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If each side of ΔACD above has length 3 and if AB has length

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If each side of ΔACD above has length 3 and if AB has length  [#permalink]

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New post Updated on: 28 Dec 2018, 05:44
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Equilateral.png
Equilateral.png [ 1.79 KiB | Viewed 6888 times ]
If each side of ΔACD above has length 3 and if AB has length 1, what is the area of region BCDE?

(A) \(\frac{9}{4}\)

(B) \(\frac{7}{4} \sqrt{3}\)

(C) \(\frac{9}{4} \sqrt{3}\)

(D) \(\frac{7}{2} \sqrt{3}\)

(E) \(6 + \sqrt{3}\)

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Originally posted by boomtangboy on 07 Apr 2012, 19:39.
Last edited by HKD1710 on 28 Dec 2018, 05:44, edited 2 times in total.
Edited the question and the diagram
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Re: If each side of ΔACD above has length 3 and if AB has length  [#permalink]

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New post 08 Apr 2012, 01:34
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boomtangboy wrote:
Attachment:
The attachment Equilateral.png is no longer available
If each side of ΔACD above has length 3 and if AB has length 1, what is the area of region BCDE?

(A) \(\frac{9}{4}\)

(B) \(\frac{7}{4} \sqrt{3}\)

(C) \(\frac{9}{4} \sqrt{3}\)

(D) \(\frac{7}{2} \sqrt{3}\)

(E) \(6 + \sqrt{3}\)

Attachment:
Equilateral.png
Equilateral.png [ 1.79 KiB | Viewed 7217 times ]

Since each side of ΔACD has length 3 then ACD is an equilateral triangle and its each angle is 60°.

Now, the are of equilateral triangle is \(area_{equilateral}=side^2*\frac{\sqrt{3}}{4}=9\frac{\sqrt{3}}{4}\) (for more check math-triangles-87197.html);

Next, since angle A is 60° then right triangle ABE is a 30°-60°-90° right triangle. Now, in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio \(1 : \sqrt{3}: 2\), the leg opposite 30° (AB) corresponds with \(1\) and the leg opposite 60° (BE) corresponds with \(\sqrt{3}\) (the hypotenuse AE corresponds with 2). So, since \(AB=1\) then \(BE=\sqrt{3}\), and the area of ABE is \(\frac{1}{2}*AB*BE=\frac{\sqrt{3}}{2}\);

The area of of region BCDE is the area of ACD minus the area of ABE: \(9\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{2}=9\frac{\sqrt{3}}{4}-2\frac{\sqrt{3}}{4}=7\frac{\sqrt{3}}{4}\).

Answer: B.

Hope it's clear.
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Re: If each side of ΔACD above has length 3 and if AB has length  [#permalink]

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New post 07 Apr 2012, 22:08
2
equilateral triangle hence Angle A = 60 deg
thus, in Triangle ABE, AB = 1 BE = 3^(1/2) and AE = 2

Area BCDE = Area ACD - Area ABE

Area ACD = 3^(1/2)/4 * (side)^2

Area ABE = 1/2 * 3^(1/2) * 1

we get 3.5/2 * 3^(1/2) multiplying numerator and denominator by 2 we get B
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Re: If each side of ΔACD above has length 3 and if AB has length  [#permalink]

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New post 31 Dec 2018, 19:51
In Triangle ABE one angle is 90, angle BAE is 60 (as ACD is equilateral) hence other angle is 30.

Sin 30 = 1/2 , Sin 60 = Root 3 / 2

Sin 30 = side opposite to the 30 degrees / Hypotenuse
Sin 60 = side opposite to the 60 degree / Hypotenuse

Solving we get Ab = 1 , AE = 2 , BE = 1

Area of ABE = 1/2 * 1 * Root 3

BCDE = ACD - ABE
solving we get option B
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Re: If each side of ΔACD above has length 3 and if AB has length  [#permalink]

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New post 02 Jan 2019, 03:09

Solution


Given:
    • In triangle ACD, each side length = 3
    • AB = 1

To find:
    • The area of quadrilateral BCDE

Approach and Working:
Area of BCDE = area of ACD – area of ABE
    • Area of ACD = \(\frac{√3 * 3^2}{4} = \frac{9√3}{4}\)
    • Area of ABE = \(\frac{1}{2} * 1 * BE = \frac{BE}{2}\)

And, we know angle A = 60 degrees. So, we can say that ABE is a 30 – 60 – 90 degrees triangle.
    • Thus, AB : BE : AE = 1 : √3 : 2
      o Implies, BE = √3

    • So, area of ABE = \(\frac{√3}{2}\)

Therefore, area of BCDE = \(\frac{9√3}{4} - \frac{√3}{2} = \frac{7√3}{4}\)

Hence, the correct answer is Option B

Answer: B

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