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INSCRIBED AND CIRCUMSCRIBED POLYGONS ON THE GMAT

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INSCRIBED AND CIRCUMSCRIBED POLYGONS ON THE GMAT

REGULAR POLYGONS AND THE IRREGULAR ONES


By Karishma, Veritas Prep.


Continuing our Geometry journey, let’s discuss polygons today.

Some years back, I used to often get confused in the polygon sum-of-the-interior-angles formula if I had to recall it after a gap of some months because I had seen two variations of it:

Sum of interior angles of a polygon = (n – 2)*180
Sum of interior angles of a polygon = (2n – 4)*90

Now, I don’t want you to judge me. Of course, in the second formula, 2 has been removed from 180 and multiplied to the first factor. It is quite simple so why would anyone get tricked here, you wonder? The problem was that after a few months, I would somehow remember (2n – 4) and 180. So I was mixing up the two and I wasn’t sure of the logic behind this formula. That is until I came across the simple explanation of this formula in our Veritas Prep Geometry book (the one which explains how you can divide every polygon with n sides into (n – 2) triangles and hence get the sum of (n – 2)*180). Now it made perfect sense! I couldn’t believe that I had not come across that explanation before and had just learned up (well, tried to!) the formula blindly. So now I ensure that all my students understand every formula that I teach them.

Usually, we are given a regular polygon and we need to find the measure of interior angles or the number of sides. But what if we are given a polygon instead, not a regular polygon. Does this formula still apply? We wouldn’t know if we didn’t understand how the formula came into being. But since we know that we obtain the formula by dividing the polygon into (n-2) triangles, we know that the sum of all interior angles of a triangle is 180 irrespective of the kind of triangle. So it doesn’t matter whether the polygon is regular or not. The sum of all interior angles will still be (n-2)*180.

Let’s look at a question to see the application of this formula in irregular polygon scenario.

Question: The measures of the interior angles in a polygon are consecutive odd integers. The largest angle measures 153 degrees. How many sides does this polygon have?
A) 8
B) 9
C) 10
D) 11
E) 12

Solution:
The interior angles are: 153, 151, 149, 147 … and so on.

Now there are two ways to approach this question – one which is straight forward but uses algebra so is time consuming, another which makes you think but doesn’t take much time. You can guess which one we are going to focus on! But before we do that let’s take a quick look at the algebraic solution too.

Method 1: Algebra

Sum of interior angles of this polygon = \(153 + 151 + 149 + … (153 – 2(n-1)) = (n – 2)*180\).

If there are n sides, there are n interior angles. The second largest angle will be 153 – 2*1. The third largest will be 153 – 2*2. The smallest will be 153 – 2*(n-1). This is an arithmetic progression.

Sum of all terms = \(\frac{(First term + Last term)}{2} * n = \frac{(153 + 153 – 2(n-1))}{2} * n\)

Equating, we get \(\frac{(153 + 153 – 2(n-1))}{2} * n = (n – 2)*180\)

Solving this you get, n = 10

But let’s figure out a solution without going through this painful calculation.

Method 2: Capitalize on what you know

Angles of the polygon: 153, 151, 149, 147, 145, 143, 141, … , (153 – 2(n-1))

The average of these angles must be equal to the measure of each interior angle of a regular polygon with n sides since the sum of all angles is the same in both the cases.

Measure of each interior angle of n sided regular polygon = Sum of all angles / n = \(\frac{(n-2)*180}{n}\)

Using the options:

Measure of each interior angle of 8 sided regular polygon = 180*6/8 = 135 degrees

Measure of each interior angle of 9 sided regular polygon = 180*7/9 = 140 degrees

Measure of each interior angle of 10 sided regular polygon = 180*8/10 = 144 degrees

Measure of each interior angle of 11 sided regular polygon = 180*9/11 = 147 degrees apprx

and so on…

Notice that the average of the given angles can be 144 if there are 10 angles.

The average cannot be higher than 144 i.e. 147 since that will give us only 7 sides (153, 151, 149, 147, 145, 143, 141 – the average is 147 is this case). But the regular polygon with interior angle measure of 147 has 11 sides. Similarly, the average cannot be less than 144 i.e. 140 either because that will give us many more sides than the required 9.

