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# In the figure above, all triangles have the same area. If the area of

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Math Expert
Joined: 02 Sep 2009
Posts: 59622
In the figure above, all triangles have the same area. If the area of  [#permalink]

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11 Nov 2017, 06:57
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15% (low)

Question Stats:

86% (01:08) correct 14% (01:07) wrong based on 44 sessions

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In the figure above, all triangles have the same area. If the area of the shaded region is 84, and the area of square PQRS is 100, what is the total area of the regions outlined by the heavy line?

(A) 16
(B) 92
(C) 108
(D) 116
(E) 132

Attachment:

2017-11-11_1745_002.png [ 53.1 KiB | Viewed 1025 times ]

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Re: In the figure above, all triangles have the same area. If the area of  [#permalink]

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11 Nov 2017, 07:55
1
Bunuel wrote:

In the figure above, all triangles have the same area. If the area of the shaded region is 84, and the area of square PQRS is 100, what is the total area of the regions outlined by the heavy line?

(A) 16
(B) 92
(C) 108
(D) 116
(E) 132

Attachment:
2017-11-11_1745_002.png

The area of the shaded region is the area of the square minus 4 of the 8 triangles with the same area. So

The (area of the shaded region) equals the (area of the square) minus ($$\frac{1}{2}$$ the area of all the congruent triangles), $$\frac{1}{2}T$$:
$$84 = 100 - \frac{1}{2}T$$
$$\frac{1}{2}T = (100 - 84) = 16$$

The area enclosed by the thick outline equals the square plus $$\frac{1}{2}$$ the area of the all the triangles:
Area = $$100 + \frac{1}{2}T$$
$$\frac{1}{2}T = 16$$

Area enclosed by thick outline = $$100 + 16 = 116$$

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Re: In the figure above, all triangles have the same area. If the area of  [#permalink]

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11 Nov 2017, 08:02
1
Bunuel wrote:

In the figure above, all triangles have the same area. If the area of the shaded region is 84, and the area of square PQRS is 100, what is the total area of the regions outlined by the heavy line?

(A) 16
(B) 92
(C) 108
(D) 116
(E) 132

Attachment:
2017-11-11_1745_002.png

Area of shaded region + Area of 4 Triangles = Area of the square

So, 84 + Area of 4 Triangles = 100

So, Area of 4 Triangles = 100 - 84

Or, Area of 4 Triangles = 16

Since, Area of each triangle = 4

Since all the triangles are of equal area , the area of the total figure = Area of the square + area of 4 trinagles

Or, Area of the total figure = 100 + 16

Or, Area of the total figure = 116, answer will be (D)

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Re: In the figure above, all triangles have the same area. If the area of  [#permalink]

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19 Nov 2017, 08:05
1
Bunuel wrote:

In the figure above, all triangles have the same area. If the area of the shaded region is 84, and the area of square PQRS is 100, what is the total area of the regions outlined by the heavy line?

(A) 16
(B) 92
(C) 108
(D) 116
(E) 132

Attachment:
2017-11-11_1745_002.png

Bunuel

Can you please explain how come the area is 132?
Is it a wrong OA or something that everyone is missing out on.

Answer clearly seems to be 116.

Math Expert
Joined: 02 Sep 2009
Posts: 59622
Re: In the figure above, all triangles have the same area. If the area of  [#permalink]

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19 Nov 2017, 08:56
GMATAspirer09 wrote:
Bunuel wrote:

In the figure above, all triangles have the same area. If the area of the shaded region is 84, and the area of square PQRS is 100, what is the total area of the regions outlined by the heavy line?

(A) 16
(B) 92
(C) 108
(D) 116
(E) 132

Attachment:
2017-11-11_1745_002.png

Bunuel

Can you please explain how come the area is 132?
Is it a wrong OA or something that everyone is missing out on.

Answer clearly seems to be 116.

The OA is D. Edited. Thank you.
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Re: In the figure above, all triangles have the same area. If the area of  [#permalink]

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22 Nov 2019, 01:12
We don't need to really get into the entire calculations for this question. The area of the shaded region is 84. => the area of the 4 triangles which are not included in the area of the shaded region is 100 - 84 = 16.

In the heavy border area, there are just those extra 4 triangles whose area is nothing but 16.

=> Total area = 100 + 16 = 116.

Ans D
Re: In the figure above, all triangles have the same area. If the area of   [#permalink] 22 Nov 2019, 01:12
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