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555-605 (Medium)|   Geometry|      
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Mastering Important Concept on Triangles – III - Question #2 11 Dec 2016, 09:51
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Mastering Important Concept on Triangles – III - Question #2


In the given figure, BD = 6 and DC =4. If angle BAD = \(30^o\) and angle ADC = \(120^o\), find the area of the triangle ADC.




Answer Choices

    a. \(6√3 cm^2\)
    b. \(8√3 cm^2\)
    c. \(10√3 cm^2\)
    d. \(12√3 cm^2\)
    e. \(18√3 cm^2\)





Key concepts on Triangles are explained in detail in the following posts:

1. Mastering Important Concepts Tested By GMAT in Triangle - I

2. Mastering Important Concepts Tested By GMAT in Triangle - II

3.Mastering Important Concepts Tested By GMAT in Triangle - III


Detailed solution will be posted soon. :)

Thanks,
Saquib
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D
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Mastering Important Concept on Triangles – III - Question #2 Updated on: 16 Dec 2016, 00:38
Originally posted by EgmatQuantExpert on 11 Dec 2016, 09:53.
Last edited by EgmatQuantExpert on 16 Dec 2016, 00:38, edited 1 time in total.
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Hey,

PFB the official solution.

It is given that angle \(BAD = 30^o\) and angle \(ADC = 120^o\)

Since we know that
    • The External Angle = Sum of Opposite Internal Angles

We can write
    • angle ADC =angle ABD + angle BAD
    • \(120^0 = ABD + 30^0\)
    • \(ABD = 90^0\)

Thus, we can conclude that triangle \(ABD\) is a right angle triangle!

To find the area of triangle \(ADC\), we need the value of the base \(DC\), which is given as 4 and we need the value of the height \(AB\).

The triangle \(ABD\) is a \(30^o-60^o-90^o\) triangle,therefore the sides \(BD : AB : AD\) are in the ratio - \(1 : √3 : 2\)

As we know the value of \(BD = 6\), we can easily find the value of \(AB\).
    • \(BD/AB = 1/√3\)
    • \(AB = 6√3\)

Thus the area of triangle \(ADC\)
    \(= 1/2 * DC * AB\)
    \(= 1/2 * 4 * 6√3\)
    \(= 12√3\)

And the correct answer option is D
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Mastering Important Concept on Triangles – III - Question #2 11 Dec 2016, 10:48
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angle ADC=120 so angle ADB=60,
IN triangle ADB, angle ADB=60, angle BAD=30, therefore angle ABD=90. Its a right angle triangle.
therefore AB/BD=tan30, AB/6=√3, AB=6√3
For triangle ADC, AB is height and DC is base,
therefore are of triangle ADC=1/2*AB*DC=1/2*6√3*4=12√3
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Mastering Important Concept on Triangles – III - Question #2 12 Dec 2016, 10:48
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Mastering Important Concept on Triangles – III - Question #2


In the given figure, BD = 6 and DC =4. If angle BAD = \(30^o\) and angle ADC = \(120^o\), find the area of the triangle ADC.




Answer Choices

    a. \(6√3 cm^2\)
    b. \(8√3 cm^2\)
    c. \(10√3 cm^2\)
    d. \(12√3 cm^2\)
    e. \(18√3 cm^2\)





[size=110]Key concepts on Triangles are explained in detail in the following posts:

Show more


since ADC=120 and BAD=30 we can find all angles in triangle ABD with ext. angle property.
thus ABD is 90-60-30 triangle with BD=6
thus we find other sides too as 2:√3:1 ratio as shown in fig as AB=6√3,BD=6 & AD=12

area of triangle ABC- ABD =area ADC

1/2*10*6√3 - 1/2*6√3*6 ==30√3 - 18√3 ==12√3

Ans D
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Mastering Important Concept on Triangles – III - Question #2 16 Dec 2016, 00:39
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Hey Everyone,

The official solution has been posted. Please go through the solution once. In case of any doubts please feel to ask :)

Thanks,
Saquib
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Mastering Important Concept on Triangles – III - Question #2 18 Feb 2020, 05:41
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Hey,

PFB the official solution.