Hence, the polygon must have 10 sides.

Answer (C).


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Re: INSCRIBED AND CIRCUMSCRIBED POLYGONS ON THE GMAT [#permalink]

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INSCRIBING POLYGONS AND CIRCLES


By Karishma, Veritas Prep.


Above we looked at regular and irregular polygons. Now, let’s try to understand how questions involving one figure inscribed in another are done. The most common example of a figure inscribed in another is a polygon inscribed in a circle or a circle inscribed in a polygon. Let’s see the various ways in which this can be done.

To inscribe a polygon in a circle, the polygon is placed inside the circle so that all the vertices of the polygon lie on the circumference of the circle.

There are a few points about inscribing a polygon in a circle that you need to keep in mind:
– Every triangle has a circumcircle so all triangles can be inscribed in a circle.
Image
– All regular polygons can also be inscribed in a circle.
Image
– Also, all convex quadrilaterals whose opposite angles sum up to 180 degrees can be inscribed in a circle.
Image

There are also a few points about inscribing a circle in a polygon that you need to keep in mind:
– All triangles have an inscribed circle (called incircle). When a circle is inscribed in a triangle, all sides of the triangle must be tangent to the circle.
Image
– All regular polygons have an inscribed circle.
Image
– Most other polygons do not have an inscribed circle.

A simple official question will help us see the relevance of these points:
Question: Which of the figures below can be inscribed in a circle?
Image
(A) I only
(B) III only
(C) I & III only
(D) II & III only
(E) I, II & III

Solution: I think it will suffice to say that the answer is (C).
This question is discussed HERE.

[Reveal] Spoiler:
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Collection of Questions:
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DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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CIRCLES AND INSCRIBED REGULAR POLYGON RELATIONS


By Karishma, Veritas Prep.


Let’s figure out the relations between the sides of various inscribed regular polygons and the radius of the circle.

We will start with the simplest regular polygon – an equilateral triangle. We will use what we already know about triangles to arrive at the required relations.

Look at the figure given below. AB, BC and AC are sides (of length ‘a’) of the equilateral triangle. OA, OB and OC are radii (of length ‘r’) of the circle.
Image
The interior angles of an equilateral triangle are 60 degrees each. Therefore, angle OBD is 30 degrees (since ABC is an equilateral triangle, BO will bisect angle ABD). So, triangle BOD is a 30-60-90 triangle.

As discussed in your geometry book, the ratio of sides in a 30-60-90 triangle is \(1:\sqrt{3}:2\) therefore, a/2 : r = \(\sqrt{3}:2\) or \(a:r = \sqrt{3}:1\)

Side of the triangle = \(\sqrt{3}\) * Radius of the circle

You don’t have to learn up this result. You can derive it if needed. Note that you can derive it using many other methods. Another method that easily comes to mind is using the altitude AD. Altitude AD of an equilateral triangle is given by \(\frac{\sqrt{3}}{2}*a\). The circum center is at a distance 2/3rd of the altitude so AO (radius) = \((\frac{2}{3})*\frac{\sqrt{3}}{2}*a = \frac{a}{\sqrt{3}}\)

Or side of the triangle = \(\sqrt{3}\) * radius of the circle

Let’s look at a square now.
Image
AB is the side of the square and AO and BO are the radii of the circle. Each interior angle of a square is 90 degrees so half of that angle will be 45 degrees. Therefore, ABO is a 45-45-90 triangle. We know that the ratio of sides in a 45-45-90 triangle is \(1:1:\sqrt{2}\).

\(r:a = 1: \sqrt{2}\)

Side of the square = \(\sqrt{2}\)*Radius of the circle

Again, no need to learn up the result. Also, there are many methods of arriving at the relation. Another one is using the diagonal of the square. The diagonal of a square is \(\sqrt{2}\) times the side of the square. The radius of the circle is half the diagonal. So the side is the square is \(\sqrt{2}\)*radius of the circle.

The case of a pentagon is more complicated since it needs the working knowledge of trigonometry which is beyond GMAT scope so we will not delve into it.

