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 \(x: x: x\sqrt{2}\)

Small triangle? (shaded region)

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)?

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

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

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

\(6+3\sqrt{2}\)

Answer D

*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°