It is given that angle \(BAD = 30^o\) and angle \(ADC = 120^o\)

Since we know that
    • The External Angle = Sum of Opposite Internal Angles

We can write
    • angle ADC =angle ABD + angle BAD
    • \(120^0 = ABD + 30^0\)
    • \(ABD = 90^0\)

Thus, we can conclude that triangle \(ABD\) is a right angle triangle!

To find the area of triangle \(ADC\), we need the value of the base \(DC\), which is given as 4 and we need the value of the height \(AB\).

The triangle \(ABD\) is a \(30^o-60^o-90^o\) triangle,therefore the sides \(BD : AB : AD\) are in the ratio - \(1 : √3 : 2\)

As we know the value of \(BD = 6\), we can easily find the value of \(AB\).
    • \(BD/AB = 1/√3\)
    • \(AB = 6√3\)

Thus the area of triangle \(ADC\)
    \(= 1/2 * DC * AB\)
    \(= 1/2 * 4 * 6√3\)
    \(= 12√3\)

And the correct answer option is D
Show more

How is AB the height of triangle ADC?

Thanks :please
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Mastering Important Concept on Triangles – III - Question #2 18 Feb 2020, 05:59
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Hey,

PFB the official solution.

It is given that angle \(BAD = 30^o\) and angle \(ADC = 120^o\)

Since we know that
    • The External Angle = Sum of Opposite Internal Angles

We can write
    • angle ADC =angle ABD + angle BAD
    • \(120^0 = ABD + 30^0\)
    • \(ABD = 90^0\)

Thus, we can conclude that triangle \(ABD\) is a right angle triangle!

To find the area of triangle \(ADC\), we need the value of the base \(DC\), which is given as 4 and we need the value of the height \(AB\).

The triangle \(ABD\) is a \(30^o-60^o-90^o\) triangle,therefore the sides \(BD : AB : AD\) are in the ratio - \(1 : √3 : 2\)

As we know the value of \(BD = 6\), we can easily find the value of \(AB\).
    • \(BD/AB = 1/√3\)
    • \(AB = 6√3\)

Thus the area of triangle \(ADC\)
    \(= 1/2 * DC * AB\)
    \(= 1/2 * 4 * 6√3\)
    \(= 12√3\)

And the correct answer option is D

How is AB the height of triangle ADC?

Thanks :please
Show more


since we find angle abc is 90deg. ab is not only height of ADC but also height of ABC triangle
height is always measured in 90 deg.
if u want your height to be measured ,u have to stand right of 90 deg. otherwise measurement would not be correct.
hope it clears ur doubt
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Mastering Important Concept on Triangles – III - Question #2 10 Nov 2020, 21:26
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Hey,

PFB the official solution.

It is given that angle \(BAD = 30^o\) and angle \(ADC = 120^o\)

Since we know that
    • The External Angle = Sum of Opposite Internal Angles

We can write
    • angle ADC =angle ABD + angle BAD
    • \(120^0 = ABD + 30^0\)
    • \(ABD = 90^0\)

Thus, we can conclude that triangle \(ABD\) is a right angle triangle!

To find the area of triangle \(ADC\), we need the value of the base \(DC\), which is given as 4 and we need the value of the height \(AB\).

The triangle \(ABD\) is a \(30^o-60^o-90^o\) triangle,therefore the sides \(BD : AB : AD\) are in the ratio - \(1 : √3 : 2\)

As we know the value of \(BD = 6\), we can easily find the value of \(AB\).
    • \(BD/AB = 1/√3\)
    • \(AB = 6√3\)

Thus the area of triangle \(ADC\)
    \(= 1/2 * DC * AB\)
    \(= 1/2 * 4 * 6√3\)
    \(= 12√3\)

And the correct answer option is D

How is AB the height of triangle ADC?

