What is the measure of angle AEC?

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What is the measure of angle AEC?

by AAPL » Thu Jan 11, 2018 2:53 pm
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Given that ABCDE ia a regular pentagon, what is the measure of ∠ACE?

A. 24³
B. 30³
C. 36³
D. 40³
E. 45³

The OA is C.

I don't have clear this PS question, I appreciate if any expert explain it for me. I know that in a regular pentagon all its sides should be equal but I don't understand how can I determine the angle ACE. Thank you so much.

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by ErikaPrepScholar » Fri Jan 12, 2018 7:40 am
Quick note at the top: it seems like your degree symbols are coming out as an exponent raising each answer to the third power.

There are a couple of important things we need to know to solve this question.

First, we need to know that a regular pentagon is one where all interior angles are the same measurement and where all sides are the same length.

Second, we need to know that the interior angle sum for any polygon is (n-2)*180, where n is the number of sides.

Since we have a hexagon (5 sides), the interior angle sum must be (5-2)*180 = 3*180 = 540 degrees. Then since all of the angles have the same measurement, and there are 5 angles, each angle must measure 540/5 = 108 degrees.

So we know that angles B and D are 108 degrees. Let's think about the triangles ABC and CDE individually. Since we also know that AB and BC as well as CD and DE are the same length, this means that the angles next to those sides (BAC and BCA as well as DCE and DEC) must also be equal. A triangle has an interior angle sum of 180 degrees. So if B and D are 108 degrees, and all of the other angles in the triangles are the same size, we can write the equation 180 = B + BAC + BCA = D + DCE + DEC = 108 + 2x, where x is the measure of BAC, BCA, DCE, and DEC. So 180 = 108 + 2x. Finally, solving for x gives us 36 degrees.

Looking at angle BCD, we see that it is composed of angles BCA, ACE, and DCE. We already know that angle BCD must equal 108 degrees, and we just found out that angles BCA and DCE are 36 degrees. So we can write the equation 108 = BCD = BCA + ACE + DCE = ACE + 2(36). So 108 = ACE + 2(36). Solving for ACE gives 36, which is answer choice C.

For problems like these, I'd recommend drawing the diagram from the problem on your own paper and labeling everything you figure out. So here, you would fill in all of the angle measurements you know. This makes building the equations used in this problem much easier.
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by GMATWisdom » Fri Jan 12, 2018 7:48 am
AAPL wrote:Image

Given that ABCDE ia a regular pentagon, what is the measure of ∠ACE?

A. 24³
B. 30³
C. 36³
D. 40³
E. 45³

The OA is C.

I don't have clear this PS question, I appreciate if any expert explain it for me. I know that in a regular pentagon all its sides should be equal but I don't understand how can I determine the angle ACE. Thank you so much.
We know that the internal angles of a regular pentagon are 108 deg. Therefore, angle ABC=108.
Now triangle ABC and triangle ECD are congruent and both are isosceles.
If angle BAC=x then angle BAC= angle BCA= angle ECD= x
Angle ACE= 108 - angleECD - angle BCA = 108 - 2x
In triangle ABC, angle BAC + angle BCA + angle CBA = 180
Or x + x + 108 = 180
Or 2x = 72
Or x=36
Hence option C

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by [email protected] » Fri Jan 12, 2018 10:59 am
Hi AAPL,

We're told that ABCDE ia a regular pentagon. We're asked for the measure of ∠ACE.

To start, since we're dealing with a REGULAR pentagon, the 5 sides are all equal length and the 5 angles are all equal. A pentagon has 540 degrees, so each of the 5 angles is...
540/5 = 108 degrees.

Next, notice how triangle ABC and triangle CDE are both the SAME ISOSCELES triangle (since they both have 2 sides from the 5 sides that make up the pentagon). Since the 'big' angle in each triangle is 108 degrees, the two smaller EQUAL angles are 72/2 = 36 degrees.

"Big" angle C is split into 3 pieces: the 36 degree angle from triangle ABC, the 36 degree angle from triangle CDE and the 'middle piece.' That middle pieces is....

108 - 36 - 36 = 36 degrees

Final Answer: C

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