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Geometry Session 6: Solutions
 
Session 6 Part A Part B Part C Homework
 
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Solutions for Session 6 Homework

See solutions for Problems: H1 | H2 | H3 | H4 | H5


Problem H1

If the length of each side is a, the height divides the triangle into two right triangles with one leg having length a / 2 (since the height bisects the side to which it is perpendicular), and the hypotenuse having length a. Let h be the length of the other leg -- i.e., the length of the height. Using the Pythagorean theorem,
we have a2 = (a / 2)2 + h2.

So h2 = a2 - (a / 2)2 = a2 - a2/4 = 3 a2/4.

.

Applying this formula, we have the following:

a. 

b. 

c. 

d. 

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Problem H2

a. 

The distance between home plate and second base is the hypotenuse of a right triangle whose legs have length 90 feet. Using the Pythagorean theorem, the distance is , or approximately 127.3 feet.

b. 

If the pitcher's mound were in the center of the diamond, its distance from home plate would be half the total distance between home plate and second base. This distance is . To the nearest inch, this is 63 feet, eight inches. This means that the pitcher's mound is three feet, two inches forward of the center of the baseball diamond.

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Problem H3

a. 

Follow the instructions.

b. 

Using the Pythagorean theorem, the length of the hypotenuse is . By assumption, the sides of the original triangle satisfy the relationship a2 + b2 = c2, so the length of the hypotenuse is .

c. 

The sides a and b are congruent by construction, and the third side is c (as shown in part (b) of the problem). So the two triangles have corresponding sides that are congruent, and by SSS congruence, they are congruent themselves.

d. 

Since congruent triangles have congruent corresponding angles, so the angle between the sides of length a and b must be a right angle for both triangles, and the second triangle is a right triangle by construction, the original triangle must also have a right angle.

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Problem H4

By definition, the altitude forms a right angle with the base, so it divides the original triangle into two right triangles. The two right triangles share one leg (the altitude) of length a and have hypotenuses of same length, h, since the original triangle is an isosceles triangle. Suppose that the altitude divides the base into two line segments of length x and y, respectively. Applying the Pythagorean theorem to the two right triangles, we have h2 = a2 + x2 and h2 = a2 + y2. So we must have a2 + x2 = a2 + y2, or x2 = y2, and x = y (since both x and y are positive, as they represent distances). So the altitude divides the base into two equal lengths; i.e., it bisects the base.

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Problem H5

Since it lies on the perpendicular bisectors of AB and AC, the point P is the same distance away from A and B and from A and C. In other words, it is the same distance away from A, B, and C. The perpendicular bisector of BC contains all points that are the same distance away from B and C, so it must contain P as well. So the perpendicular bisector of BC must go through P. In this case, the converse is also true and easily verified.

Now think of a circle whose center is P and whose radius is the length of PA. Since P is the same distance away from A, B, and C, this means that PA, PB, and PC will all have the same length. The circle with center P that passes through A will also pass through B and C. This circle is called the circumcircle of triangle ABC, and P is called the circumcenter.

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