). So we can get the screen area by multiplying the monitor area, , by . The screen area is .

To get the diagonal length for the screen, recall once again that . So the screen has side length . And the diagonal again creates a  right triangle, so the hypotenuse is .

While the math looks a little ugly to start, it cleans up nicely (as usual on the GMAT). Since the longest distance between two points on a square is its diagonal, if that distance for the current plot of land is , then that means that the length of each side of that smaller square is . And since we know that the width of the new property is the same but the length is double, that makes the length of the new property . When combined, the width stays the same at  and the new length becomes , and then to find the area just multiply them together. The roots in the denominator will multiply out to just , so the fraction is

 for a total of .