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Find the ratio of areas of the Triangle ABC and ADB
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JusTLucK04
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Find the ratio of areas of the Triangle ABC and ADB [#permalink] New post  Updated on: 25 Apr 2014, 23:19
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Find the ratio of areas of the triangle ADB and ABC. If BD:DC = 3:5

A. 3:5
B. 9:25
C. 3:8
D. 5:9
E. 9:16

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Originally posted by JusTLucK04 on 25 Apr 2014, 00:30.
Last edited by JusTLucK04 on 25 Apr 2014, 23:19, edited 5 times in total.
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Bunuel
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Re: Find the ratio of areas of the Triangle ABC and ADB [#permalink] New post  06 May 2014, 08:22
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JusTLucK04 wrote:
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Find the ratio of areas of the triangle ADB and ABC. If BD:DC = 3:5

A. 3:5
B. 9:25
C. 3:8
D. 5:9
E. 9:16

Kudos If you liked the Question


BD:DC = 3:5 --> BD:(BD+DC) = 3:(3+5) --> BD:BC = 3:8.

The area of triangle ADB = 1/2*(height from A to DB)*(DB).
The area of triangle ABC = 1/2*(height from A to BC)*(BC).

Notice that (height from A to DB) = (height from A to BC).

Therefore the ration of the areas of ADB and ABC = DB:BC = 3:8.

Answer: C.
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Re: Find the ratio of areas of the Triangle ABC and ADB [#permalink] New post  06 May 2014, 11:03
Hi Bunuel,
Excuse me if i do not have clear concept in Geometry.

So I do not understand how below statement is true.

(height from A to DB) = (height from A to BC).

Thanks for your help.


Bunuel wrote:
JusTLucK04 wrote:
Image
Find the ratio of areas of the triangle ADB and ABC. If BD:DC = 3:5

A. 3:5
B. 9:25
C. 3:8
D. 5:9
E. 9:16

Kudos If you liked the Question


BD:DC = 3:5 --> BD:(BD+DC) = 3:(3+5) --> BD:BC = 3:8.

The area of triangle ADB = 1/2*(height from A to DB)*(DB).
The area of triangle ABC = 1/2*(height from A to BC)*(BC).

Notice that (height from A to DB) = (height from A to BC).

Therefore the ration of the areas of ADB and ABC = DB:BC = 3:8.

Answer: C.
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Bunuel
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Re: Find the ratio of areas of the Triangle ABC and ADB [#permalink] New post  07 May 2014, 01:08
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sunita123 wrote:
Hi Bunuel,
Excuse me if i do not have clear concept in Geometry.

So I do not understand how below statement is true.

(height from A to DB) = (height from A to BC).

Thanks for your help.


Bunuel wrote:
JusTLucK04 wrote:
Image
Find the ratio of areas of the triangle ADB and ABC. If BD:DC = 3:5

A. 3:5
B. 9:25
C. 3:8
D. 5:9
E. 9:16

Kudos If you liked the Question


BD:DC = 3:5 --> BD:(BD+DC) = 3:(3+5) --> BD:BC = 3:8.

The area of triangle ADB = 1/2*(height from A to DB)*(DB).
The area of triangle ABC = 1/2*(height from A to BC)*(BC).

Notice that (height from A to DB) = (height from A to BC).

Therefore the ration of the areas of ADB and ABC = DB:BC = 3:8.

Answer: C.


Line segments DB and BC are on the same line, thus the perpendicular from point A to DB is the same as the perpendicular from point A to BC.
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Re: Find the ratio of areas of the Triangle ABC and ADB [#permalink] New post  10 Jun 2014, 14:02
Bunuel wrote:
JusTLucK04 wrote:
Image
Find the ratio of areas of the triangle ADB and ABC. If BD:DC = 3:5

A. 3:5
B. 9:25
C. 3:8
D. 5:9
E. 9:16

Kudos If you liked the Question


BD:DC = 3:5 --> BD:(BD+DC) = 3:(3+5) --> BD:BC = 3:8.

The area of triangle ADB = 1/2*(height from A to DB)*(DB).
The area of triangle ABC = 1/2*(height from A to BC)*(BC).

Notice that (height from A to DB) = (height from A to BC).

Therefore the ration of the areas of ADB and ABC = DB:BC = 3:8.

Answer: C.

Hi Bunuel,

in a lot of questions, we get to the bottom line of: ratio of area = (ratio of side)^2.

Is this not true? Is this true only in specific cases?
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Bunuel
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Re: Find the ratio of areas of the Triangle ABC and ADB [#permalink] New post  11 Jun 2014, 13:07
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ronr34 wrote:
Bunuel wrote:
JusTLucK04 wrote:
Image
Find the ratio of areas of the triangle ADB and ABC. If BD:DC = 3:5

A. 3:5
B. 9:25
C. 3:8
D. 5:9
E. 9:16

Kudos If you liked the Question


BD:DC = 3:5 --> BD:(BD+DC) = 3:(3+5) --> BD:BC = 3:8.

The area of triangle ADB = 1/2*(height from A to DB)*(DB).
The area of triangle ABC = 1/2*(height from A to BC)*(BC).

Notice that (height from A to DB) = (height from A to BC).

Therefore the ration of the areas of ADB and ABC = DB:BC = 3:8.

Answer: C.

Hi Bunuel,

in a lot of questions, we get to the bottom line of: ratio of area = (ratio of side)^2.

Is this not true? Is this true only in specific cases?


This is true for similar triangles.

If two similar triangles have sides in the ratio \(\frac{x}{y}\), then their areas are in the ratio \(\frac{x^2}{y^2}\).
OR in another way: in two similar triangles, the ratio of their areas is the square of the ratio of their sides: \(\frac{AREA}{area}=\frac{SIDE^2}{side^2}\).
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Re: Find the ratio of areas of the Triangle ABC and ADB [#permalink] New post  20 Apr 2016, 18:27
oh damn..i was on my laptop..and the screen is small...i did not see the image..was going crazy to identify how the triangle might look like...
BD= suppose 3x
DC = 5x
BC=8x
the height is the same for the both triangles, so basically, we are asked for the ratio of the bases. BC/BD = 3x/8x = 3/8
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Find the ratio of areas of the Triangle ABC and ADB [#permalink] New post  24 Mar 2021, 13:31
Bunuel

I am confused. So this current problem, the triangles are not similar?
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Re: Find the ratio of areas of the Triangle ABC and ADB [#permalink] New post  24 Mar 2021, 13:38
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CEdward wrote:
Bunuel

I am confused. So this current problem, the triangles are not similar?


No, there are no similar triangles in the figure.
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Re: Find the ratio of areas of the Triangle ABC and ADB [#permalink] New post  25 Mar 2021, 09:49
I too fell in the trap thinking that these are similar triangles.... and I was making a mistake of drawing the 2 perpendicular lines 1 from D and the other from C

But perpendicular from A makes more sense. Answer is indeed 3/8

Bunuel Thanks a lot
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Re: Find the ratio of areas of the Triangle ABC and ADB [#permalink]
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