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# In the accompanying diagram ABCD is a rectangle. The area of isosceles

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In the accompanying diagram ABCD is a rectangle. The area of isosceles [#permalink]

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26 Oct 2017, 02:57
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In the accompanying diagram ABCD is a rectangle. The area of isosceles right triangle ABE = 7, and EC = 3(BE). The area of ABCD is

(A) 21
(B) 28
(C) 42
(D) 56
(E) 84

[Reveal] Spoiler:
Attachment:

2017-10-26_1355.png [ 5.52 KiB | Viewed 847 times ]
[Reveal] Spoiler: OA

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Re: In the accompanying diagram ABCD is a rectangle. The area of isosceles [#permalink]

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26 Oct 2017, 03:20
Area of triangle
1/2 bh = 7
1/2 (BE)(AB) = 7.

Area of rectangle
BC * CD = 4 BE * AB = 4*14 = 56 (OA)
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In the accompanying diagram ABCD is a rectangle. The area of isosceles [#permalink]

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26 Oct 2017, 09:17
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Bunuel wrote:

In the accompanying diagram ABCD is a rectangle. The area of isosceles right triangle ABE = 7, and EC = 3(BE). The area of ABCD is

(A) 21
(B) 28
(C) 42
(D) 56
(E) 84

[Reveal] Spoiler:
Attachment:
The attachment 2017-10-26_1355.png is no longer available

Attachment:

aaaaa.png [ 10.73 KiB | Viewed 600 times ]

I have a hard time tracking on too many letters in side and segment names.
Let $$x$$ = side of isosceles triangle (AB and BE)
Let $$3x$$ = EC, which is three times $$x$$(BE)

1) Find side length of isosceles triangle (= rectangle width) from triangle's area

Area of isosceles right triangle** is $$\frac{s^2}{2}$$
Given area = $$7$$. Side length:

$$\frac{x^2}{2} = 7$$

$$x^2 = 14$$

$$x = \sqrt{14}$$ = triangle side length = width of rectangle

2) Length of rectangle:
Length = $$x + 3x$$.
$$3x + x = 4x = 4\sqrt{14}$$

3) Area of rectangle:
$$(L * W) =$$
$$(4x * x) = (4\sqrt{14} * \sqrt{14}) = (4 * 14) = 56$$

**Or use: Area of triangle =
$$\frac{bh}{2} = \frac{(x*x)}{2}$$
$$7 = \frac{x^2}{2}$$
$$x^2 = 14$$
$$x = \sqrt{14}$$

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In the accompanying diagram ABCD is a rectangle. The area of isosceles [#permalink]

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26 Oct 2017, 10:16
Let the length BE be X.
EC = 3X
Area of triangle = 1/2 * BE * AB
7 = 1/2 * X * X
X^2 = 14

Area of rectangle = length * breadth = BC * AB
= 4X * X = 4X^2 = 4 * 14 = 56.

Option D

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Re: In the accompanying diagram ABCD is a rectangle. The area of isosceles [#permalink]

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30 Oct 2017, 13:53
Bunuel wrote:

In the accompanying diagram ABCD is a rectangle. The area of isosceles right triangle ABE = 7, and EC = 3(BE). The area of ABCD is

(A) 21
(B) 28
(C) 42
(D) 56
(E) 84

[Reveal] Spoiler:
Attachment:
2017-10-26_1355.png

We can first determine side AB = BE, which we can label as n.

(n^2)/2 = 7

n^2 = 14

n = √14

Thus, EC = 3√14 and BC = BE + EC = √14 + 3√14 = 4√14.

Thus, the area of rectangle ABCD is AB x BC = √14 x 4√14 = 4 x 14 = 56.

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Re: In the accompanying diagram ABCD is a rectangle. The area of isosceles   [#permalink] 30 Oct 2017, 13:53
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# In the accompanying diagram ABCD is a rectangle. The area of isosceles

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