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In the coordinate axis above, line segment AC is three times

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wastedyouth
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In the coordinate axis above, line segment AC is three times [#permalink] New post   Updated on: 15 Mar 2019, 04:32
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In the coordinate axis above, line segment AC is three times as long as line segment AB. In addition, segment AC is perpendicular to segment BD. What is the area of triangle DOB?

A. 4
B. \(4\sqrt{2}\)
C. 6
D. 8
E. It cannot be determined from the information given.


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D

Originally posted by wastedyouth on 09 Dec 2013, 11:08.
Last edited by Bunuel on 15 Mar 2019, 04:32, edited 2 times in total.
Edited the question.
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KarishmaB
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Re: In the coordinate axis above, line segment AC is three times [#permalink] New post   09 Dec 2013, 22:39
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wastedyouth wrote:
In the coordinate axis above, line segment AC is three times as long as line segment AB. In addition, segment AC is perpendicular to segment BD. What is the area of triangle DOB?


4

4\sqrt{2}

6

8

It cannot be determined from the information given.


You can use similar triangles concept here.
Drop a perpendicular AE on the y axis.

Attachment:
Ques3.jpg
Ques3.jpg [ 9.95 KiB | Viewed 12238 times ]


Now, triangle BEA is similar to triangle BOC (by the AA rule. Angle ABE = Angle CBO since they are vertically opposite and both triangles have right angles)
Since AB/BC = 1/2, so BE/BO = AE/CO = 1/2
Since AE = 2, CO = 4. Also since EO is 6, BE = 2 and BO = 4.
So BOC is isosceles right triangle so both angles OBC and OCB are 45. This means angle OBD is 45 (since angle DBC is 90) making triangle DOB isosceles as well. Since BO = 4, DO = 4 as well.

Area of triangle DOB = (1/2)*4*4 = 8
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Re: In the coordinate axis above, line segment AC is three times [#permalink] New post   10 Dec 2013, 11:08
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wastedyouth wrote:
Attachment:
The attachment dia.png is no longer available
In the coordinate axis above, line segment AC is three times as long as line segment AB. In addition, segment AC is perpendicular to segment BD. What is the area of triangle DOB?

A. 4
B. \(4\sqrt{2}\)
C. 6
D. 8
E. It cannot be determined from the information given.


The given line segment AC is divided by B in the ratio 1:2

In general, the co-ordinates of Point B that divides a line segment in the ratio of m:n is

X-coordinate of B : \(\frac{(m*x2+n*x1)}{m+n}\)

Y-coordinate of B : \(\frac{(m*y2+n*y1)}{m+n}\)

Back to out problem,

1. we know that B lies on the Y-axis.
2.m:n = 1:2

Hence, the Y -coordinate of B: \(\frac{(m*x2+n*x1)}{m+n} \to \frac{(1*0+2*6)}{1+2} = 4\). Thus, B(0,4)

Again, slope of AB : \(\frac{6-4}{-2-0} = -1\), hence slope of BD where D has the coordinates as (p,0) =\(\frac{4-0}{0-p} = 1 \to p=-4\).

Thus, the area of \(\triangle ADB = \frac{4*4}{2} = 8.\)

D.
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mikemcgarry
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Re: In the coordinate axis above, line segment AC is three times [#permalink] New post   Updated on: 06 Jun 2016, 10:24
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wastedyouth wrote:
In the coordinate axis above, line segment AC is three times as long as line segment AB. In addition, segment AC is perpendicular to segment BD. What is the area of triangle DOB?

(A) 4
(B) 4\sqrt{2}
(C) 6
(D) 8
(E) It cannot be determined from the information given.

I'm happy to help. :-)

This is tricky because the diagram is not to scale, but we can use proportional reasoning. If 3*AB = BC, that means (BC)/(AC) = 2/3, that means from the origin to B must be (2/3) of the distance from the x-axis up to point A --- (2/3) of 6 is 4, so B = (0, 4). That's a 45 degree line, so C = (4, 0), and D = (-4, 0). Here's a scaled diagram:
Attachment:
scaled triangle.JPG
scaled triangle.JPG [ 22.86 KiB | Viewed 11971 times ]

DOB is a 45-45-90 triangle, with DO = OB = 4, so area = 8. Answer = (D).

