If AE = ED and AB = BC = CD, what is the value of x in the figure abov

Question

https://gmatclub.com/forum/download/file.php?id=61185 If AE = ED and AB = BC = CD, what is the value of x in the figure above? A. 105 B. 130 C. 135 D. 140 E. 145 1.png

sujoykrdatta · Accepted Answer

This is a much shorter approach: In the figure, O is the center of the circle. IMG_20210429_100810__01.jpg Let me know if you have any questions. (Note: think why I made the angles x/2)

BharatTj · Answer

since AE = ED we will be having the same 75 angle for D
Sum of interior angle of polygon = (n-2)180 n= number of sides of polygon = 5 (from the diagram)
 =(5-2)180
 = 540

AB= BC= CD = X

540 = AB+BC+CD+DE+EA
 = X+X+X+75+75
540 = 150+3X
3X = 540-150
 3X = 390
 X = 130

beeblebrox · Answer

Hi  Bunuel , can you please post the solution for this?

Thanks,
Regards.

MBAB123 · Answer

1. There are 2 sets of sides that are equal and another 3 sets of sides that are equal. As the we are also given one of the angles (75 degree), that fixes are figure to a certain extent.
2. Now, 2 sets of equal sides would have equal angles and the other 3 would have equal angles.
3. 75 falls in the 2 set side, so the 2 sets of equal angles would be 75 each (total 150) and then you can find the X. X= (540-150)/3 = 130.
4. Note that the other 75 degree angle could either be angle A or angle D. Either way, the answer would not change.

^the above approach is not mechanical but I used this and got the right answer. The only formula I used is the  sum of angles formula , which I'm sure most of us would memorize at some stage of our prep. No other concepts here, but while taking the test it might help to visualize if you're not certain of a formula/property based approach. Of course, the more properties you know the better!

beeblebrox · Answer

Hi  Brian123  &  BharatTj ,

Wasn't satisfied with your proposed solution. Here is my take on this problem. Took me some time to get there though.

I have used concepts from arcs to solve this particular problem. Please ignore clumsy hand!

Fdambro294 · Answer

As a general proposition, whenever you see an irregular polygon such as this inscribed in a circle, the first thing that should pop in the mind is the Inscribed Angle Theorem.

Basically 3 main moves to get there:

(1) draw in the central angle by connecting a radius from the Center (draw in point O) to Points A and D on the perimeter of the circle.

The central angle subtended by Arc ABCD will be (2) (75) ——-> which is twice the inscribed angle of 75 degrees subtended by the same arc.

(2nd move) connect a Radii from the center to each of the 5 vertices. It will cut this beast up into 5 triangles:

3 congruent triangles up top (AOB——-BOC——-COD)

In which each of these Isosceles triangles will have two angles of equal measure across from the Radii Sides (call it A) and a 50 degree angle at the center of the triangle (A-A-A postulate: each of the 3 triangles have 2 equal angles that we called A.......this means the 3rd angle at the center will be congruent as well. The 150 degree central angle will be cut up equally 3 ways, one angle for each congruent triangle )

And 2 congruent triangles near the 75 degree angle (AOE———DOE)

(Last step)

Angle chase and fill in the equal angles opposite the equal sides of length = radius R

At which point you will see that OE (one of the 5 radii drawn) bisects the 75 degree angle.

Angle EOA and angle EOD will each equal 37.5 degrees, as well EDO and EAO.

Fill in the rest of the angles in each of the 3 congruent triangles up top near each of the vertices as A degrees.

Entire degree measure of a pentagon = 540 degrees

Angle at Vertex C is composed of (2) (A)

75 + (37.5) + (37.5) + (6) (A) = 540 degrees

Solving for 2A = Angle at vertex C = 130 deg

Posted from my mobile device

MBAB123 · Answer

beeblebrox , this definitely seems like the perfect way to solve this problem! I agree my solution to this question was not the best. However, I was just making the point that if this question came on this exam, it's possible to visualize the shape and then answer. Of course, if the solution suggested by you is way better, but, it might not be possible to come up with such or more complex solutions in the exam under timed condition. My post wasn't about the solution per se, but more about the approach if you're stuck and can't think of properties, methods, etc.

beeblebrox · Answer

Hi sujoykrdatta why have you made the angles x/2?

sujoykrdatta · Answer

Triangles AOB, BOC, COD are isosceles and congruent. Does that help?

Posted from my mobile device

bumpbot · Answer

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If AE = ED and AB = BC = CD, what is the value of x in the figure abov

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If AE = ED and AB = BC = CD, what is the value of x in the figure abov

Prep Toolkit

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