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The area of square ABDE in the figure above is 100. If BC = CD, what

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Joined: 02 Sep 2009
Posts: 42305

Kudos [?]: 133072 [0], given: 12403

The area of square ABDE in the figure above is 100. If BC = CD, what [#permalink]

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11 Nov 2017, 06:41
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The area of square ABDE in the figure above is 100. If BC = CD, what is the perimeter of ∆ BCD?

(A) 5√2 + 10
(B) 20
(C) 10√2 + 10
(D) 30
(E) 20√2 + 10

[Reveal] Spoiler:
Attachment:

2017-11-10_1035_001.png [ 4.73 KiB | Viewed 246 times ]
[Reveal] Spoiler: OA

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Kudos [?]: 133072 [0], given: 12403

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Joined: 22 May 2016
Posts: 998

Kudos [?]: 347 [0], given: 594

The area of square ABDE in the figure above is 100. If BC = CD, what [#permalink]

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11 Nov 2017, 08:44
Bunuel wrote:

The area of square ABDE in the figure above is 100. If BC = CD, what is the perimeter of ∆ BCD?

(A) 5√2 + 10
(B) 20
(C) 10√2 + 10
(D) 30
(E) 20√2 + 10

[Reveal] Spoiler:
Attachment:
2017-11-10_1035_001.png

Area of square = s$$^2$$ = 100
s = 10
BD = 10

∆ BCD has a right angle and two equals sides (BC = CD).
∆ BCD is a right isosceles triangle with side lengths in ratio $$x: x: x\sqrt{2}$$

Sides BC and CD correspond with $$x$$
Side BD corresponds with $$x\sqrt{2}$$

$$x\sqrt{2} = 10$$

$$x =\frac{10}{\sqrt{2}}$$

$$x=\frac{10}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}}=\frac{10\sqrt{2}}{2}=5\sqrt{2}$$

$$x=5\sqrt{2}$$ = BC = CD

Perimeter P of ∆ BCD = BC + CD + BD

P = $$5\sqrt{2} + 5\sqrt{2} + 10 =$$
P = $$10\sqrt{2} + 10$$

Kudos [?]: 347 [0], given: 594

The area of square ABDE in the figure above is 100. If BC = CD, what   [#permalink] 11 Nov 2017, 08:44
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