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# In the square above, M, N, P and Q are midpoints of the sides. If the

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In the square above, M, N, P and Q are midpoints of the sides. If the  [#permalink]

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09 Nov 2017, 23:43
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84% (00:49) correct 16% (01:25) wrong based on 64 sessions

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In the square above, M, N, P and Q are midpoints of the sides. If the area of the square region is K, what is the area of the shaded region?

(A) K/3
(B) K/2
(C) 2K/3
(D) 3K/4
(E) 7K/8

Attachment:

2017-11-10_1034_001.png [ 4.72 KiB | Viewed 883 times ]

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Re: In the square above, M, N, P and Q are midpoints of the sides. If the  [#permalink]

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10 Nov 2017, 00:25
Bunuel wrote:

In the square above, M, N, P and Q are midpoints of the sides. If the area of the square region is K, what is the area of the shaded region?

(A) K/3
(B) K/2
(C) 2K/3
(D) 3K/4
(E) 7K/8

Attachment:
2017-11-10_1034_001.png

Consider K=16

Side will be 4, Since M,N,P,Q are midpoints . sides of triangle will be 2

Hence area of two triangles will be 1/2(2)(2)=2+2 4(Area of two triangles)

Total area of shaded region will be 16-4=12

Sub K=16 in options then D gives us 12.

Hence D
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Re: In the square above, M, N, P and Q are midpoints of the sides. If the  [#permalink]

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10 Nov 2017, 00:30
These problems are easier to work with provided you select an arbitrary value for K.

Let the area of the square(K) be 16. Therefore the side of the square is 4.
The triangles formed using the mid points M,N,P, and Q have side 2 and are isosceles in nature.

The area of the triangle will be $$\frac{1}{2}*2*2 = 2$$
Hence, the area of the two triangles(which are congruent) are 4
This is $$\frac{1}{4}$$th the area of the square, making the area of shaded potion, $$\frac{3}{4}$$th of the area of the square

The area of the shaded portion is $$\frac{3K}{4}$$(Option D)
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Re: In the square above, M, N, P and Q are midpoints of the sides. If the  [#permalink]

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10 Nov 2017, 03:33
Bunuel wrote:

In the square above, M, N, P and Q are midpoints of the sides. If the area of the square region is K, what is the area of the shaded region?

(A) K/3
(B) K/2
(C) 2K/3
(D) 3K/4
(E) 7K/8

Attachment:
2017-11-10_1034_001.png

Say the side of the square is 2 so that half the side is 1. So MN will be $$\sqrt{2}$$.
Also area of this square is 2^2 = 4

Consider the shaded region as a square formed by joining mid points + two triangles (out of the leftover 4 congruent triangles)

So area of inscribed smaller square will be $$\sqrt{2}^2 = 2$$

Leftover are of 4 triangles is 2 so area of 2 triangles will be 1.

Shaded area $$= (2+1)/4 = 3/4$$th the area of original square.

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Re: In the square above, M, N, P and Q are midpoints of the sides. If the  [#permalink]

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10 Nov 2017, 23:49
Bunuel wrote:

In the square above, M, N, P and Q are midpoints of the sides. If the area of the square region is K, what is the area of the shaded region?

(A) K/3
(B) K/2
(C) 2K/3
(D) 3K/4
(E) 7K/8

Attachment:
2017-11-10_1034_001.png

Side of the square region = $$\sqrt{(K)}$$
Area of right isosceles triangle = 1/2 * $$\sqrt{(K)}$$ /2 * $$\sqrt{(K)}$$ /2 = K/8
Area of shaded region = Total area of square - 2* area of right isosceles triangle = K - K/4 = 3K/4

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Re: In the square above, M, N, P and Q are midpoints of the sides. If the  [#permalink]

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14 Nov 2017, 07:12
Bunuel wrote:

In the square above, M, N, P and Q are midpoints of the sides. If the area of the square region is K, what is the area of the shaded region?

(A) K/3
(B) K/2
(C) 2K/3
(D) 3K/4
(E) 7K/8

Attachment:
2017-11-10_1034_001.png

Since the area of the square is K, each side is √K. Thus, the legs of each triangle are √K/2. Thus:

Shaded region = K - [(√K/2 x √K/2 x 1/2) + (√K/2 x √K/2 x 1/2)]

Shaded region = K - [K/8 + K/8] = 8K/8 - 2K/8 = 6K/8 = 3K/4

Alternate Solution:

Let’s assume that the length of a side of the square is 2; thus, the area of the square is 2^2 = 4.

Look at the unshaded triangle with hypotenuse MN. The base and height of that triangle are both 1, and so the area of that triangle is A = (bh)/2 = 1 x 1 /2 = ½.

Likewise, the unshaded triangle with hypotenuse QP has the same area A = ½.

The two unshaded regions have a total area of ½ + ½ = 1. Since the entire square has an area of 4, the shaded region has an area of 4 – 1 = 3, and thus the shaded region’s area is ¾ that of the entire square.
Since the square has a total region of K, we thus can say that the shaded region is (¾)K, or 3K/4.

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Re: In the square above, M, N, P and Q are midpoints of the sides. If the &nbs [#permalink] 14 Nov 2017, 07:12
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