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# What is the area of triangle BDC?

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What is the area of triangle BDC?  [#permalink]

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07 Dec 2019, 19:22
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GMATBusters’ Quant Quiz Question -10

What is the area of triangle BDC?
1) BD = 6
2) AC= 20

Attachment:

12222.jpg [ 19.12 KiB | Viewed 955 times ]

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Re: What is the area of triangle BDC?  [#permalink]

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07 Dec 2019, 21:59

Area=BD*CD/2

1) BD=6.no CD . Not sufficient
2)AC=20 Not sufficient

Both together also not sufficient.
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Re: What is the area of triangle BDC?  [#permalink]

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07 Dec 2019, 22:17
What is the area of triangle BDC?
1) BD = 6
2) AC= 20

Attachment:
12222.jpg

To find the area of a triangle, we require altitude and corresponding base in most of the triangles, unless it is equilateral etc where just one side is enough.
Here, we have BD and AC given and we can show that the three triangles formed are similar
ABC is similar to BDC is similar to ADB
x(20-x)=36......x^2-20x+36=0....(x-18)(x-2)=0
So x=18 or 2, that is CD could be 2 and 18. Although by looks we can say that AD >BD, and area should be 6, but we cannot be sure about it
as area = 1/2*6*2 =6 or 1/2*18*6=54
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Re: What is the area of triangle BDC?  [#permalink]

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07 Dec 2019, 22:25
What is the area of triangle BDC?

(Statement1): BD = 6
In order to find the area, there should be at least two info about the triangle.
Clearly insufficient.

(Statement2) AC= 20
In order to find the area, there should be at least two info about the triangle.
Clearly insufficient.

Taken together 1&2, $$BC^{2}+ AB^{2}=400$$

$$BC^{2}= 36+ DC^{2}$$
$$AB^{2}= 36+ (20-DC)^2$$
--> $$BC^{2}+ AB^{2}= 36+ DC^{2}+ 36+ (20-DC)^2$$
$$400= 72+ DC^{2}+ 400 -40DC+DC^{2}$$
--> $$2DC^{2} -40DC+72=0$$
$$DC^{2} - 20DC+ 36=0$$
(DC-2)(DC-18)

--> DC=2 and DC=18

if DC=2, then the area of $$BDC=\frac{6*2}{2}= 6$$
if DC=18, then the area of $$BDC= \frac{6*18}{2}= 54$$
--> TWO VALUES
Insufficient

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Re: What is the area of triangle BDC?  [#permalink]

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08 Dec 2019, 02:21
Ans:C
(1) and(2) individually insufficient.
Since the triangle is right angled it can be considered to be inscribed in acircle with AC as diameter.(refer to photo attached).
Hence OC=OB=10
from pythagoras theorem OD can be found.
Hence DC can be found.
BD is given 6.
Hence the area of BDC can be found.
Attachments

webcam-toy-photo1.jpg [ 34.62 KiB | Viewed 798 times ]

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Re: What is the area of triangle BDC?  [#permalink]

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Updated on: 08 Dec 2019, 20:07
What is the area of triangle BDC?
1) BD = 6
2) AC= 20

#1
BD= 6 ; angle DBC not know
insufficient
#2
AC=20 AB = 16 and BC = 12 ; area of ∆ ABC = 1/2 *12* 16 ; 72
area of BDC not know
From 1 & 2 no info
Insufficient
IMO E

Originally posted by Archit3110 on 08 Dec 2019, 02:35.
Last edited by Archit3110 on 08 Dec 2019, 20:07, edited 1 time in total.
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Re: What is the area of triangle BDC?  [#permalink]

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08 Dec 2019, 03:16
Option 1 alone cannot help us to determine the area of Triangle BDC

Option 2 alone cannot help to determine the area of triangle BDC

option 1 and 2 together help us to get the area of triangle ABC but not of BDC
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Re: What is the area of triangle BDC?  [#permalink]

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08 Dec 2019, 04:10
1

EXPLANATION:

We need to find area of BDC. Hence, we need to find BD, DC, and BC. Or BC and altitude of BDC (which is a perpendicular dropped from D towards BC). Let's analyse the options:

From A:
We know BD = 6. But we have no idea in what proportion it divides AC. Infact, we have no clue about AC. So we can't find the area of BDC. Eliminate AD.

From B:
We know about AC = 20. If DC = x, then AD = 20-x. We still don't have value of BD, which is needed to get the value of BC. So we can't find the area of BDC. Eliminate B.

From both A and B:
We know, BD = 6, AC = 20. If DC = x, then AD = 20-x.

Now, from Pythagoras theorem:
$$AB^2 = BD^2 + AD^2$$

Substituting values:
$$AB^2 = 36 + (20-x)^2$$

Similarly, $$BC^2 = BD^2 + DC^2$$

Substituting values:
$$BC^2 = 36 + x^2$$

Now, let's combine it to the bigger triangle.
Using pythagoras theorem again:
$$AB^2 + BC^2 = AC^2$$

Substituting values that we got earlier: (Note: AC = 20, given).
$$36 + (20-x)^2 + 36 + x^2 = 400$$
$$72 + x^2 + 400 - 40x + x^2 = 400$$
$$2x^2 - 40x + 72 = 0$$
$$x^2 - 20x + 36 = 0$$

We get two values of X = 18, and X = 2.

Hence, we get two values of DC. Either DC = 2, or DC = 18.
From both of these values, we get two different lengths of BC.