We will look at a hexagon though.
Image
Notice that the interior angle of a regular hexagon is 120 degrees so half of that will be 60 degrees. Therefore, both angles OAB and OBA will be 60 degrees each. This means that triangle OAB is an equilateral triangle with all angles 60 degrees and all sides equal. Hence,

Side of the regular hexagon = Radius of the circle.

The higher order regular polygons and more complicated and we will not take them up. We will discuss a circle inscribed in a polygon in the next post.

[Reveal] Spoiler:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: INSCRIBED AND CIRCUMSCRIBED POLYGONS ON THE GMAT [#permalink]

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New post 03 Jun 2015, 03:47

AND NOW THE OTHER WAY


By Karishma, Veritas Prep.


In this post will work with circles inscribed in regular polygons.

We begin by considering an equilateral triangle whose each side is of length ‘a’. Recall that every triangle has an incircle i.e. a circle can be inscribed in every triangle. The diagram given below shows the circle of radius ‘r’ inscribed in an equilateral triangle.
Image
How can we find the relation between ‘r’ and ‘a’? Every angle of an equilateral triangle is 60 degrees. Since it is an equilateral triangle, due to the symmetry, angle OBD = angle OBA = 30 degrees. So we see that triangle BOD is a 30-60-90 triangle. So the ratio of the sides OD:BD:OB = \(1: \sqrt{3} = r : \frac{a}{2}\).

Therefore, \(a = 2\sqrt{3} * r\)

Side of the triangle = \(2\sqrt{3}\) * Radius of the circle

As discussed last week, there are many other methods of getting this result. We can use the altitude method.

Altitude of an equilateral triangle is given by \(\frac{\sqrt{3}}{2}*a\). The incenter is at a distance 2/3rd of the altitude so OD (radius) = \((\frac{1}{3})*\frac{\sqrt{3}}{2}*a = \frac{a}{2\sqrt{3}}\)

Or Side of the triangle = \(2\sqrt{3}\) * Radius of the circle

Now we will look at a square.
Image
The figure itself shows us that r = a/2

Side of the square = 2 * Radius of the circle

There is no need to delve deeper into it. Though, here is something for you to think about: Can you have a circle inscribed in a rectangle?

Now let’s consider a circle inscribed in a regular hexagon.
Image
We know that the interior angle of a regular hexagon is 120 degrees. OA will bisect that angle making angle OAD = 60 degrees. Since AB is tangent to the circle, OD will be perpendicular to AB. Hence OAD is a 30-60-90 triangle. Therefore, \(\frac{a}{2} : r = 1: \sqrt{3}\)

Hence, \(a = \frac{2r}{\sqrt{3}}\)

Side of the hexagon = \(\frac{2}{\sqrt{3}}\) * Radius of the circle

Again, remember, you are not expected to ‘know’ these results so don’t try to learn them up. You can always derive any relation you want once you know some basic tricks. The intent of these posts is to familiarize you with those tricks.

In the next post, we will look at some interesting Geometry questions based on these concepts!

[Reveal] Spoiler:
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New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: INSCRIBED AND CIRCUMSCRIBED POLYGONS ON THE GMAT [#permalink]

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New post 03 Jun 2015, 04:05

QUESTIUONS ON POLYGONS INSCRIBED IN CIRCLES


By Karishma, Veritas Prep.


In this post, I have two questions for you – both on polygons inscribed in a circle. You must go through the previous post based on this topic before trying these questions.

Question 1: Four points that form a polygon lie on the circumference of the circle. What is the area of the polygon ABCD?
Statement I: The radius of the circle is 3 cm.
Statement II: ABCD is square.
Image
Solution:
Notice that you have been given that angles B and D are right angles. Does that imply that the polygon is a square? No. You haven’t been given that the polygon is a regular polygon. The diagonal AC is the diameter since arc ADC subtends a right angle ABC. Hence arc ADC and arc ABC are semi-circles. But the sides of the polygon (AB, BC, CD, DA) may not be equal. Look at the diagram given below:
Image
Statement I: The radius of the circle is 3 cm.
This statement alone is not sufficient. Look at the two figures given above. The area in the two cases will be different depending on the length of the sides. Just knowing the diagonal AC is not enough. Hence this statement alone is not sufficient.

