Geometry: An Interactive Journey to Mastery

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This question, of course, is contrived: You can’t match the materials of a model ship with those of a real ship,
for example, so weights don’t scale with volume in this setting as claimed. Nonetheless, this is a useful exercise
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We are told that the scale factor between the two objects is k ͼ௘IURPPRGHOWRUHDOVKLS௘ͽRUk 1001
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D௘ͽ $UHDVFDOHVDVk^2 = 10,000. Thus, 2 × 10,000 = 20,000 gallons of paint are needed for the real ship.
E௘ͽ 9ROXPHͼ௘DQGKHQFHZHLJKW௘ͽVFDOHVDV k^3 = 1,000,000. The weight of the real ship is 50 million pounds.
F௘ͽ /HQJWKVFDOHVDVk 1001 , so the length of the model’s mast is 50 §· ̈ ̧©¹ 1001 0.5 feet.
G௘ͽ $UHDVFDOHVDVk^2 §· ̈ ̧©¹ 1001 2 , so the area of the matching window on the model is 60 §· ̈ ̧©¹ 1001 2 0.006
square feet.
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The altitude of a right triangle from the vertex of the right angle
to the hypotenuse of the right triangle divides the hypotenuse into
two sections of lengths x and y, as shown in )LJXUH6KRZ
that the length h of the altitude is given by hxy.
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Label the vertices of the triangle A, B, and C, as shown
in )LJXUH, and label the point at the base of the
altitude P.
If ‘A has measure aƒWKHQ‘ ABP 90 a,
‘  PBC 90 90 a a, and ‘ BCA 90 a.
By the AA principle, it follows that +ABP is similar
to +BCP.
Thus, matching sides in these two triangles come in the same ratio. In particular, hyxh , from which it follows
that hxy.

h

xy

Figure 24.3

h

a x

ía a

A ía

B

PCy

Figure 24.4