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# In the figure above, the perimeter of ∆MNP is how much greater than th

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Senior Manager
Joined: 18 Jun 2018
Posts: 256
In the figure above, the perimeter of ∆MNP is how much greater than th  [#permalink]

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08 Aug 2018, 06:07
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25% (medium)

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78% (01:51) correct 22% (02:15) wrong based on 76 sessions

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In the figure above, the perimeter of ∆MNP is how much greater than the perimeter of the shaded region?

A) $$2 + \sqrt{2}$$
B) $$6$$
C) $$8\sqrt{2}$$
D) $$6+3\sqrt{2}$$
E) $$6+8\sqrt{2}$$
Manager
Joined: 23 May 2017
Posts: 231
Concentration: Finance, Accounting
WE: Programming (Energy and Utilities)
Re: In the figure above, the perimeter of ∆MNP is how much greater than th  [#permalink]

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08 Aug 2018, 07:01
It is D.

Perimeter of larger triangle = 16 +8$$\sqrt{2}$$
Perimeter of shaded triangle = 10 + 5$$\sqrt{2}$$

difference = 6 + 3$$\sqrt{2}$$
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Joined: 22 May 2016
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In the figure above, the perimeter of ∆MNP is how much greater than th  [#permalink]

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08 Aug 2018, 08:31
1
Bismarck wrote:
Attachment:
Untitled.png

In the figure above, the perimeter of ∆MNP is how much greater than the perimeter of the shaded region?

A) $$2 + \sqrt{2}$$
B) $$6$$
C) $$8\sqrt{2}$$
D) $$6+3\sqrt{2}$$
E) $$6+8\sqrt{2}$$

The shaded region and ∆MNP are both right isosceles triangles with angle measures 45-45-90. By definition, if a right angled triangle (given) has legs of equal length (given), it is a 45-45-90 triangle.

The length of the hypotenuse of each triangle is missing but very easy to find.

45-45-90 triangles have sides opposite those angles in the ratio $$x: x: x\sqrt{2}$$

Legs, opposite the 45° angles, correspond with $$x$$ in the ratio
Each leg length = $$5$$, so in the ratio, $$x=5$$

The hypotenuse of the small triangle is opposite the 90° angle.
The hypotenuse corresponds with $$x\sqrt{2}$$. Therefore its length = $$5\sqrt{2}$$

Or use Pythagorean theorem:
$$5^2+5^2=h^2$$
$$(25+25)=50=h^2$$
$$\sqrt{h^2}=\sqrt{25*2}$$
$$h=5\sqrt{2}$$

Perimeter = Leg + Leg + Hypotenuse
$$(5+5+5\sqrt{2})=10+5\sqrt{2}$$

Large triangle
$$x=8=$$ length of each leg
Hypotenuse*, $$h=x\sqrt{2}=8\sqrt{2}$$
Perimeter: $$(8+8+8\sqrt{2})=16+8\sqrt{2}$$

Perimeter of ∆MNP is how much greater than the perimeter of shaded region (the small triangle)?

$$(16+8\sqrt{2} - (10+5\sqrt{2}))=$$
$$16+8\sqrt{2} - 10-5\sqrt{2}=$$
$$(16-10)+(8\sqrt{2}-5\sqrt{2})=$$

$$6+3\sqrt{2}$$

*If you do not catch the 45-45-90 special triangle, again, just use the Pythagorean theorem to find the length of the hypotenuse
$$8^2+8^2=h^2$$
$$(64+64)=h^2$$
$$\sqrt{h^2}=\sqrt{64*2}$$
$$h=8\sqrt{2}$$
You do have to notice that the triangles are similar by AAA. Angles opposite equal leg lengths are equal. Each triangle has a 90° angle. In both triangles, the other two congruent angles must each = 45° because (90° + 2x) = 180°. Solve, and each x = 45°

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In the figure above, the perimeter of ∆MNP is how much greater than th   [#permalink] 08 Aug 2018, 08:31
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