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# In the figure above, the vertices of rectangle ABCD lie on a circle wi

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Math Expert
Joined: 02 Sep 2009
Posts: 47077
In the figure above, the vertices of rectangle ABCD lie on a circle wi [#permalink]

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21 Nov 2017, 23:30
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In the figure above, the vertices of rectangle ABCD lie on a circle with center O. If BD = 8, CD = 4 and x = 120, what is the perimeter of the shaded region?

(A) 8π/3 + 2√3
(B) 8π/3 + 4√2
(C) 8π/3 + 4√3
(D) 8π + 2√3
(E) 8π + 4√3

Attachment:

2017-11-22_1022_003.png [ 8.37 KiB | Viewed 698 times ]

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Re: In the figure above, the vertices of rectangle ABCD lie on a circle wi [#permalink]

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22 Nov 2017, 00:32

We can calculate the Length of arc from the relationship between the angle subtended at the centre and the net circumference of the circle.

If angle X. Then the perimeter of arc is (X/360) X 2*pi*r.

We know that O is the centre and BD is 8. CD is 4.
There are 2 ways to find out the length of BC. Using trigonometry in Triangle OBC or using pythagorean theorem in Triangle BCD.

Arc length + BC is the perimeter.
Which is 8*pi*1/3 + 4 sqrt3.

Hence C

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Re: In the figure above, the vertices of rectangle ABCD lie on a circle wi [#permalink]

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22 Nov 2017, 00:47
Perimeter of shaded region = length of arc BC + line BC
length of arc BC = 120/360 * 2*P*r=1/3*2*P*4=8/3*P
line BC = (8^2-4^2)^1/2=4*3^1/2
substituting:
Perimeter of shaded region = length of arc BC + line BC = 8/3*P + 4*3^1/2
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In the figure above, the vertices of rectangle ABCD lie on a circle wi [#permalink]

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27 Nov 2017, 08:06
1
Bunuel wrote:

In the figure above, the vertices of rectangle ABCD lie on a circle with center O. If BD = 8, CD = 4 and x = 120, what is the perimeter of the shaded region?

(A) 8π/3 + 2√3
(B) 8π/3 + 4√2
(C) 8π/3 + 4√3
(D) 8π + 2√3
(E) 8π + 4√3

Attachment:
2017-11-22_1022_003.png

(Sector arc length) + (rectangle length BC)

Sector Arc Length

$$\frac{SectorAngle}{360}=\frac{Part}{Whole}=\frac{ArcLength}{Circumference}$$

Sector angle = x = 120°
r = 1/2 diagonal = 8/2 = 4
Circumference = $$2\pi r = 8\pi$$

$$\frac{120}{360}=\frac{1}{3}=\frac{ArcLength}{8\pi}$$

Arc Length$$*(3) = (8\pi)$$
Arc Length = $$\frac{8\pi}{3}$$

Length BC of rectangle

Length of BC = x
$$BC^2 + x^2 = AC^2$$
$$16 + x^2 = 64$$
$$x^2 = 48$$
$$\sqrt{x^2} =\sqrt{16*3}$$
$$x = 4\sqrt{3}$$

Perimeter = $$\frac{8\pi}{3} + 4\sqrt{3}$$

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Re: In the figure above, the vertices of rectangle ABCD lie on a circle wi [#permalink]

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27 Nov 2017, 12:37
Bunuel wrote:

In the figure above, the vertices of rectangle ABCD lie on a circle with center O. If BD = 8, CD = 4 and x = 120, what is the perimeter of the shaded region?

(A) 8π/3 + 2√3
(B) 8π/3 + 4√2
(C) 8π/3 + 4√3
(D) 8π + 2√3
(E) 8π + 4√3

Attachment:
2017-11-22_1022_003.png

We see that the perimeter of the shaded region consists of side BC of the rectangle and arc BC of the circle. Let’s first determine the length of side BC of the rectangle.

Using the Pythagorean theorem, we have:

4^2 + b^2 = 8^2

16 + b^2 = 64

b^2 = 48

b = √48 = √16 x √3 = 4√3 = side BC

Since the diagonal BD = 8 = diameter of the circle, the circumference of the circle = 8π. Since arc BC corresponds to a central angle of 120 degrees, the length of arc BC = (120/360) x 8π = (1/3) x 8π = 8π/3.

Thus, the perimeter of the shaded region = 8π/3 + 4√3.

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Re: In the figure above, the vertices of rectangle ABCD lie on a circle wi [#permalink]

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27 Nov 2017, 21:25
genxer123 wrote:
Bunuel wrote:

In the figure above, the vertices of rectangle ABCD lie on a circle with center O. If BD = 8, CD = 4 and x = 120, what is the perimeter of the shaded region?

(A) 8π/3 + 2√3
(B) 8π/3 + 4√2
(C) 8π/3 + 4√3
(D) 8π + 2√3
(E) 8π + 4√3

Attachment:
2017-11-22_1022_003.png

(Sector arc length) + (rectangle length BC)

Sector Arc Length

$$\frac{SectorAngle}{360}=\frac{Part}{Whole}=\frac{ArcLength}{Circumference}$$

Sector angle = x = 120°
r = 1/2 diagonal = 8/2 = 4
Circumference = $$2\pi r = 8\pi$$

$$\frac{120}{360}=\frac{1}{3}=\frac{ArcLength}{8\pi}$$

Arc Length$$*(3) = (8\pi)$$
Arc Length = $$\frac{8\pi}{3}$$

Length BC of rectangle

Length of BC = x
$$BC^2 + x^2 = AC^2$$
$$16 + x^2 = 64$$
$$x^2 = 48$$
$$\sqrt{x^2} =\sqrt{16*3}$$
$$x = 4\sqrt{3}$$

Perimeter = $$\frac{8\pi}{3} + 4\sqrt{3}$$

I like the way you explain things stepwise and with proper formatting-sign of a good teacher, this makes students understand and learn rather than scratch their heads.
Re: In the figure above, the vertices of rectangle ABCD lie on a circle wi   [#permalink] 27 Nov 2017, 21:25
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