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Find the ratio of areas of the Triangle ABC and ADB

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Find the ratio of areas of the Triangle ABC and ADB  [#permalink]

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Updated on: 25 Apr 2014, 23:19
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Difficulty:

65% (hard)

Question Stats:

60% (01:49) correct 40% (02:07) wrong based on 176 sessions

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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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Joined: 02 Sep 2009
Posts: 51121
Re: Find the ratio of areas of the Triangle ABC and ADB  [#permalink]

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06 May 2014, 08:22
2
1
JusTLucK04 wrote:

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.

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Re: Find the ratio of areas of the Triangle ABC and ADB  [#permalink]

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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).

Bunuel wrote:
JusTLucK04 wrote:

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.

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Re: Find the ratio of areas of the Triangle ABC and ADB  [#permalink]

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07 May 2014, 01:08
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).

Bunuel wrote:
JusTLucK04 wrote:

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.

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]

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10 Jun 2014, 14:02
Bunuel wrote:
JusTLucK04 wrote:

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.

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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Posts: 51121
Re: Find the ratio of areas of the Triangle ABC and ADB  [#permalink]

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11 Jun 2014, 13:07
ronr34 wrote:
Bunuel wrote:
JusTLucK04 wrote:

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.

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]

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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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Re: Find the ratio of areas of the Triangle ABC and ADB  [#permalink]

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09 Nov 2017, 00:31
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Re: Find the ratio of areas of the Triangle ABC and ADB &nbs [#permalink] 09 Nov 2017, 00:31
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