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In the figure attached, the length of line segment CD is twice the len

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Bunuel
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In the figure attached, the length of line segment CD is twice the len [#permalink] New post   11 Jun 2019, 01:44
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In the figure attached, the length of line segment CD is twice the length of line segment BC. The ratio of the area of ΔABC to the area of ΔABD is:


A. \(1:\sqrt{2}\)

B. \(1:\sqrt{3}\)

C. 1:2

D. 1:3

E. 1:4

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D
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Lampard42
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Re: In the figure attached, the length of line segment CD is twice the len [#permalink] New post   11 Jun 2019, 02:24
Bunuel wrote:

In the figure attached, the length of line segment CD is twice the length of line segment BC. The ratio of the area of ΔABC to the area of ΔABD is:


A. \(1:\sqrt{2}\)

B. \(1:\sqrt{3}\)

C. 1:2

D. 1:3

E. 1:4

Show Spoiler
Attachment:
216155.17.g500054img01.gif


Area of right angled triangle = 1/2*leg1*leg2; Leg1 is same for both triangles (y) and Leg 2 is X for ABC and ABD=3X(X+2X); So area is 1:3 IMO Option D
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Re: In the figure attached, the length of line segment CD is twice the len [#permalink] New post   11 Jun 2019, 06:25
Bunuel wrote:

In the figure attached, the length of line segment CD is twice the length of line segment BC. The ratio of the area of ΔABC to the area of ΔABD is:


A. \(1:\sqrt{2}\)

B. \(1:\sqrt{3}\)

C. 1:2

D. 1:3

E. 1:4

Show Spoiler
Attachment:
216155.17.g500054img01.gif


Let BC = x --> CD = 2x
BD = BC + CD = x + 2x = 3x
Let AB = y

--> Area of ΔABC = 1/2*AB*BC = 1/2*y*x = xy/2
--> Area of ΔABD = 1/2*AB*BD = 1/2*y*3x = 3xy/2

--> ratio of the area of ΔABC to the area of ΔABD = (xy/2)/(3xy/2) = 1:3

Option D

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Re: In the figure attached, the length of line segment CD is twice the len [#permalink] New post   12 Jun 2019, 03:11
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Solution

Given:
• In the figure attached, the length of line segment CD is twice the length of line segment BC

To Find:
• The ratio of the area of ΔABC to the area of ΔABD

Approach & Working Out:
    • The ratio of the area of ΔABC to the area of ΔABD = the ratio of length of BC to length of BD = \(\frac{BC}{(BC + CD)} = \frac{BC}{3BC} = \frac{1}{3}\)

Hence, the correct answer is Option D
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In the figure attached, the length of line segment CD is twice the len [#permalink] New post   20 Jun 2019, 17:09
If its a right triangle, we can solve it like this:

3 : 4 : 5 Triangle

Base: x= 4 and 2x = 8, total base: 12 (4x3)

So height must be 3x3 = 9

\(\frac{(9x4)}{2}=18\)

\(\frac{(9x12)}{2}=54\)

\(\frac{18}{54}=1:3\)

Kind regards!
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In the figure attached, the length of line segment CD is twice the len [#permalink]
20 Jun 2019, 17:09
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