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# In the figure above, if AC = 18, what is the area of ∆ BCD?

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In the figure above, if AC = 18, what is the area of ∆ BCD? [#permalink]

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24 Aug 2017, 00:54
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In the figure above, if AC = 18, what is the area of ∆ BCD?

(A) 15
(B) 30
(C) 54
(D) 60
(E) It cannot be determined

[Reveal] Spoiler:
Attachment:

2017-08-24_1249.png [ 7.77 KiB | Viewed 617 times ]
[Reveal] Spoiler: OA

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Re: In the figure above, if AC = 18, what is the area of ∆ BCD? [#permalink]

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24 Aug 2017, 01:51
Bunuel wrote:

In the figure above, if AC = 18, what is the area of ∆ BCD?

(A) 15
(B) 30
(C) 54
(D) 60
(E) It cannot be determined

[Reveal] Spoiler:
Attachment:
2017-08-24_1249.png

$$AB = \sqrt{100-64} = \sqrt{36} = 6$$
Area of ABC = 1/2 * 6 * 18 = 54
Area of ABD = 1/2 * 6 * 8 = 24
Area of BCD = 54-24 = 30

B
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Re: In the figure above, if AC = 18, what is the area of ∆ BCD? [#permalink]

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24 Aug 2017, 04:00

AB^2 + 8^2 = 100

We get AB = 6

Area of triangle ABC - Area of triangle ABD = Area of triangle DBC

We get area = 30

Kudos [?]: 7 [0], given: 36

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Re: In the figure above, if AC = 18, what is the area of ∆ BCD? [#permalink]

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24 Aug 2017, 04:10

In right angled ∆BAD, AB $$= \sqrt{100 - 64} = \sqrt{36} = 6$$
The area of the ∆BAC = $$\frac{1}{2} * AB * AC = \frac{1}{2} * 6 * 18 = 54$$
The triangles have the same height, but their bases are in the ratio $$9:4(18:8)$$
Hence, the areas of the two triangles also have to be in the same ratio.

So the area of the ∆BCD must be $$\frac{9-4}{9}*54 = 30$$(Option B)
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Re: In the figure above, if AC = 18, what is the area of ∆ BCD? [#permalink]

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24 Aug 2017, 12:09
Bunuel wrote:

In the figure above, if AC = 18, what is the area of ∆ BCD?

(A) 15
(B) 30
(C) 54
(D) 60
(E) It cannot be determined

[Reveal] Spoiler:
Attachment:
2017-08-24_1249.png

$$Area = \frac{b*h}{2}$$

The height AB of ∆ BCD is the short leg of a 3-4-5 triangle, ∆ ABD.

Sides here have ratio of 6: 8: 10 (multiplier is x = 2, so 3x is 3*2 = 6)

If you are given a right triangle, and two of the three sides correspond with the 3x: 4x: 5x ratio, by definition the third side also corresponds.

Base CD of ∆ BCD is 10

(18 - 8) = 10

$$\frac{(6 * 10)}{2}$$ = 30

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Re: In the figure above, if AC = 18, what is the area of ∆ BCD? [#permalink]

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29 Aug 2017, 15:18
Bunuel wrote:

In the figure above, if AC = 18, what is the area of ∆ BCD?

(A) 15
(B) 30
(C) 54
(D) 60
(E) It cannot be determined

Since triangle ABD is a right triangle, we see that it’s a 6-8-10 right triangle with a height of 6. Thus, the area of triangle ABD is (8 x 6)/2 = 24.

Since triangle ABC is also a right triangle, the area of triangle ABC is [18 x 6]/2 = 54.

Since the area of triangle BCD is the difference between the areas of triangle ABC and triangle ABD, the area of triangle BCD is 54 - 24 = 30.

Alternate Solution:

We see that |DC| = |AC| - |AD| = 18 - 8 = 10. We also note that the height belonging to this base is the side AB, which is found to have a length of 6 from the ABD right triangle. Thus, the area of BCD is (10 x 6)/2 = 30.

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Re: In the figure above, if AC = 18, what is the area of ∆ BCD?   [#permalink] 29 Aug 2017, 15:18
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