Thanks :please


since we find angle abc is 90deg. ab is not only height of ADC but also height of ABC triangle
height is always measured in 90 deg.
if u want your height to be measured ,u have to stand right of 90 deg. otherwise measurement would not be correct.
hope it clears ur doubt
Show more


I am still unclear how we know that AB is height of ADC. If it's because of the 90 deg angle, what about BD? Doesn't that also touch the 90 deg angle? I'm a bit confused on how to identify height for these types of problems, where it doesn't seem obvious.
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Mastering Important Concept on Triangles – III - Question #2 11 Nov 2020, 03:04
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EgmatQuantExpert
Mastering Important Concept on Triangles – III - Question #2


In the given figure, BD = 6 and DC =4. If angle BAD = \(30^o\) and angle ADC = \(120^o\), find the area of the triangle ADC.




Answer Choices

    a. \(6√3 cm^2\)
    b. \(8√3 cm^2\)
    c. \(10√3 cm^2\)
    d. \(12√3 cm^2\)
    e. \(18√3 cm^2\)





Key concepts on Triangles are explained in detail in the following posts:

1. Mastering Important Concepts Tested By GMAT in Triangle - I

2. Mastering Important Concepts Tested By GMAT in Triangle - II

3.Mastering Important Concepts Tested By GMAT in Triangle - III


Detailed solution will be posted soon. :)

Thanks,
Saquib
Quant Expert
e-GMAT

Show more


Hi EgmatQuantExpert ,

I have a doubt.

Why can't we draw a perpendicular bisector from D to AC (example DE) and thus split △ADC into 2 30-60-90 triangles.?

If we do so then, we get AE as 2√3 and EC as 2√3. DE as 2.
Therefore, AC = 4√3 becomes our base and DE = 2 becomes our height. Thus giving us area of 4√3.

What is wrong in this approach?

I don't see anything wrong.. please let me know if I'm missing anything.

Tagging others just in case
Bunuel chetan2u GMATinsight IanStewart ScottTargetTestPrep yashikaaggarwal VeritasKarishma egmat EMPOWERgmatRichC

Thank you :)

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Mastering Important Concept on Triangles – III - Question #2 11 Nov 2020, 06:15
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Nups1324

I have a doubt.

Why can't we draw a perpendicular bisector from D to AC (example DE) and thus split △ADC into 2 30-60-90 triangles.?
Show more

When you draw a height in a triangle (which is what you're doing, when you draw your line DE), you're only ever dividing the triangle perfectly in half if your triangle is isosceles (or equilateral), and you're drawing the height between the two equal sides. In this question, AD and DC are not the same length, so the line DE you're trying to draw will not be a bisector -- you won't get two 30-60-90 triangles when you draw that line. Rounding off to integers, it turns out you'll actually get one 14-76-90 triangle, and one 46-44-90 triangle when you draw the line DE (I used trigonometry to work that out, which you'd never need to know for a GMAT question).
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Mastering Important Concept on Triangles – III - Question #2 11 Nov 2020, 21:17
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Nups1324

I have a doubt.

Why can't we draw a perpendicular bisector from D to AC (example DE) and thus split △ADC into 2 30-60-90 triangles.?

When you draw a height in a triangle (which is what you're doing, when you draw your line DE), you're only ever dividing the triangle perfectly in half if your triangle is isosceles (or equilateral), and you're drawing the height between the two equal sides. In this question, AD and DC are not the same length, so the line DE you're trying to draw will not be a bisector -- you won't get two 30-60-90 triangles when you draw that line. Rounding off to integers, it turns out you'll actually get one 14-76-90 triangle, and one 46-44-90 triangle when you draw the line DE (I used trigonometry to work that out, which you'd never need to know for a GMAT question).
Show more

Hi! Could you help explain how we know that AB is height of ADC. If it's because of the 90 deg angle, what about BD? Doesn't that also touch the 90 deg angle? I'm a bit confused on how to identify height for these types of problems, where it doesn't seem obvious.
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Mastering Important Concept on Triangles – III - Question #2 11 Nov 2020, 23:52
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Nups1324

I have a doubt.

