Bismarck wrote:
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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) \(6+8\sqrt{2}\)
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°
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