Does all this make sense?
Mike :-)

Originally posted by mikemcgarry on 09 Dec 2013, 14:57.
Last edited by mikemcgarry on 06 Jun 2016, 10:24, edited 1 time in total.
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Re: In the coordinate axis above, line segment AC is three times [#permalink] New post   10 Dec 2013, 07:38
Hi wastedyouth,

Excellent question.

I marked E because I was unable to determine the area of the triangle DOB.
But, after reading the various explanations, I am convinced with the answer.
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Re: In the coordinate axis above, line segment AC is three times [#permalink] New post   28 Dec 2013, 09:12
wastedyouth wrote:
Attachment:
dia.png
In the coordinate axis above, line segment AC is three times as long as line segment AB. In addition, segment AC is perpendicular to segment BD. What is the area of triangle DOB?

A. 4
B. \(4\sqrt{2}\)
C. 6
D. 8
E. It cannot be determined from the information given.


Good question, let' solve

1. Draw the figure w/labels
2. now distance from A to B is X, distance from B to C is 2X
3. Since B will have x coordinate 0 then we only need to 'y' coordinate. Now since all the height for segment AC is 6 then AB will only have height 2, then 6-2 = 4, so B (0,4)
4. Now, right triangle BOD is an isosceles right triangle with angles 45-45-90. Angle 45 corresponds to 4. So area is 4*4/2 = 8

Answer is 8

Is this clear?

Cheers!
J :)
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Re: In the coordinate axis above, line segment AC is three times [#permalink] New post   03 Jun 2016, 18:26
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hey,
a slight correction. You have calculated B to be (4,0). It should be (0,4) as it lies on y axis where X=0


mikemcgarry wrote:
wastedyouth wrote:
In the coordinate axis above, line segment AC is three times as long as line segment AB. In addition, segment AC is perpendicular to segment BD. What is the area of triangle DOB?

(A) 4
(B) 4\sqrt{2}
(C) 6
(D) 8
(E) It cannot be determined from the information given.

I'm happy to help. :-)

This is tricky because the diagram is not to scale, but we can use proportional reasoning. If 3*AB = BC, that means (BC)/(AC) = 2/3, that means from the origin to B must be (2/3) of the distance from the x-axis up to point A --- (2/3) of 6 is 4, so B = (4, 0). That's a 45 degree line, so C = (4, 0), and D = (-4, 0). Here's a scaled diagram:
Attachment:
scaled triangle.JPG

DOB is a 45-45-90 triangle, with DO = OB = 4, so area = 8. Answer = (D).

Does all this make sense?
Mike :-)
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Re: In the coordinate axis above, line segment AC is three times [#permalink] New post   06 Jun 2016, 11:03
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tallyho_88 wrote:
hey,
a slight correction. You have calculated B to be (4,0). It should be (0,4) as it lies on y axis where X=0

Dear tallyho_88,
Thank you for pointing that out, my friend.
Mike :-)
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Re: In the coordinate axis above, line segment AC is three times [#permalink] New post   14 Jul 2016, 02:00
VeritasPrepKarishma wrote:
In the coordinate axis above, line segment AC is three times as long as line segment AB. In addition, segment AC is perpendicular to segment BD. What is the area of triangle DOB?

4
4\sqrt{2}
6
8

It cannot be determined from the information given.

You can use similar triangles concept here.
Drop a perpendicular AE on the y axis.

Attachment:
Ques3.jpg


Now, triangle BEA is similar to triangle BOC (by the AA rule. Angle ABE = Angle CBO since they are vertically opposite and both triangles have right angles)
Since AB/BC = 1/2, so BE/BO = AE/CO = 1/2
Since AE = 2, CO = 4. Also since EO is 6, BE = 2 and BO = 4.
So BOC is isosceles right triangle so both angles OBC and OCB are 45. This means angle OBD is 45 (since angle DBC is 90) making triangle DOB isosceles as well. Since BO = 4, DO = 4 as well.

Area of triangle DOB = (1/2)*4*4 = 8


Quote:
Hi Karishma:
Typically, the definition of similarity between two triangles is as follows:

Triangles are similar if:
AAA (angle angle angle)
All three pairs of corresponding angles are the same.
See Similar Triangles AAA.
SSS in same proportion (side side side)
All three pairs of corresponding sides are in the same proportion
See Similar Triangles SSS.
SAS (side angle side)
Two pairs of sides in the same proportion and the included angle equal.
See Similar Triangles SAS.

