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# In the diagram above, if arc ABC is a semicircle, what is

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In the diagram above, if arc ABC is a semicircle, what is [#permalink]

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26 Aug 2011, 01:24
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In the diagram above, if arc ABC is a semicircle, what is the length of AC?

(2) DC = 10

[Reveal] Spoiler:
Attachment:

cricle.jpg [ 23.94 KiB | Viewed 13007 times ]
[Reveal] Spoiler: OA

Last edited by Berbatov on 26 Aug 2011, 01:59, edited 1 time in total.
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26 Aug 2011, 02:41
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In triangle ABD, tan <bad = 5/2.5=2
Hence we know the value of <bad
=2.5^2+5^2= 31.25
AB =$$\sqrt{31.25}$$
In triangle ABC, <ABC is right angle.
We know one side (AB) and one angel (<bad) in triangle ABC. Sufficient to find AC.

Stmt2: By similar reason, in triangle BDC we can find out <bcd
Also, BC^2=5^2+10^=125
BC = $$\sqrt{125}$$
In triangle ABC we know one side (BC) and one angel (<bcd). Sufficient to find AC.

OA D.
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03 Nov 2011, 10:45
D. It can be visualized if you split the triangle into 2; then notice you have a proportion with 3 numbers missing 1.
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14 Jan 2012, 16:37
I'm not able to understand the ans to this q.. could someone elaborate please?
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14 Jan 2012, 17:27
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karthiksms wrote:
I'm not able to understand the ans to this q.. could someone elaborate please?
Attachment:

12.JPG [ 15.74 KiB | Viewed 14173 times ]
In the diagram above, if arc ABC is a semicircle, what is the length of AC?

You should know the following properties to solve this question:
A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle.

So, as given that AC is a diameter then angle ABC is a right angle.

Perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.

Thus, the perpendicular BD divides right triangle ABC into two similar triangles ADB and BDC (which are also similar to big triangle ABC). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles). For example: AB/AC=AD/AB=BD/BC. This property (sometimes along with Pythagoras) will give us the following: if we know ANY 2 values from AB, AD, AC, BC, BD, CD then we'll be able to find the other 4. We are given that BD=5 thus to find AC we need to know the length of any other line segment.

Also in such kind of triangles might be useful to equate the areas to find the length of some line segment, for example area of ABC=1/2*AC*BD=1/2*AB*BC (for more check: triangles-106177.html, geometry-problem-106009.html, mgmat-ds-help-94037.html, help-108776.html)

(2) DC = 10. Sufficient.

Hope it helps.
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Re: In the diagram above, if arc ABC is a semicircle, what is [#permalink]

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22 May 2013, 03:40
Bumping for review and further discussion.
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24 May 2014, 14:16
Bunuel wrote:

Thus, the perpendicular BD divides right triangle ABC into two similar triangles ADB and BDC (which are also similar to big triangle ABC). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles). For example: AB/AC=AD/AB=BD/BC. This property (sometimes along with Pythagoras) will give us the following: if we know ANY 2 values from AB, AD, AC, BC, BD, CD then we'll be able to find the other 4.We are given that BD=5 thus to find AC we need to know the length of any other line segment.

Hope it helps.

Hi Bunuel,

I have a theoretical question based on the highlighted statement above:

It seems as though there is always overlap of sides(since we have similar triangles) so your highlighted statement looks to be true. For the sake of saving time -- i was trying to get a deeper understanding of this method -- Can there ever be a case when they give us the length of two sides, but since they are ONLY of one triangle, we can't make the leap onto others(minus the obvious correlations between the ratio) Meaning, In this same problem, if Statement 1 and 2 were given but if we weren't given a value of BD, then we wouldn't have been able to solve it. Correct?

Is it safe to say that two sides will ALWAYS create sufficiency, even if the two sides are on the same triangle?
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24 May 2014, 14:29
russ9 wrote:
Bunuel wrote:

Thus, the perpendicular BD divides right triangle ABC into two similar triangles ADB and BDC (which are also similar to big triangle ABC). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles). For example: AB/AC=AD/AB=BD/BC. This property (sometimes along with Pythagoras) will give us the following: if we know ANY 2 values from AB, AD, AC, BC, BD, CD then we'll be able to find the other 4.We are given that BD=5 thus to find AC we need to know the length of any other line segment.

Hope it helps.