Why can't we draw a perpendicular bisector from D to AC (example DE) and thus split △ADC into 2 30-60-90 triangles.?

When you draw a height in a triangle (which is what you're doing, when you draw your line DE), you're only ever dividing the triangle perfectly in half if your triangle is isosceles (or equilateral), and you're drawing the height between the two equal sides. In this question, AD and DC are not the same length, so the line DE you're trying to draw will not be a bisector -- you won't get two 30-60-90 triangles when you draw that line. Rounding off to integers, it turns out you'll actually get one 14-76-90 triangle, and one 46-44-90 triangle when you draw the line DE (I used trigonometry to work that out, which you'd never need to know for a GMAT question).
Show more

Oh yes. I overlooked the fact that AD and DC are not equal.

Thank you so much :)

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Mastering Important Concept on Triangles – III - Question #2 11 Nov 2020, 23:58
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Nups1324

I have a doubt.

Why can't we draw a perpendicular bisector from D to AC (example DE) and thus split △ADC into 2 30-60-90 triangles.?

When you draw a height in a triangle (which is what you're doing, when you draw your line DE), you're only ever dividing the triangle perfectly in half if your triangle is isosceles (or equilateral), and you're drawing the height between the two equal sides. In this question, AD and DC are not the same length, so the line DE you're trying to draw will not be a bisector -- you won't get two 30-60-90 triangles when you draw that line. Rounding off to integers, it turns out you'll actually get one 14-76-90 triangle, and one 46-44-90 triangle when you draw the line DE (I used trigonometry to work that out, which you'd never need to know for a GMAT question).

Hi! Could you help explain how we know that AB is height of ADC. If it's because of the 90 deg angle, what about BD? Doesn't that also touch the 90 deg angle? I'm a bit confused on how to identify height for these types of problems, where it doesn't seem obvious.
Show more

Hi,

If we take BD as the height then we must take AB as the base, since the height must be perpendicular to the base. Taking BD and AB will give us the area of triangle ABD.

Hope it's clear.

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Mastering Important Concept on Triangles – III - Question #2 12 Nov 2020, 21:05
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Quote:
Hi! Could you help explain how we know that AB is height of ADC. If it's because of the 90 deg angle, what about BD? Doesn't that also touch the 90 deg angle? I'm a bit confused on how to identify height for these types of problems, where it doesn't seem obvious.
Show more

Quote:
Hi,

If we take BD as the height then we must take AB as the base, since the height must be perpendicular to the base. Taking BD and AB will give us the area of triangle ABD.

Hope it's clear.

Posted from my mobile device
Show more

Thank you for your response! My confusion is that EgmatQuantExpert said:
"To find the area of triangle ADC, we need the value of the base DC, which is given as 4 and we need the value of the height AB."

I don't see how AB is the height of triangle ADC. As far as I can tell, I don't see the connection between AB and triangle ADC. Am I missing something really obvious here?
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Mastering Important Concept on Triangles – III - Question #2 15 Nov 2020, 19:44
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Hey,

PFB the official solution.

It is given that angle \(BAD = 30^o\) and angle \(ADC = 120^o\)

Since we know that
    • The External Angle = Sum of Opposite Internal Angles

We can write
    • angle ADC =angle ABD + angle BAD
    • \(120^0 = ABD + 30^0\)
    • \(ABD = 90^0\)

Thus, we can conclude that triangle \(ABD\) is a right angle triangle!

To find the area of triangle \(ADC\), we need the value of the base \(DC\), which is given as 4 and we need the value of the height \(AB\).

