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# In rectangle ADEH above, BG and CF are parallel to AH. Given lengths A

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In rectangle ADEH above, BG and CF are parallel to AH. Given lengths A  [#permalink]

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04 Oct 2017, 00:37
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15% (low)

Question Stats:

91% (01:26) correct 9% (01:52) wrong based on 44 sessions

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In rectangle ADEH above, BG and CF are parallel to AH. Given lengths AB = 2, BC = 1, CD = 3, and the length of DH, not shown, is equal to 10, what is the area of the small rectangle BCFG?

(A) 8
(B) 10
(C) 16
(D) 24
(E) 48

Attachment:

2017-10-04_1122_001.png [ 5.04 KiB | Viewed 720 times ]

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Re: In rectangle ADEH above, BG and CF are parallel to AH. Given lengths A  [#permalink]

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04 Oct 2017, 05:26
DH = diagonal = 10
AB+BC+CD=2+1+3=6

100-36=64
ah =8
8*1=8
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In rectangle ADEH above, BG and CF are parallel to AH. Given lengths A  [#permalink]

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04 Oct 2017, 06:21
Bunuel wrote:

In rectangle ADEH above, BG and CF are parallel to AH. Given lengths AB = 2, BC = 1, CD = 3, and the length of DH, not shown, is equal to 10, what is the area of the small rectangle BCFG?

(A) 8
(B) 10
(C) 16
(D) 24
(E) 48

Attachment:
2017-10-04_1122_001.png

The area of small rectangle BCFG is L*W. Its width, BC = 1

Find its length from the given length of the diagonal of the larger rectangle. DH = 10

∆ ADH is a right triangle. Side AD = 6 (2 + 1 + 3). Its diagonal DH = 10

∆ ADH has side lengths
6: AH: 10
∆ ADH is a 3x-4x-5x right triangle with side length ratio 6: 8: 10

Rule: If given the lengths of a diagonal and one leg of a right triangle, and the two known values conform to the 3-4-5 ratio, the triangle is a 3-4-5 right triangle. The third side, AH = 8.*

AH = 8 = length of rectangle BCFG

L = 8, W = 1, area = LW

8 * 1 = 8 = Area of BCFG

*Or use Pythagorean theorem. Let the two known values = $$a$$ and $$c$$

$$a^2 + AH^2 = c^2$$
$$6^2 + AH^2 = 10^2$$
$$36 + AH^2 = 100$$
$$AH^2 = 64$$
$$AH = 8$$
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Re: In rectangle ADEH above, BG and CF are parallel to AH. Given lengths A  [#permalink]

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05 Oct 2017, 01:26
Side AD= AB+BC+CD= 6 units.
Diagnol DH= 10 units.

So, ADE is a 6-8-10 Right Angled Triangle, with Side DE=CF= 8 units.

Therefore, Area of Rectangle BCFG= BC x CF = 1x8 = 8 Sq. Units.

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Re: In rectangle ADEH above, BG and CF are parallel to AH. Given lengths A  [#permalink]

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06 Oct 2017, 10:51
Bunuel wrote:

In rectangle ADEH above, BG and CF are parallel to AH. Given lengths AB = 2, BC = 1, CD = 3, and the length of DH, not shown, is equal to 10, what is the area of the small rectangle BCFG?

(A) 8
(B) 10
(C) 16
(D) 24
(E) 48

We can see that the length of AH = DE = GB = FC. If we denote any of these lengths by b, using the Pythagorean theorem, we have:

6^2 + b^2 = 10^2

b^2 = 64

b = 8

Thus, the area of rectangle BCFG is 8 x 1 = 8.

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Re: In rectangle ADEH above, BG and CF are parallel to AH. Given lengths A &nbs [#permalink] 06 Oct 2017, 10:51
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# In rectangle ADEH above, BG and CF are parallel to AH. Given lengths A

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