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

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Math Expert
Joined: 02 Sep 2009
Posts: 44285
What is the area of this figure? ABDC is a rectangle and BDE is an iso [#permalink]

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24 Oct 2017, 01:46
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Difficulty:

15% (low)

Question Stats:

96% (00:38) correct 4% (01:01) wrong based on 52 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 [ 4.82 KiB | Viewed 592 times ]
[Reveal] Spoiler: OA

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GMAT 1: 730 Q49 V41
Re: What is the area of this figure? ABDC is a rectangle and BDE is an iso [#permalink]

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24 Oct 2017, 01:53
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.
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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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24 Oct 2017, 07:57
Bunuel wrote:

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: $$b(a + \frac{b}{2})$$

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What is the area of this figure? ABDC is a rectangle and BDE is an iso   [#permalink] 24 Oct 2017, 07:57
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