The triangle \(ABD\) is a \(30^o-60^o-90^o\) triangle,therefore the sides \(BD : AB : AD\) are in the ratio - \(1 : √3 : 2\)

As we know the value of \(BD = 6\), we can easily find the value of \(AB\).
    • \(BD/AB = 1/√3\)
    • \(AB = 6√3\)

Thus the area of triangle \(ADC\)
    \(= 1/2 * DC * AB\)
    \(= 1/2 * 4 * 6√3\)
    \(= 12√3\)

And the correct answer option is D
Show more


Hi EgmatQuantExpert! I have a question on this part: "To find the area of triangle ADC, we need the value of the base DC, which is given as 4 and we need the value of the height AB."

I don't see how AB is the height of triangle ADC. As far as I can tell, I don't see the connection between AB and triangle ADC. Am I missing something really obvious here?
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Mastering Important Concept on Triangles – III - Question #2 15 Nov 2020, 23:58
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Quote:
Hi! Could you help explain how we know that AB is height of ADC. If it's because of the 90 deg angle, what about BD? Doesn't that also touch the 90 deg angle? I'm a bit confused on how to identify height for these types of problems, where it doesn't seem obvious.

Quote:
Hi,

If we take BD as the height then we must take AB as the base, since the height must be perpendicular to the base. Taking BD and AB will give us the area of triangle ABD.

Hope it's clear.

Posted from my mobile device

Thank you for your response! My confusion is that EgmatQuantExpert said:
"To find the area of triangle ADC, we need the value of the base DC, which is given as 4 and we need the value of the height AB."

I don't see how AB is the height of triangle ADC. As far as I can tell, I don't see the connection between AB and triangle ADC. Am I missing something really obvious here?
Show more

Triangle ADC is an obtuse triangle.
Think about how you get the area of an obtuse triangle:
Attachment:
drawingobtusetriangleheight.png
drawingobtusetriangleheight.png [ 2.74 KiB | Viewed 6664 times ]

The altitude may not fall on the base because of the obtuse angle but nevertheless, the altitude has to be the perpendicular as shown in the diagram above.

Area = (1/2)*base*altitude
We have found that AB is perpendicular so it is the altitude (you can redraw the diagram according to given values and see that angle ABD is a right triangle).
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Mastering Important Concept on Triangles – III - Question #2 16 Nov 2020, 22:45
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Quote:
Thank you for your response! My confusion is that EgmatQuantExpert said:
"To find the area of triangle ADC, we need the value of the base DC, which is given as 4 and we need the value of the height AB."

I don't see how AB is the height of triangle ADC. As far as I can tell, I don't see the connection between AB and triangle ADC. Am I missing something really obvious here?
Show more

Quote:
Triangle ADC is an obtuse triangle.
Think about how you get the area of an obtuse triangle:
Attachment:
The attachment drawingobtusetriangleheight.png is no longer available

The altitude may not fall on the base because of the obtuse angle but nevertheless, the altitude has to be the perpendicular as shown in the diagram above.

Area = (1/2)*base*altitude
We have found that AB is perpendicular so it is the altitude (you can redraw the diagram according to given values and see that angle ABD is a right triangle).
Show more

Thank you VeritasKarishma! What if triangle ABD looked like the following where angle BAD is 90 degree instead? Which segment would then be the height of obtuse triangle ADC? Apologies for the rough drawing - I don't have the proper tools to make it look better.
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Mastering Important Concept on Triangles – III - Question #2 21 Nov 2020, 21:58
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Quote:
Thank you for your response! My confusion is that EgmatQuantExpert said:
"To find the area of triangle ADC, we need the value of the base DC, which is given as 4 and we need the value of the height AB."

I don't see how AB is the height of triangle ADC. As far as I can tell, I don't see the connection between AB and triangle ADC. Am I missing something really obvious here?

Quote:
Triangle ADC is an obtuse triangle.
Think about how you get the area of an obtuse triangle:
Attachment:
The attachment drawingobtusetriangleheight.png is no longer available

The altitude may not fall on the base because of the obtuse angle but nevertheless, the altitude has to be the perpendicular as shown in the diagram above.

