Geometry - GRE Quantitative Reasoning

To find the perimeter, add up the sides, here 5 + 12 + 13 = 30. To find the area, multiply the two legs together and divide by 2, here (5 * 12)/2 = 30.

If B is a midpoint of AC, then we know AB is 12. Moreover, triangles ACE and ABD share angle DAB and have right angles which makes them similar triangles. Thus, their sides will all be proportional, and BD is 4. 1/2bh gives us 1/2 * 12 * 4, or 24.

Area = 1/2(base)(height). You could use Pythagorean theorem to find the height or, if you know the special right triangles, recognize the 5-12-13. The area = 1/2(12)(5) = 30.

First rearrange the equation so that it is in slope-intercept form, resulting in y=2/3 x + 14/9. The slope of this line is 2/3, so the slope of the line perpendicular will have the opposite reciprocal as a slope, which is -3/2.

Quantity A: area = base * height / 2 = 10 * 24 / 2 = 120

Quantity B: 2 * area = 2 * base * height / 2 = base * height = 5 * 12 = 60

Therefore Quantity A is greater.

Perpendicular lines have slopes that are the opposite of the reciprocal of each other. In this case, the slope of the first line is -2. The reciprocal of -2 is -1/2, so the opposite of the reciprocal is therefore 1/2.

distance2 = (_x_2 – _x_1)2 + (_y_2 – _y_1)2

Looking at the two order pairs given, _x_1 = 1, _y_1 = 1, _x_2 = 7, _y_2 = 9.

distance2 = (7 – 1)2 + (9 – 1)2 = 62 + 82 = 100

distance = 10

Quantity A: Area = 1/2 * b * h = 1/2 * 6 * 7 = 42/2 = 21

Quantity B: Area = 1/2 * (_b_1 + _b_2) * h = 1/2 * (6 + 10) * 6 = 48

Half of the area = 48/2 = 24

Quantity B is greater.

Since the vertical coordinates of each point are given by y, solve each equation for y and plug in 6 for x, as follows:

Taking the difference of the resulting y -values give the vertical distance between the points (6,27) and (6,48), which is 21.

Since we have –2x and +2x in the equations, it makes sense to add the equations together to give 8y = 16 yielding y = 2. Then we substitute y = 2 into one of the original equations to get x = –2. So the solution to the system of equations is (–2, 2)

When the line crosses the x-axis, the y-coordinate is 0. Substitute 0 into the equation for y and solve for x.

.25(x – 20) + 12 = 0

.25_x_ – 5 = –12

.25_x_ = –7

x = –28

The answer is the point (–28,0).

The base of the original cylinder would have been πr2, and the outer face would have been 2πrh, where h is the height of the cylinder.

Let's represent the original area with A, the original radius with r, and the new radius with R: therefore, we know πR2 = 4A, or πR2 = 4πr2. Solving for R, we get R = 2r; therefore, the new outer face of the cylinder will have an area of 2πRh or 2π2rh or 4πrh, which is double the original face area; thus the percentage of increase is 100%. (Don't be tricked into thinking it is 200%. That is not the percentage of increase.)

This problem is simple if we remember the surface area formula!

If the area of the square is 9, then s2 = 9 and s = 3. If the sides thus equal 3, we can calculate the diagonals (either CB or AD) by using the 45-45-90 triangle ratio. For a side of 3, the diagonal will be 3√(2). Note that since the square is inscribed in the circle, this diagonal is also the diameter of the circle. If it is such, the radius is one half of that or 1.5√(2).

Based on that value, we can computer the circle’s area:

A = πr2 = π(1.5√(2))2 = (2.25 * 2)π = 4.5π

We know the triangle inscribed within the circle must be isosceles, as it contains a 90-degree angle and fixed radii. As such, the opposite angles must be equal. Therefore we can use a simplified version of the Pythagorean Theorem,

a2 + a2 = c2 → 2r2 = 16 → r2 = 8; r = √8 < 3. (since we know √9 = 3, we know √8 must be less); therefore, Quantity B is greater.

A radius of 5 means we need a distance of 5 from the center to any points on the circle. We need 52 = (1 – _x_2)2 + (2 – _y_2)2. Let's start with the first point, (3,4). (1 – 3)2 + (2 – 4)2 ≠ 25. Next let's try (4,6). (1 – 4)2 + (2 – 6)2 = 25, so (4,6) is our answer. The same can be done for the other three points to prove they are incorrect answers, but this is something to do ONLY if you have enough time.

We need the formula for the surface area of a cylinder: SA = 2_πr_2 + 2_πrh_. This formula has π in it, but the answer choices don't. This means we must approximate π. None of the answers are too close to each other so we could really even use 3 here, but it is safest to use 3.14 as an approximate value of π.

Then SA = 2 * 3.14 * 172 + 2 * 3.14 * 17 * 3 ≈ 2137

Quantity A: SA of a cylinder = 2_πr_2 + 2_πrh_ = 2_π *_ 16 + 2_π_ * 4 * 2 = 48_π_

Quantity B: SA of a rectangular solid = 2_ab_ + 2_bc_ + 2_ac_ = 2 * 3 * 2 + 2 * 2 * 4 + 2 * 3 * 4 = 52

48_π_ is much larger than 52, because π is approximately 3.14.

surface area of a cylinder

= 2_πr_2 + 2_πrh_

= 2_π_ * 62 + 2_π_ * 6 *9

= 180_π_

O is the center of the circle above.

The length of is .

Quantity A: The area of the circle.

Quantity B:

Do not be tricked by this question. It is true that can be split into halves, each of which are in length. These halves are not, however, radii to the circle. Since this does not go through the center of the circle, its length is shorter than the diameter. This means that the radius of the circle must be greater than . Now, if it were , the area would be . Since it is larger than , the area must be larger than . Quantity A is larger than quantity B.