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

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Question Stats: 32% (01:32) correct 68% (01:29) wrong based on 115 sessions

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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 1583 times ]

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

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Area=BD*CD/2

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

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

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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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1
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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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 1380 times ]

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

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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, 01:35.
Last edited by Archit3110 on 08 Dec 2019, 19:07, edited 1 time in total.
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Re: What is the area of triangle BDC?  [#permalink]

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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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GMAT 1: 610 Q43 V31 Re: What is the area of triangle BDC?  [#permalink]

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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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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, 05:59.
Last edited by ITGMAT on 09 Dec 2019, 23:02, edited 1 time in total.
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Re: What is the area of triangle BDC?  [#permalink]

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1
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.
Attachments 20191208_200208.jpg [ 761.15 KiB | Viewed 1290 times ] 20191208_200241.jpg [ 1.8 MiB | Viewed 1287 times ]

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

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What is the area of triangle BDC?

1) BD = 6
2) AC= 20

Refer the attachment for explanation.

Imo. E
Attachments IMG_20191208_200610.jpg [ 2.31 MiB | Viewed 1296 times ]

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

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(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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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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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 1076 times ]

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

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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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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]

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GMATBusters wrote:

GMATBusters’ Quant Quiz Question -10

What is the area of triangle BDC?
1) BD = 6
2) AC= 20 Attachment:
12222.jpg

Asked: What is the area of triangle BDC?
Area of triangle BDC = 1/2 * BD * CD

1) BD = 6
No information is provided for CD
NOT SUFFICIENT

2) AC= 20
No information is provided for BD & CD
NOT SUFFICIENT

(1) + (2)
1) BD = 6
2) AC= 20
There is no information regarding CD
NOT SUFFICIENT

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

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there is no way by which we can find value of cd so we cant find are of triangle , e is correct option Re: What is the area of triangle BDC?   [#permalink] 08 Mar 2020, 19:49

# What is the area of triangle BDC?  