If DC = 2, $$BC = \sqrt{ 36 + 4 }$$, which equals approx 6.3.

If DC = 18, $$BC = \sqrt{ 36 + 324 }$$, which equals approx 19.

Hence, we would get two different area of triangle. Therefore, the answer is E.

ALTERNATE APPROACH:
Quote:
Right Triangle Altitude Theorem: The measure of the altitude drawn from the vertex of the right angle of a right triangle to its hypotenuse is the geometric mean between the measures of the two segments of the hypotenuse.

This means, $$BD^2 = AD*DC$$
So, 36 = x*(20-x) (assuming DC = x, and AD = 20-x).

If we solve for x, we will again get two values of x, as 18 and 2, and we would get to the same conclusion as before.

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What is the area of triangle BDC?  [#permalink]

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Updated on: 10 Dec 2019, 00:02
to calculate area of a triangle need to know height and base dimension,
st 1: only height is given - not sufficient
st 2: only combined base is given - not sufficient

combined, we get height and base (AD+CD) , but we dont get specific value of CD so not possible to calculate area of triangle BDC. hence Answer is E.

Originally posted by ITGMAT on 08 Dec 2019, 06:59.
Last edited by ITGMAT on 10 Dec 2019, 00:02, edited 1 time in total.
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Re: What is the area of triangle BDC?  [#permalink]

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08 Dec 2019, 07:35
Let AC be the diameter of a circle. B will lie on the arc ABC and B = 90 DEG
(1) For BD = 6, the diameter of this circle, AC can be varied for various circles. So, the area of the triangle BDC will vary. We do not have a unique triangle. Clearly, not sufficient.
(2) For AC = 20, B can be any point on the arc ABC, thus area of triangle BDC will vary depending on the height of BD. Thus, not sufficient.

Combining two statements, we get, AC = 20 and BD = 6. We still have two different triangles with different area of triangle BDC.
Therefore, choice E is the answer.
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20191208_200208.jpg [ 761.15 KiB | Viewed 712 times ]

20191208_200241.jpg [ 1.8 MiB | Viewed 709 times ]

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Re: What is the area of triangle BDC?  [#permalink]

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08 Dec 2019, 07:39
What is the area of triangle BDC?

1) BD = 6
2) AC= 20

Refer the attachment for explanation.

Imo. E
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IMG_20191208_200610.jpg [ 2.31 MiB | Viewed 717 times ]

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Re: What is the area of triangle BDC?  [#permalink]

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09 Dec 2019, 13:20
(1) BD=6
Since AC is not given. Area of triangle BDC can not be determined.
Not sufficient.

(2) AC=20
Since BD is not given. Area of triangle BDC can not be determined.
Not sufficient.

(1)+(2)
With AC=20, BD=6 and angle B = 90 deg., exactly two triangles can be drawn. Since the length of AD is not fixed, area of triangle BDC can not be determined.

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Re: What is the area of triangle BDC?  [#permalink]

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11 Dec 2019, 01:56
I have one question. Might be silly for few mates!

By seeing the notions of right angle we assumed that BD is perpendicular to AC. Similarly why couldn't understand that the BD is bisecting the angle B as BD is dividing the right angle symbol into two halfs.

In many problems, we take the symbol in consideration while solving the problem as we did here.
I don't think that I will see such question in actual GMAT in which I can take benefit of symbols or representations.
Whether it is right angle or centre of circle or any specific detail which may help test takers to solve the problem should be mentioned clearly.

That's my view!
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What is the area of triangle BDC?  [#permalink]

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11 Dec 2019, 02:23
A rectangle symbol drawn at the point of intersection of 2 lines means PERPENDICULARITY. It is an acceptable convention.

Angle bisector cannot be inferred from a line seemingly bisecting the angle.
In fact, this is not even a convention.

If BD, had been angle bisector, some other property will be provided.

For example, as rightly, pointed out by chetan2u, D is seemingly bisecting AC but we cannot take it that way, in fact on solving it is clear that D is not bisecting AC.

Never assume figure drawn to scale, unless written explicitly.

gvij2017 wrote:
I have one question. Might be silly for few mates!

By seeing the notions of right angle we assumed that BD is perpendicular to AC. Similarly why couldn't understand that the BD is bisecting the angle B as BD is dividing the right angle symbol into two halfs.

In many problems, we take the symbol in consideration while solving the problem as we did here.
I don't think that I will see such question in actual GMAT in which I can take benefit of symbols or representations.
Whether it is right angle or centre of circle or any specific detail which may help test takers to solve the problem should be mentioned clearly.

That's my view!

Attachment:

symbol.png [ 5.21 KiB | Viewed 456 times ]

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Re: What is the area of triangle BDC?  [#permalink]

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21 Dec 2019, 13:07
gmatbusters

If Angle BAC = angle ACB

Would then BD bisect Line AC into two halves

I am trying to understand what property has to be given for us to believe than BD will bisect line AC into two halves

Thank you !
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Re: What is the area of triangle BDC?  [#permalink]

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21 Dec 2019, 20:58
Hi

If Angle BAC = Angle ACB
Then, it is a case of an isosceles right angled triangle. Hence by symmetry,
BD would bisect AC.

jabhatta@umail.iu.edu wrote:
gmatbusters

If Angle BAC = angle ACB

Would then BD bisect Line AC into two halves

I am trying to understand what property has to be given for us to believe than BD will bisect line AC into two halves

Thank you !

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Re: What is the area of triangle BDC?   [#permalink] 21 Dec 2019, 20:58
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