Statement II: ABCD is square.
This tells us that the first figure is valid i.e. the polygon is actually a square. But this statement alone doesn’t give us the measure of any side/diagonal. Hence this statement alone is not sufficient.

Using both statements together, we know that ABCD is a square with a diagonal of length 6 cm. This means that the side of the square is \(\frac{6}{\sqrt{2}}\) cm giving us an area of \((\frac{6}{\sqrt{2}})^2 = 18\) cm^2.

Answer (C). This question is discussed HERE.

Let’s look at a more complicated question now.

Question 2: A regular polygon is inscribed in a circle. How many sides does the polygon have?
Statement I: The length of the diagonal of the polygon is equal to the length of the diameter of the circle.
Statement II: The ratio of area of the polygon to the area of the circle is less than 2:3.

Solution:
In this question, we know that the polygon is a regular polygon i.e. all sides are equal in length. As the number of sides keeps increasing, the area of the circle enclosed in the regular polygon keeps increasing till the number of sides is infinite (i.e. we get a circle) and it overlaps with the original circle. The diagram given below will make this clearer.
Image
Let’s look at each statement:

Statement I: The length of one of the diagonals of the polygon is equal to the length of the diameter of the circle.
Do we get the number of sides of the polygon using this statement? No. The diagram below tells you why.
Image
Regular polygons with even number of sides will be symmetrical around their middle diagonal and hence the diagonal will be the diameter. Hence the polygon could have 4/6/8/10 etc sides. Hence this statement alone is not sufficient.

Statement II: The ratio of area of the polygon to the area of the circle is less than 2:3.

Let’s find the fraction of area enclosed by a square.

In the previous post we saw that
Side of the square = \(\sqrt{2}\) * Radius of the circle
Area of the square = \(Side^2 = 2*Radius^2\)
Area of the circle = \(\pi*Radius^2 = 3.14*Radius^2\)
Ratio of area of the square to area of the circle is 2/3.14 i.e. slightly less than 2/3.

So a square encloses less than 2/3 of the area of the circle. This means a triangle will enclose even less area. Hence, we see that already the number of sides of the regular polygon could be 3 or 4. Hence this statement alone is not sufficient.

Using both statements together, we see that the polygon has 4/6/8 etc sides but the area enclosed should be less than 2/3 of the area of the circle. Hence the regular polygon must have 4 sides. Since the area of a square is a little less than 2/3rd the area of the circle, we can say with fair amount of certainty that the area of a regular hexagon will be more than 2/3rd the area of the circle. But just to be sure, you can do this:

Side of the regular hexagon = Radius of the circle

Area of a regular hexagon = 6*Area of each of the 6 equilateral triangles = \(6*(\frac{\sqrt{3}}{4})*Radius^2 = 2.6*Radius^2\)

2.6/3.14 is certainly more than 2/3 so the regular polygon cannot be a hexagon. The regular polygon must have 4 sides only.

Answer (C). This question is discussed HERE.


[Reveal] Spoiler:
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New to the Math Forum?
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Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: INSCRIBED AND CIRCUMSCRIBED POLYGONS ON THE GMAT [#permalink]

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QUESTIONS ON CIRCLES INSCRIBED IN POLYGONS


By Karishma, Veritas Prep.


Above we looked at questions on polygons inscribed in a circle. This week, let’s look at questions on circles inscribed in regular polygons. As noted earlier, it’s important to keep in mind that regular polygons are symmetrical figures. You need very little information to solve for anything in a symmetrical figure.

Question 1: A circle is inscribed in a regular hexagon. A regular hexagon is inscribed in this circle. Another circle is inscribed in the inner regular hexagon and so on. What is the area of the tenth such circle?
Image
Statement I: The length of the side of the outermost regular hexagon is 6 cm.
Statement II: The length of a diagonal of the outermost regular hexagon is 12 cm.

Solution: Thankfully, in DS questions, we don’t need to calculate the answer. We just need to establish the sufficiency of the given data. Note that we have found that there is a defined relation between the sides of a regular hexagon and the radius of an inscribed circle and there is also a defined relation between the radius of a circle and the side of an inscribed regular hexagon.

