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# What is the area of the rectangle ABCD? (1) BD = 5 (2) AC = 5/3*BC

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Posts: 50700
What is the area of the rectangle ABCD? (1) BD = 5 (2) AC = 5/3*BC  [#permalink]

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06 Sep 2018, 23:47
00:00

Difficulty:

55% (hard)

Question Stats:

78% (01:45) correct 22% (01:45) wrong based on 36 sessions

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What is the area of the rectangle ABCD?

(1) BD = 5

(2) AC = 5/3*BC

Attachment:

image016.jpg [ 1.45 KiB | Viewed 415 times ]

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Re: What is the area of the rectangle ABCD? (1) BD = 5 (2) AC = 5/3*BC  [#permalink]

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07 Sep 2018, 00:58
Bunuel wrote:

What is the area of the rectangle ABCD?

(1) BD = 5

(2) AC = 5/3*BC

Attachment:
image016.jpg

QUESTION: What is the area of the rectangle ABCD?

Statement 1: BD = 5

Wit the help of diagonals, we can't calculate the area of the rectangle because the angle between the diagonals is unknown hence

NOT SUFFICIENT

Statement 2: AC = 5/3*BC

This statement gives us the ratio of the lengths of sides of rectangle no exact dimension hence

NOT SUFFICIENT

Combining the two statements
Let, BC = x
AC = (5/3)x
i.e. $$AB = √{(25x^2/9)-(x^2)]$$

Also, AC = BD = (5/3)x = 5

i.e. x is unknown which can get us the exacl length and width as well hence we can calculate the area of the rectangle hence

SUFFICIENT

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Re: What is the area of the rectangle ABCD? (1) BD = 5 (2) AC = 5/3*BC  [#permalink]

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08 Sep 2018, 10:41
Bunuel wrote:

What is the area of the rectangle ABCD?

(1) BD = 5

(2) AC = 5/3*BC

Attachment:
image016.jpg

S1 - BD=5
Insufficient.

S2 - AC = 5/3*BC.
We don't have any values. Insufficient.

Combining,
AC=BD ; $$AB^2+BC^2=AC^2 = BD^2 = 25$$
BC = 3/5 AC.
We can find AB and hence the area.
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Re: What is the area of the rectangle ABCD? (1) BD = 5 (2) AC = 5/3*BC &nbs [#permalink] 08 Sep 2018, 10:41
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