Area = (1/2)*base*altitude
We have found that AB is perpendicular so it is the altitude (you can redraw the diagram according to given values and see that angle ABD is a right triangle).

Thank you VeritasKarishma! What if triangle ABD looked like the following where angle BAD is 90 degree instead? Which segment would then be the height of obtuse triangle ADC? Apologies for the rough drawing - I don't have the proper tools to make it look better.
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Hi VeritasKarishma, wanted to follow up on this one. What if triangle ABD looked like the following where angle BAD is 90 degree instead? Which segment would then be the height of obtuse triangle ADC? Apologies again for the rough drawing.
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KarishmaB
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Mastering Important Concept on Triangles – III - Question #2 23 Nov 2020, 22:15
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Quote:
Thank you for your response! My confusion is that EgmatQuantExpert said:
"To find the area of triangle ADC, we need the value of the base DC, which is given as 4 and we need the value of the height AB."

I don't see how AB is the height of triangle ADC. As far as I can tell, I don't see the connection between AB and triangle ADC. Am I missing something really obvious here?

Quote:
Triangle ADC is an obtuse triangle.
Think about how you get the area of an obtuse triangle:
Attachment:
drawingobtusetriangleheight.png

The altitude may not fall on the base because of the obtuse angle but nevertheless, the altitude has to be the perpendicular as shown in the diagram above.

Area = (1/2)*base*altitude
We have found that AB is perpendicular so it is the altitude (you can redraw the diagram according to given values and see that angle ABD is a right triangle).

Thank you VeritasKarishma! What if triangle ABD looked like the following where angle BAD is 90 degree instead? Which segment would then be the height of obtuse triangle ADC? Apologies for the rough drawing - I don't have the proper tools to make it look better.

Hi VeritasKarishma, wanted to follow up on this one. What if triangle ABD looked like the following where angle BAD is 90 degree instead? Which segment would then be the height of obtuse triangle ADC? Apologies again for the rough drawing.
Show more


I hope you understand that the figure you have given cannot be the figure of this question because BDC needs to be a straight line.

The figure you have presented is a quadrilateral. If we were to find the area of triangle ADC in this figure, we would extend DC to the left and make an altitude fall from A as shown in my previous figure by blue dotted lines. Or we could make an altitude drop from D on to AC (whatever works as per data given in the question).
The measure of angle BAD is irrelevant to get the area of triangle ADC.
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Mastering Important Concept on Triangles – III - Question #2 04 Dec 2020, 02:58
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Mastering Important Concept on Triangles – III - Question #2


In the given figure, BD = 6 and DC =4. If angle BAD = \(30^o\) and angle ADC = \(120^o\), find the area of the triangle ADC.




Answer Choices

    a. \(6√3 cm^2\)
    b. \(8√3 cm^2\)
    c. \(10√3 cm^2\)
    d. \(12√3 cm^2\)
    e. \(18√3 cm^2\)









Hi EgmatQuantExpert ,

I have a doubt.

Why can't we draw a perpendicular bisector from D to AC (example DE) and thus split △ADC into 2 30-60-90 triangles.?

If we do so then, we get AE as 2√3 and EC as 2√3. DE as 2.
Therefore, AC = 4√3 becomes our base and DE = 2 becomes our height. Thus giving us area of 4√3.

What is wrong in this approach?

I don't see anything wrong.. please let me know if I'm missing anything.



Thank you :)

Posted from my mobile device
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Solution:

We can indeed draw a perpendicular from D to AC or draw the bisector of the side AC; however, the perpendicular and the bisector will not coincide since triangle ACD is not an isosceles triangle. Further, notice that neither angle DAC nor angle ACD is 30 degrees; thus, it is impossible to create two 30-60-90 triangles by drawing some line from the point D onto side AC.
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