E

Close enough - you took 1 minutes and 29 seconds to answer this question.

Incorrect.

the perimeter of a shapes is always a line enclosing the shape on all sides - think of a wall surrounding a castle from all sides.

Thus, the perimeter of the semicircle is comprised of the diameter AB and of arc AB. If you denote the radius of C as r, then the diameter is 2r, and the arc is half of the circumference: 2πr/2.

Thus, the perimeter of the semicircle is 2r + 2πr/2 = 4+4π

B

Incorrect.

the perimeter of a shapes is always a line enclosing the shape on all sides - think of a wall surrounding a castle from all sides.

Thus, the perimeter of the semicircle is comprised of the diameter AB and of arc AB. If you denote the radius of C as r, then the diameter is 2r, and the arc is half of the circumference: 2πr/2.

Thus, the perimeter of the semicircle is 2r + 2πr/2 = 4+4π

A

Incorrect.

the perimeter of a shapes is always a line enclosing the shape on all sides - think of a wall surrounding a castle from all sides.

Thus, the perimeter of the semicircle is comprised of the diameter AB and of arc AB. If you denote the radius of C as r, then the diameter is 2r, and the arc is half of the circumference: 2πr/2.

Thus, the perimeter of the semicircle is 2r + 2πr/2 = 4+4π

D

Incorrect.

the perimeter of a shapes is always a line enclosing the shape on all sides - think of a wall surrounding a castle from all sides.

Thus, the perimeter of the semicircle is comprised of the diameter AB and of arc AB. If you denote the radius of C as r, then the diameter is 2r, and the arc is half of the circumference: 2πr/2.

Thus, the perimeter of the semicircle is 2r + 2πr/2 = 4+4π

C

Correct.

--> 2r + πr = 4+4π

--> r(2+π) = 4+4π

--> r = (4+4π)/(2+π) E

You grossly underestimated the time this question took you. You actually solved it in 6 minutes and 43 seconds.

Correct.

Numbers in the answer choices and a specific question ("How many candies...") call for Plugging In The Answers. You may feel like jotting down an equation or more. This is just your algebraic urge, which is another stop sign for Reverse PI problems.

Start with answer choice C - assume the amount in the answer choice is the number of candies Rebecca has now, and then follow the story in the problem. If everything fits - stop. Pick it. Otherwise - POE and move on, until you find an answer that works.

Start with answer choice C. If Rebecca has 54 candies, then she had 54-24=30 candies to begin with. Since the question states that Tina and Rebecca had the same number of candies originally, this means that Tina also had 30 candies. Therefore, after giving 24 candies to Rebecca, Tina should now have 30-24=6 left and Rebecca 54, but that means that Rebecca has 9 times more than Tina, not 5 (as the question stipulates). So you should eliminate C. Which direction should you go now?

The answer is 'not sure', so test either way. Check answer choice B. If Rebecca has 48 candies now, she and Tina had 24 each to begin with, so now Rebecca has all the candies and Tina has none, which doesn't fit the question. POE B and A as well, because A would result in even less candies for Tina. Plug in D or E to check which is correct.

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The correct answer is E.

If Rebecca has 60 candies, it means that she and Tina had 36 candies. Now Tina has 36-24=12, exactly a fifth of the amount Rebecca has.

Note: knowing that the number of Rebecca's candies is a multiple of 5, you can also automatically eliminate all the answer choices which are not divisible by 5. Noticing that makes your life a whole lot easier, as it leaves you with D or E. B

Remember the ratio of the sides of a 45, 45, 90 triangle is 1, 1, √2.

Correct.

Plug in 1 for the side of the smallest triangle so the hypotenuse is √2. Since the leg of the second triangle is the hypotenuse of the first, the legs of the second triangle are √2, √2 and the hypotenuse is 2. Likewise, the legs and hypotenuse of the third triangle are 2, 2 and 2√2. Lastly, the legs and the hypotenuse of the largest triangle are 2√2, 2√2 and 4. Hence the hypotenuse of the largest triangle is 4 times the leg of the smallest triangle. A

Incorrect.

This is a question about the measure of an arc in angles.

Draw the circle and go on to indicate ABCD and the rest of the information on it.

Point B bisects minor arc AC, so minor arc AB and minor arc BC are 80° each.

Minor arc BD of 140° is the total of minor arcs BC (=80°) and CD, therefore, minor arc CD=60°.

Calculate arc AD, then find minor arc AD. Recall that the minor of an arc larger than 180° is the arc that completes it to 360°.

C

Correct.

This is a question about the measure of an arc in angles.

Draw the circle and go on to indicate ABCD and the rest of the information on it.

Point B bisects minor arc AC, so minor arc AB and minor arc BC are 80° each.

Minor arc BD of 140° is the total of minor arcs BC (=80°) and CD, therefore, minor arc CD=60°.

Finally, arc AD is the total of minor arcs AB, BC and CD, i.e., 80+80+60=220°. Obviously, this is the measure of major arc AD. The minor arc completes the major arc to 360°, so minor AD is 360−220=140°. D

You underestimated the time this question took you. You actually solved it in 1 minutes and 54 seconds.

Correct.

The perimeter of the shaded region is composed of the length of arc BC, and the length of the sides AC and AB.

To find AC, connect B and C to create the ABC recycled 30:60:90 triangle. ABC is a recycled 30:60:90 triangle because it has a 30 degree angle from the original figure, and a 90 degree angle, ∠ACB, is an inscribed angle on the diameter.

Use the 30:60:90 ratio: the hypotenuse, which is also the diameter, is 2r=12. Thus the little leg, CB, is 12/2 = 6 and the big leg, AC is 6√3.

Now we need the minor arc BC. Recall that The measure of an arc is twice the measure of an inscribed angle that defines it. Angle CAB is a 30 degrees inscribed angle, so the measure of minor arc CB is 2·30=60°. 60 is 1/6 of 360, so small arc BC is 1/6 of the circle's circumference. Plug in the radius=6 into the circumference formula to get

1/6×(2πr) = 1/6×(2×π×6) = 2π.

Thus, the total perimeter of the shaded region is (minor arc CB)+AB+AC = 2π+12+6√3. ;