When the circle is inscribed in a regular hexagon,

Radius of the inscribed circle = \(\frac{\sqrt{3}}{2}\)* Side of the hexagon

When a regular hexagon is inscribed in a circle,

Side of the inscribed regular hexagon = Radius of the circle

So all we need is the side of any one regular hexagon or the radius of any one circle and we will know the length of the sides of all hexagons and the radii of all circles.

Statement I: The length of the side of the outermost regular hexagon is 6 cm.
If length of the side of the outermost regular hexagon is 6 cm, the radius of the inscribed circle is \((\sqrt{3}/2)*6 = 3\sqrt{3}\) cm

In that case, the side of the regular hexagon inscribed in this circle is also \(3\sqrt{3}\) cm. Now we can get the radius of the circle inscribed in this second hexagon and go on the same lines till we reach the tenth circle. This statement alone is sufficient.

Statement II: The length of a diagonal of the outermost regular hexagon is 12 cm.

Note that a hexagon has diagonals of two different lengths. The diagonals that connect vertices with one vertex between them are smaller than the diagonals that connect vertices with two vertices between them. Length of AC will be shorter than length of AD. Given the length of a diagonal, we do not know which diagonal it is. Is AC = 12 or is AD = 12? The length of the side will be different in the two cases. So this statement alone is not sufficient.

Answer (A). This question is discussed HERE.

Keep in mind that you don’t actually need to solve for an answer is DS; in fact, in some questions you will not be able to solve for the answer under the given time constraints. All you need to do is ensure that given unlimited time, you will get a unique answer.

Question 2: Four identical circles are drawn in a square such that each circle touches two sides of the square and two other circles (as shown in the figure below). If the side of the square is of length 20 cm, what is the area of the shaded region?
Image
(A) 400 – 100π
(B) 200 – 50π
(C) 100 – 25π
(D) 8π
(E) 4π

Solution: First let’s recall that squares and circles are symmetrical figures. The given figure is symmetrical.

We don’t know any formula that will help us get the area of the curved shaded grey shape in the center. In such cases, very often what you need is to find the area of one region and subtract the area of another out of it. Here, if we subtract the area of the four circles out of the area of the square, the leftover area includes the shaded region but it includes other regions (around the corners etc) too. This is where symmetry helps us.
Image
Notice that we can split the figure into four equal regions to get four smaller squares. Now focus on the diagram give below which shows you one such smaller square. The area around the four corners of the smaller squares is equal i.e. the area of the red region = area of the blue region = area of the yellow region = area of the green region.
Image
Our shaded grey region has four such equal areas so

Area of the shaded grey region = Area of the smaller square – Area of one circle

Area of the shaded grey region = \((10)^2 – \pi(5)^2 = 100 – 25\pi\)

Answer (C). This question is discussed HERE.


[Reveal] Spoiler:
Attachment:
GeometryPost12Ques1.jpg
GeometryPost12Ques1.jpg [ 9.24 KiB | Viewed 34255 times ]

Attachment:
GeometryPost12Ques2.jpg
GeometryPost12Ques2.jpg [ 7.89 KiB | Viewed 30558 times ]

Attachment:
GeometryPost12Ques2Fig2.jpg
GeometryPost12Ques2Fig2.jpg [ 8.58 KiB | Viewed 30536 times ]

Attachment:
GeometryPost12Ques2Fig3.jpg
GeometryPost12Ques2Fig3.jpg [ 3.78 KiB | Viewed 30762 times ]

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Re: INSCRIBED AND CIRCUMSCRIBED POLYGONS ON THE GMAT [#permalink]

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New post 03 Jun 2015, 04:29

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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: INSCRIBED AND CIRCUMSCRIBED POLYGONS ON THE GMAT [#permalink]

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Re: INSCRIBED AND CIRCUMSCRIBED POLYGONS ON THE GMAT [#permalink]

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New post 08 Jul 2015, 02:46
Bunuel wrote:

INSCRIBED AND CIRCUMSCRIBED POLYGONS ON THE GMAT

REGULAR POLYGONS AND THE IRREGULAR ONES


By Karishma, Veritas Prep.


Continuing our Geometry journey, let’s discuss polygons today.

