Plane 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.

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 r by plugging our given area into the equation for the area of a circle:

A = pi $r^2$

25pi = pi $r^2$

25 = $r^2$

Then calculate circumference:

C = 2pi r

C = 2pi times 5 = 10pi approx 31.4 inches (approximating pi as 3.14)

To find the perimeter of the square, we can use P = 4s , where P is the perimeter and s is the side length:

31.4>28, 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!

Since we know that the area of Square is , we know , where is the length of one of its sides. From this, we can solve for by taking the square root of both sides. You will have to do this by estimating upward. Therefore, you know that is . By careful guessing, you can quickly see that is . From this, you know that the diameter of your circle must be half of , or (because it is circumscribed). Therefore, you can draw:

The circumference of this circle is defined as:

or, for your values:

(You could also compute this from the diameter, but many students just memorize the formula above.)

Here, you need to “solve backward” from the data you have been given. We know that 0.25C = 5.5; therefore, C = 22. In order to solve for the area, we will need the radius of the circle. This can be obtained by recalling that C = 2πr. Replacing 22 for C, we get 22 = 2πr.

Solve for r: r = 22 / 2π = 11 / π.

Now, we solve for the area: A = πr2. Replacing 11 / π for r: A = π (11 / π)2 = (121π) / (π2) = 121 / π.

There are two steps to this problem: determining the area of the circle and determining the area of the square. The area of the circle is πr2 which is π(2/1)2 or π. AD is a diameter of circle O and creates two isosceles right triangles with ACD and ABD. The relationship between sides of an isosceles right triangle is 1 : 1 : √2. Thus the sides of square ABCD are √2 and the area is 2. The area of the shaded region is the area of the circle minus the area of the square, or π – 2.

Therefore the 16 inch pizza is the better deal.

To find the radius of circle B, use the circumference formula (c = πd = 2πr):

2πr = 36π

Divide each side by 2π: r = 18

Now, if circle A has a radius half the length of that of B, A's radius is 18 / 2 = 9.

The area of a circle is πr2. Therefore, for A, it is π*92 = 81π.

Since the answers are in terms of pi, simply find the area of each circle in terms of x and ∏:

Smaller: ∏(5)2 = 25∏

Larger: ∏x2

We must subtract the inner circle from the outer circle; this translates to ∏x2-25∏.