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# In the figure above, ΔABC and ΔBCD are right-angled at points B and C

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Joined: 14 Sep 2015
Posts: 65
Location: India
GMAT 1: 700 Q45 V40
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In the figure above, ΔABC and ΔBCD are right-angled at points B and C  [#permalink]

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Updated on: 23 May 2017, 12:18
2
00:00

Difficulty:

55% (hard)

Question Stats:

55% (02:02) correct 45% (02:45) wrong based on 50 sessions

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In the figure below, ΔABC and ΔBCD are right-angled at points B and C respectively. If AC = $$3\sqrt{2}$$, ∠ACB = $$45^∘$$ and ∠CBD = $$30^∘$$, what is the area of ΔBCD?

A. $$\frac{9}{4}$$

B. $$\frac{3}{2}\sqrt{3}$$

C. $$\frac{27}{4}$$

D. $$\frac{9}{2}\sqrt{3}$$

E. 9

Attachment:

QS430.png [ 23.52 KiB | Viewed 1034 times ]

Originally posted by niteshwaghray on 23 May 2017, 12:04.
Last edited by Bunuel on 23 May 2017, 12:18, edited 2 times in total.
Edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 52161
In the figure above, ΔABC and ΔBCD are right-angled at points B and C  [#permalink]

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23 May 2017, 12:35
1
niteshwaghray wrote:
In the figure below, ΔABC and ΔBCD are right-angled at points B and C respectively. If AC = $$3\sqrt{2}$$, ∠ACB = $$45^∘$$ and ∠CBD = $$30^∘$$, what is the area of ΔBCD?

A. $$\frac{9}{4}$$ B. $$\frac{3}{2}\sqrt{3}$$ C. $$\frac{27}{4}$$ D. $$\frac{9}{2}\sqrt{3}$$ E. 9
Attachment:
QS430.png

∠ACB = 45° implies that ACB is 45°-45°-90° triangle. The sides are always in the ratio 1:1:√2. With the √2 being the hypotenuse (longest side).

Since, $$AC = 3\sqrt{2}$$, then AB = BC = 3.

Next, ∠CBD = 30° implies that CBD is 30°-60°-90°. The sides are always in the ratio 1:√3:2. Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).

Since BC = 3, then CD (the smallest side opposite the smallest angle of 30°) is √3.

Finally, the area of BCD is BC*CD/2 = 3*√3/2.

_________________
In the figure above, ΔABC and ΔBCD are right-angled at points B and C &nbs [#permalink] 23 May 2017, 12:35
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