Some years back, I used to often get confused in the polygon sum-of-the-interior-angles formula if I had to recall it after a gap of some months because I had seen two variations of it:

Sum of interior angles of a polygon = (n – 2)*180
Sum of interior angles of a polygon = (2n – 4)*90

Now, I don’t want you to judge me. Of course, in the second formula, 2 has been removed from 180 and multiplied to the first factor. It is quite simple so why would anyone get tricked here, you wonder? The problem was that after a few months, I would somehow remember (2n – 4) and 180. So I was mixing up the two and I wasn’t sure of the logic behind this formula. That is until I came across the simple explanation of this formula in our Veritas Prep Geometry book (the one which explains how you can divide every polygon with n sides into (n – 2) triangles and hence get the sum of (n – 2)*180). Now it made perfect sense! I couldn’t believe that I had not come across that explanation before and had just learned up (well, tried to!) the formula blindly. So now I ensure that all my students understand every formula that I teach them.

Usually, we are given a regular polygon and we need to find the measure of interior angles or the number of sides. But what if we are given a polygon instead, not a regular polygon. Does this formula still apply? We wouldn’t know if we didn’t understand how the formula came into being. But since we know that we obtain the formula by dividing the polygon into (n-2) triangles, we know that the sum of all interior angles of a triangle is 180 irrespective of the kind of triangle. So it doesn’t matter whether the polygon is regular or not. The sum of all interior angles will still be (n-2)*180.

Let’s look at a question to see the application of this formula in irregular polygon scenario.

Question: The measures of the interior angles in a polygon are consecutive odd integers. The largest angle measures 153 degrees. How many sides does this polygon have?
A) 8
B) 9
C) 10
D) 11
E) 12

Solution:
The interior angles are: 153, 151, 149, 147 … and so on.

Now there are two ways to approach this question – one which is straight forward but uses algebra so is time consuming, another which makes you think but doesn’t take much time. You can guess which one we are going to focus on! But before we do that let’s take a quick look at the algebraic solution too.

Method 1: Algebra

Sum of interior angles of this polygon = \(153 + 151 + 149 + … (153 – 2(n-1)) = (n – 2)*180\).

If there are n sides, there are n interior angles. The second largest angle will be 153 – 2*1. The third largest will be 153 – 2*2. The smallest will be 153 – 2*(n-1). This is an arithmetic progression.

Sum of all terms = \(\frac{(First term + Last term)}{2} * n = \frac{(153 + 153 – 2(n-1))}{2} * n\)

Equating, we get \(\frac{(153 + 153 – 2(n-1))}{2} * n = (n – 2)*180\)

Solving this you get, n = 10

But let’s figure out a solution without going through this painful calculation.

Method 2: Capitalize on what you know

Angles of the polygon: 153, 151, 149, 147, 145, 143, 141, … , (153 – 2(n-1))

The average of these angles must be equal to the measure of each interior angle of a regular polygon with n sides since the sum of all angles is the same in both the cases.

Measure of each interior angle of n sided regular polygon = Sum of all angles / n = \(\frac{(n-2)*180}{n}\)

Using the options:

Measure of each interior angle of 8 sided regular polygon = 180*6/8 = 135 degrees

Measure of each interior angle of 9 sided regular polygon = 180*7/9 = 140 degrees

Measure of each interior angle of 10 sided regular polygon = 180*8/10 = 144 degrees

Measure of each interior angle of 11 sided regular polygon = 180*9/11 = 147 degrees apprx

and so on…

Notice that the average of the given angles can be 144 if there are 10 angles.

The average cannot be higher than 144 i.e. 147 since that will give us only 7 sides (153, 151, 149, 147, 145, 143, 141 – the average is 147 is this case). But the regular polygon with interior angle measure of 147 has 11 sides. Similarly, the average cannot be less than 144 i.e. 140 either because that will give us many more sides than the required 9.

Hence, the polygon must have 10 sides.

Answer (C).



Im confused about method 2.
Please explain in detail.
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Re: INSCRIBED AND CIRCUMSCRIBED POLYGONS ON THE GMAT [#permalink]

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Re: INSCRIBED AND CIRCUMSCRIBED POLYGONS ON THE GMAT [#permalink]

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New post 27 Nov 2017, 19:31
Bunuel wrote:

QUESTIONS ON CIRCLES INSCRIBED IN POLYGONS


By Karishma, Veritas Prep.