Hi Bunuel,

I have a theoretical question based on the highlighted statement above:

It seems as though there is always overlap of sides(since we have similar triangles) so your highlighted statement looks to be true. For the sake of saving time -- i was trying to get a deeper understanding of this method -- Can there ever be a case when they give us the length of two sides, but since they are ONLY of one triangle, we can't make the leap onto others(minus the obvious correlations between the ratio) Meaning, In this same problem, if Statement 1 and 2 were given but if we weren't given a value of BD, then we wouldn't have been able to solve it. Correct?

Is it safe to say that two sides will ALWAYS create sufficiency, even if the two sides are on the same triangle?

If we were not give the length of BD, then the answer would be C: 2.5/BD = BD/10 --> BD = 5 --> we can find AC.
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17 Jun 2014, 20:09
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Bunuel wrote:
karthiksms wrote:
I'm not able to understand the ans to this q.. could someone elaborate please?
Attachment:
12.JPG
In the diagram above, if arc ABC is a semicircle, what is the length of AC?

You should know the following properties to solve this question:
A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle.

So, as given that AC is a diameter then angle ABC is a right angle.

Perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.

Thus, the perpendicular BD divides right triangle ABC into two similar triangles ADB and BDC (which are also similar to big triangle ABC). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles). For example: AB/AC=AD/AB=BD/BC. This property (sometimes along with Pythagoras) will give us the following: if we know ANY 2 values from AB, AD, AC, BC, BD, CD then we'll be able to find the other 4. We are given that BD=5 thus to find AC we need to know the length of any other line segment.

Also in such kind of triangles might be useful to equate the areas to find the length of some line segment, for example area of ABC=1/2*AC*BD=1/2*AB*BC (for more check: triangles-106177.html, geometry-problem-106009.html, mgmat-ds-help-94037.html, help-108776.html)

(2) DC = 10. Sufficient.

Hope it helps.

Hi Bunuel, AB/AC=AD/AB=BD/BC means that AB and AC are corresponding , and also AD and AB are corresponding and BD and BC are coresponding angles. Please tell which angles are they opposite to, because AB is opposite to angle BCA, and AC is opposite to angle ABC. How are these two angles corresponding. Same doubt for other corresponding angles also. Please clarify.
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18 Jun 2014, 01:58
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thoufique wrote:
Bunuel wrote:

In the diagram above, if arc ABC is a semicircle, what is the length of AC?

You should know the following properties to solve this question:
A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle.

So, as given that AC is a diameter then angle ABC is a right angle.

Perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.

Thus, the perpendicular BD divides right triangle ABC into two similar triangles ADB and BDC (which are also similar to big triangle ABC). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles). For example: AB/AC=AD/AB=BD/BC. This property (sometimes along with Pythagoras) will give us the following: if we know ANY 2 values from AB, AD, AC, BC, BD, CD then we'll be able to find the other 4. We are given that BD=5 thus to find AC we need to know the length of any other line segment.

Also in such kind of triangles might be useful to equate the areas to find the length of some line segment, for example area of ABC=1/2*AC*BD=1/2*AB*BC (for more check: triangles-106177.html, geometry-problem-106009.html, mgmat-ds-help-94037.html, help-108776.html)

(2) DC = 10. Sufficient.

Hope it helps.

Hi Bunuel, AB/AC=AD/AB=BD/BC means that AB and AC are corresponding , and also AD and AB are corresponding and BD and BC are coresponding angles. Please tell which angles are they opposite to, because AB is opposite to angle BCA, and AC is opposite to angle ABC. How are these two angles corresponding. Same doubt for other corresponding angles also. Please clarify.

In similar triangles corresponding sides are the sides opposite the same angles.

In triangle ABC: AB is opposite blue angle, AC is opposite right angle.
In triangle ADB: AD is opposite blue angle, AB is opposite right angle.
In triangle BDC: BD is opposite blue angle, BC is opposite right angle.

Since the above three triangles are similar then the ratio of these sides must be the same.

Hope it's clear.
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18 Jun 2014, 07:06
Thanks a lot bunuel , its clear now.
Can the ratios be taken using red angles also. In that case we get BC/AC = BD/AB = CD/BC.
If yes, then depending upon the unknown, we can consider appropriate corresponding angles.
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18 Jun 2014, 08:40
thoufique wrote:
Thanks a lot bunuel , its clear now.
Can the ratios be taken using red angles also. In that case we get BC/AC = BD/AB = CD/BC.
If yes, then depending upon the unknown, we can consider appropriate corresponding angles.

