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.

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.

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).

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.

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

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.

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.

Now, we know that the circumference of a circle is:

or

This means that the diameter of our circle is must be . Given this, we know that the must be shorter than , for the diameter is the longer than any chord that does not pass through the center of the circle. Quantity B is larger than quantity A.

We know that the area of a circle can be expressed: a = πr2

If we know that the area is 36π, we can substitute this into said equation and get: 36π = πr2

Solving for r, we get: 36 = r2; (after taking the square root of both sides:) 6 = r

Now, we know that the circuference of a circle is expressed: c = πd. Since we know that d = 2r (two radii, placed one after the other, make a diameter), we can rewrite the circumference equation to be: c = 2πr

Since we have r, we can rewrite this to be: c = 2π*6 = 12π

Based on our information, we know that the 121π = πr2; 121 = r2; r = 11.

Our other circle with half the radius of A has a diameter equal to the radius of A. Therefore, the circumference of this circle is 11π. Half of this is 5.5π. However, since this is a semi circle, it is enclosed and looks like this:

Therefore, we have to include the diameter in the perimeter. Therefore, the total perimeter of the semi-circle is 5.5π + 11.

Starting with the circle, we need to find the radius in order to get the circumference. Find by plugging our given area into the equation for the area of a circle:

Then calculate circumference:

(approximating as 3.14)

To find the perimeter of the square, we can use , where is the perimeter and is the side length:

, so the circle's circumference is greater.

Let's compute each value separately. We know that the radii are positive numbers that are greater than or equal to . This means that we do not need to worry about the fact that the area could represent a square of a decimal value like .

Quantity A

Since , we know:

Quantity B

If the diameter is one-fourth the radius of A, we know:

Thus, the radius must be half of that, or .

Now, we need to compute the area of this circle. We know:

Therefore,

Now, notice that if , Quantity A is larger.

However, if we choose a value like , we have:

Quantity A:

Quantity B:

Therefore, the relation cannot be determined!