Above we looked at questions on polygons inscribed in a circle. This week, let’s look at questions on circles inscribed in regular polygons. As noted earlier, it’s important to keep in mind that regular polygons are symmetrical figures. You need very little information to solve for anything in a symmetrical figure.

Question 1: A circle is inscribed in a regular hexagon. A regular hexagon is inscribed in this circle. Another circle is inscribed in the inner regular hexagon and so on. What is the area of the tenth such circle?
Image
Statement I: The length of the side of the outermost regular hexagon is 6 cm.
Statement II: The length of a diagonal of the outermost regular hexagon is 12 cm.

Solution: Thankfully, in DS questions, we don’t need to calculate the answer. We just need to establish the sufficiency of the given data. Note that we have found that there is a defined relation between the sides of a regular hexagon and the radius of an inscribed circle and there is also a defined relation between the radius of a circle and the side of an inscribed regular hexagon.

When the circle is inscribed in a regular hexagon,

Radius of the inscribed circle = \(\frac{\sqrt{3}}{2}\)* Side of the hexagon

When a regular hexagon is inscribed in a circle,

Side of the inscribed regular hexagon = Radius of the circle

So all we need is the side of any one regular hexagon or the radius of any one circle and we will know the length of the sides of all hexagons and the radii of all circles.

Statement I: The length of the side of the outermost regular hexagon is 6 cm.
If length of the side of the outermost regular hexagon is 6 cm, the radius of the inscribed circle is \((\sqrt{3}/2)*6 = 3\sqrt{3}\) cm

In that case, the side of the regular hexagon inscribed in this circle is also \(3\sqrt{3}\) cm. Now we can get the radius of the circle inscribed in this second hexagon and go on the same lines till we reach the tenth circle. This statement alone is sufficient.

Statement II: The length of a diagonal of the outermost regular hexagon is 12 cm.

Note that a hexagon has diagonals of two different lengths. The diagonals that connect vertices with one vertex between them are smaller than the diagonals that connect vertices with two vertices between them. Length of AC will be shorter than length of AD. Given the length of a diagonal, we do not know which diagonal it is. Is AC = 12 or is AD = 12? The length of the side will be different in the two cases. So this statement alone is not sufficient.

Answer (A). This question is discussed HERE.

Keep in mind that you don’t actually need to solve for an answer is DS; in fact, in some questions you will not be able to solve for the answer under the given time constraints. All you need to do is ensure that given unlimited time, you will get a unique answer.

Question 2: Four identical circles are drawn in a square such that each circle touches two sides of the square and two other circles (as shown in the figure below). If the side of the square is of length 20 cm, what is the area of the shaded region?
Image
(A) 400 – 100π
(B) 200 – 50π
(C) 100 – 25π
(D) 8π
(E) 4π

Solution: First let’s recall that squares and circles are symmetrical figures. The given figure is symmetrical.

We don’t know any formula that will help us get the area of the curved shaded grey shape in the center. In such cases, very often what you need is to find the area of one region and subtract the area of another out of it. Here, if we subtract the area of the four circles out of the area of the square, the leftover area includes the shaded region but it includes other regions (around the corners etc) too. This is where symmetry helps us.
Image
Notice that we can split the figure into four equal regions to get four smaller squares. Now focus on the diagram give below which shows you one such smaller square. The area around the four corners of the smaller squares is equal i.e. the area of the red region = area of the blue region = area of the yellow region = area of the green region.
Image
Our shaded grey region has four such equal areas so

Area of the shaded grey region = Area of the smaller square – Area of one circle

Area of the shaded grey region = \((10)^2 – \pi(5)^2 = 100 – 25\pi\)

Answer (C). This question is discussed HERE.


[Reveal] Spoiler:
Attachment:
GeometryPost12Ques1.jpg

Attachment:
GeometryPost12Ques2.jpg

Attachment:
GeometryPost12Ques2Fig2.jpg

Attachment:
GeometryPost12Ques2Fig3.jpg


Thank you Karishma, excellent post.

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Re: INSCRIBED AND CIRCUMSCRIBED POLYGONS ON THE GMAT   [#permalink] 27 Nov 2017, 19:31
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