BC/AC = BD/AB = CD/BC is correct. And yes, you can equate ratios of any pair of corresponding angles.
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18 Jun 2014, 09:57
Bunuel wrote:
thoufique wrote:
Thanks a lot bunuel , its clear now.
Can the ratios be taken using red angles also. In that case we get BC/AC = BD/AB = CD/BC.
If yes, then depending upon the unknown, we can consider appropriate corresponding angles.

BC/AC = BD/AB = CD/BC is correct. And yes, you can equate ratios of any pair of corresponding angles.

Thanks a ton bunuel, was struggling with this for a long time.
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Re: In the diagram above, if arc ABC is a semicircle, what is [#permalink]

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27 Jun 2015, 13:08
Hello from the GMAT Club BumpBot!

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Re: In the diagram above, if arc ABC is a semicircle, what is [#permalink]

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14 Jun 2016, 02:20
Bunuel wrote:
thoufique wrote:
Bunuel wrote:
[/img]
In the diagram above, if arc ABC is a semicircle, what is the length of AC?

You should know the following properties to solve this question:
A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle.

So, as given that AC is a diameter then angle ABC is a right angle.

Perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.

Thus, the perpendicular BD divides right triangle ABC into two similar triangles ADB and BDC (which are also similar to big triangle ABC). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles). For example: AB/AC=AD/AB=BD/BC. This property (sometimes along with Pythagoras) will give us the following: if we know ANY 2 values from AB, AD, AC, BC, BD, CD then we'll be able to find the other 4. We are given that BD=5 thus to find AC we need to know the length of any other line segment.

Also in such kind of triangles might be useful to equate the areas to find the length of some line segment, for example area of ABC=1/2*AC*BD=1/2*AB*BC \

(2) DC = 10. Sufficient.

Hope it helps.

Hi Bunuel, AB/AC=AD/AB=BD/BC means that AB and AC are corresponding , and also AD and AB are corresponding and BD and BC are coresponding angles. Please tell which angles are they opposite to, because AB is opposite to angle BCA, and AC is opposite to angle ABC. How are these two angles corresponding. Same doubt for other corresponding angles also. Please clarify.

In similar triangles corresponding sides are the sides opposite the same angles.

In triangle ABC: AB is opposite blue angle, AC is opposite right angle
.
In triangle ADB: AD is opposite blue angle, AB is opposite right angle.
In triangle BDC: BD is opposite blue angle, BC is opposite right angle.

Since the above three triangles are similar then the ratio of these sides must be the same.

Hope it's clear.

Bunuel, I understand this . However i don't understand this. How can you know AD is opposite blue and not red angle?

Thanks,
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Re: In the diagram above, if arc ABC is a semicircle, what is [#permalink]

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14 Jun 2016, 02:26
plalud wrote:
Bunuel, I understand this . However i don't understand this. How can you know AD is opposite blue and not red angle?

Thanks,

Please make a post so that it's clear what you are asking, quoting an image etc.

We k ow which angles are equal to each other, blue = blue and red = red. Corresponding sides are opposite equal angles.
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Re: In the diagram above, if arc ABC is a semicircle, what is [#permalink]

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14 Jun 2016, 03:23
Bunuel I cannot insert any U r l.
My question is:
In ABC we have a right, a blue and a red angle.
In ADB, we have a right, a blue and a red angle. The corresponding right one is obviously the right one. But how can we know which angle in ADB corresponds to the blue angle in ABC?

I hope i am clear.
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Re: In the diagram above, if arc ABC is a semicircle, what is [#permalink]

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14 Jun 2016, 07:07
plalud wrote:
Bunuel I cannot insert any U r l.
My question is:
In ABC we have a right, a blue and a red angle.
In ADB, we have a right, a blue and a red angle. The corresponding right one is obviously the right one. But how can we know which angle in ADB corresponds to the blue angle in ABC?

I hope i am clear.

$$\angle BAC + \angle ACB = 90°$$;
$$\angle BAC + \angle ABD = 90°$$;
Thus $$\angle ACB=\angle ABD$$.

Similarly:

$$\angle BAC + \angle ACB = 90°$$;
$$\angle CBD + \angle ACB = 90°$$;
Thus $$\angle BAC =\angle CBD$$.
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Re: In the diagram above, if arc ABC is a semicircle, what is   [#permalink] 14 Jun 2016, 07:07
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