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What is the area of this figure? ABDC is a rectangle and BDE is an iso

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What is the area of this figure? ABDC is a rectangle and BDE is an iso [#permalink]

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New post 24 Oct 2017, 00:46
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A
B
C
D
E

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  15% (low)

Question Stats:

96% (00:40) correct 4% (01:01) wrong based on 47 sessions

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What is the area of this figure? ABDC is a rectangle and BDE is an isosceles right triangle.

(A) ab
(B) ab^2
(C) b(a + b/2)
(D) cab
(E) bc/2


[Reveal] Spoiler:
Attachment:
2017-10-24_1216_002.png
2017-10-24_1216_002.png [ 4.82 KiB | Viewed 468 times ]
[Reveal] Spoiler: OA

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Kudos [?]: 135406 [0], given: 12692

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Kudos [?]: 193 [0], given: 83

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Re: What is the area of this figure? ABDC is a rectangle and BDE is an iso [#permalink]

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New post 24 Oct 2017, 00:53
The area of the rectangle = a*b
the area of the isosceles triangle = 1/2*b*b = 2b^2/2

Total area = ab + 1/2b^2 = b(a + b/2)
Ans: C.

Kudos [?]: 193 [0], given: 83

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Joined: 22 May 2016
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Kudos [?]: 398 [0], given: 640

What is the area of this figure? ABDC is a rectangle and BDE is an iso [#permalink]

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New post 24 Oct 2017, 06:57
Bunuel wrote:
Image
What is the area of this figure? ABDC is a rectangle and BDE is an isosceles right triangle.

(A) ab
(B) ab^2
(C) b(a + b/2)
(D) cab
(E) bc/2

[Reveal] Spoiler:
Attachment:
2017-10-24_1216_002.png

The area of the figure = (Area of rectangle) + (Area of triangle)

Area of rectangle = \((LW) = (ab)\)

Area of isosceles right triangle:
It helps to know that the area of an isosceles right triangle = \(\frac{s^2}{2} =
\frac{b^2}{2}\)


If not, area of the isosceles right triangle, where side b is both Base and height =
\(\frac{B*h}{2}=\frac{b*b}{2}=\frac{b^2}{2}\)

Total area: \(ab + \frac{b^2}{2}\)
Looking at choices, factor out \(b\)
Total area: \(b(a + \frac{b}{2})\)

Answer C

Kudos [?]: 398 [0], given: 640

What is the area of this figure? ABDC is a rectangle and BDE is an iso   [#permalink] 24 Oct 2017, 06:57
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