Here is the source of information:
https://www.mathopenref.com/similartriangles.html

My Question:
Instead of 3 angles (AAA), can we just consider two angles (AA) to conclude that two triangles are similar?
Please share your thoughts.
Thanks in advance.

Regards,
Yosita
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Re: In the coordinate axis above, line segment AC is three times [#permalink] New post   18 Jul 2016, 23:34
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yosita18 wrote:
VeritasPrepKarishma wrote:
In the coordinate axis above, line segment AC is three times as long as line segment AB. In addition, segment AC is perpendicular to segment BD. What is the area of triangle DOB?

4
4\sqrt{2}
6
8

It cannot be determined from the information given.

You can use similar triangles concept here.
Drop a perpendicular AE on the y axis.

Attachment:
Ques3.jpg


Now, triangle BEA is similar to triangle BOC (by the AA rule. Angle ABE = Angle CBO since they are vertically opposite and both triangles have right angles)
Since AB/BC = 1/2, so BE/BO = AE/CO = 1/2
Since AE = 2, CO = 4. Also since EO is 6, BE = 2 and BO = 4.
So BOC is isosceles right triangle so both angles OBC and OCB are 45. This means angle OBD is 45 (since angle DBC is 90) making triangle DOB isosceles as well. Since BO = 4, DO = 4 as well.

Area of triangle DOB = (1/2)*4*4 = 8


Quote:
Hi Karishma:
Typically, the definition of similarity between two triangles is as follows:

Triangles are similar if:
AAA (angle angle angle)
All three pairs of corresponding angles are the same.
See Similar Triangles AAA.
SSS in same proportion (side side side)
All three pairs of corresponding sides are in the same proportion
See Similar Triangles SSS.
SAS (side angle side)
Two pairs of sides in the same proportion and the included angle equal.
See Similar Triangles SAS.

Here is the source of information:
https://www.mathopenref.com/similartriangles.html

My Question:
Instead of 3 angles (AAA), can we just consider two angles (AA) to conclude that two triangles are similar?
Please share your thoughts.
Thanks in advance.

Regards,
Yosita


Sum of all 3 angles of a triangle is always 180. If 2 pairs of angles are equal, the third pair always has to be equal too. So AA is enough.
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Re: In the coordinate axis above, line segment AC is three times [#permalink] New post   28 Jul 2017, 05:51
VeritasPrepKarishma wrote:
wastedyouth wrote:
In the coordinate axis above, line segment AC is three times as long as line segment AB. In addition, segment AC is perpendicular to segment BD. What is the area of triangle DOB?


4

4\sqrt{2}

6

8

It cannot be determined from the information given.


You can use similar triangles concept here.
Drop a perpendicular AE on the y axis.

Attachment:
Ques3.jpg


Now, triangle BEA is similar to triangle BOC (by the AA rule. Angle ABE = Angle CBO since they are vertically opposite and both triangles have right angles)
Since AB/BC = 1/2, so BE/BO = AE/CO = 1/2
Since AE = 2, CO = 4. Also since EO is 6, BE = 2 and BO = 4.
So BOC is isosceles right triangle so both angles OBC and OCB are 45. This means angle OBD is 45 (since angle DBC is 90) making triangle DOB isosceles as well. Since BO = 4, DO = 4 as well.

Area of triangle DOB = (1/2)*4*4 = 8



Superb explanation!! Thanks :)
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Re: In the coordinate axis above, line segment AC is three times [#permalink] New post   31 Jul 2019, 04:47
wastedyouth wrote:

In the coordinate axis above, line segment AC is three times as long as line segment AB. In addition, segment AC is perpendicular to segment BD. What is the area of triangle DOB?

A. 4
B. \(4\sqrt{2}\)
C. 6
D. 8
E. It cannot be determined from the information given.


Show Spoiler
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The attachment Untitled.png is no longer available



Please see solution in attached image below

IMO D
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Re: In the coordinate axis above, line segment AC is three times [#permalink] New post   07 Jun 2022, 03:16
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Re: In the coordinate axis above, line segment AC is three times [#permalink]
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