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Triangle ACE is equilateral with side lengths of 8. Points B
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24 Jan 2020, 06:39
24 Jan 2020, 06:39
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#greprepclub Triangle ACE is equilateral with side lengths of 8.jpg [ 12.07 KiB | Viewed 4027 times ]
Triangle ACE is equilateral with side lengths of 8. Points B and D are the midpoints of line segments AC and CE respectively.
Line segment BD is a diameter of the circle with center F. What is the area of the shaded region?
A. \(8 \sqrt{2} - 4\pi\)
B. \(12 \sqrt{3} - 2\pi
\)
C. \(12 \sqrt{3} -4\pi\)
D. \(16 \sqrt{3} - 2\pi\)
E. \(16 \sqrt{2} -4\pi\)
Kudos for the right answer and explanation
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Re: Triangle ACE is equilateral with side lengths of 8. Points B
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24 Jan 2020, 07:20
24 Jan 2020, 07:20
1
Solution:
We know that BD is the diameter (F is the center of the circle)
Let G be the mid-point of the triangle AEC; we join BG and DG
We can conclude (think why) that triangles ABG, DGE, BDG and CBD are congruent, and hence, of equal area
Area of equilateral triangle AEC (of side 8) = \(\sqrt{3}\)/4 * \(8^2\) = \(16\sqrt{3}\)
=> Areas of the triangles ABG, DGE, BDG and CBD = \((1/4) * 16\sqrt{3}\) = \(4\sqrt{3}\)
Thus, area of ABDE = sum of the areas of ABG, BGD and GDE = \(3 * 4\sqrt{3}\) = \(12\sqrt{3}\)
From ABDE, if we remove the area of the semicircle, we will obtain the area of the shaded region as asked in the question.
Since triangles ABG, DGE, BDG and CBD are congruent, we have: BD = AG = GE = 1/2 * AE = 4 = Diameter of the circle
=> Radius of the circle = 4/2 = 2
=> Area of the semicircle (lower half) = \((1/2) * π * 2^2\) = \(2π\)
Thus, area of the shaded (grey) region = \(12\sqrt{3} - 2π\)
Answer B
We know that BD is the diameter (F is the center of the circle)
Let G be the mid-point of the triangle AEC; we join BG and DG
We can conclude (think why) that triangles ABG, DGE, BDG and CBD are congruent, and hence, of equal area
Area of equilateral triangle AEC (of side 8) = \(\sqrt{3}\)/4 * \(8^2\) = \(16\sqrt{3}\)
=> Areas of the triangles ABG, DGE, BDG and CBD = \((1/4) * 16\sqrt{3}\) = \(4\sqrt{3}\)
Thus, area of ABDE = sum of the areas of ABG, BGD and GDE = \(3 * 4\sqrt{3}\) = \(12\sqrt{3}\)
From ABDE, if we remove the area of the semicircle, we will obtain the area of the shaded region as asked in the question.
Since triangles ABG, DGE, BDG and CBD are congruent, we have: BD = AG = GE = 1/2 * AE = 4 = Diameter of the circle
=> Radius of the circle = 4/2 = 2
=> Area of the semicircle (lower half) = \((1/2) * π * 2^2\) = \(2π\)
Thus, area of the shaded (grey) region = \(12\sqrt{3} - 2π\)
Answer B
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Triangle ACE is equilateral with side lengths of 8. Points B
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14 Aug 2021, 09:34
14 Aug 2021, 09:34
1
first 20 seconds strategy just by looking and applying concepts: BD is a diameter, and it is half of the side of CAE, since triangles CBD and CAE are similar, two equal sides, common angle, etc. To find the requested we need to select half of the circle, which is below it's diameter BD with center F (also, check triangle-circle intercepted points B and D), and find the area of CBD. Subtract both areas of half of the circle and triangle CBD from the area of triangle CAE to find the requested.
sometimes typing everything and someone else's reading is more time consuming than one's intuitive thinking and strategic development, this must be a click in mind
now calculate everything
Diameter BD is half the side 8 and equals to 4. The circle area is \(3.14*2^2\), where radius is BD/2=4/2=2. Half of the circle area is just 2*3.14 or 2pi
Area of equilateral triangle CBD with each side=4 equals to \(4^2*sq.root(3)/4=4*sq.root(3)\)
Area of equilateral triangle CAE equals to \(8^2*sq.root(3)/4=16*sq.root(3)\)
The required answers is \(16*sq.root(3)\) - \(4*sq.root(3)\) - \(2pi\) = \(12*sq.root(3)- 2pi\)
Answer is B
Triangle ACE is equilateral with side lengths of 8. Points B and D are the midpoints of line segments AC and CE respectively.
Line segment BD is a diameter of the circle with center F. What is the area of the shaded region?
A. \(8 \sqrt{2} - 4\pi\)
B. \(12 \sqrt{3} - 2\pi
\)
C. \(12 \sqrt{3} -4\pi\)
D. \(16 \sqrt{3} - 2\pi\)
E. \(16 \sqrt{2} -4\pi\)
Kudos for the right answer and explanation
sometimes typing everything and someone else's reading is more time consuming than one's intuitive thinking and strategic development, this must be a click in mind
now calculate everything
Diameter BD is half the side 8 and equals to 4. The circle area is \(3.14*2^2\), where radius is BD/2=4/2=2. Half of the circle area is just 2*3.14 or 2pi
Area of equilateral triangle CBD with each side=4 equals to \(4^2*sq.root(3)/4=4*sq.root(3)\)
Area of equilateral triangle CAE equals to \(8^2*sq.root(3)/4=16*sq.root(3)\)
The required answers is \(16*sq.root(3)\) - \(4*sq.root(3)\) - \(2pi\) = \(12*sq.root(3)- 2pi\)
Answer is B
Carcass wrote:
Attachment:
#greprepclub Triangle ACE is equilateral with side lengths of 8.jpg
Triangle ACE is equilateral with side lengths of 8. Points B and D are the midpoints of line segments AC and CE respectively.
Line segment BD is a diameter of the circle with center F. What is the area of the shaded region?
A. \(8 \sqrt{2} - 4\pi\)
B. \(12 \sqrt{3} - 2\pi
\)
C. \(12 \sqrt{3} -4\pi\)
D. \(16 \sqrt{3} - 2\pi\)
E. \(16 \sqrt{2} -4\pi\)
Kudos for the right answer and explanation
