## Learning Math: Geometry

# The Pythagorean Theorem Homework

## Session 6, Homework

**Problem H1**

Find the height of an equilateral triangle with the following side lengths:

a. |
1 centimeter |

b. |
2 centimeters |

c. |
10 centimeters |

d. |
n centimeters |

**Problem H2
**

a. |
A baseball diamond is really a square 90 feet on a side. How far is second base from home plate? |

b. |
Baseball rules specify that the pitcher’s mound must be 60 feet, six inches from home plate, in the direction of second base. Is the pitcher’s mound in the center of the diamond? If not, is it closer to home plate or second base? |

**Converse**

One common mistake in mathematics is assuming that if a statement is true, the converse of the statement is also true. To form the converse of a statement, you switch the “if” and “then” parts of the statement. Here’s an example where the converse is clearly not true:

Statement: If you live in San Francisco, then you live in California. Converse: If you live in California, then you live in San Francisco.

You can write the Pythagorean theorem as an “if-then” statement as well:

The Pythagorean theorem: If, given a triangle, the square built on the hypotenuse is equal to the sum of the sqares built on the other two sides.

Converse: If the square built on the hypotenuse is equal to the sum of the squares built on the other two sides, then you have a right triangle. (If a^{2} + b^{2} = c^{2} for some triangle, then it must be a right triangle.)

**Problem H3**

Below is an outline for a proof of the Pythagorean theorem’s converse. Complete the proof by filling in the reasons.

Suppose you have some triangle ABC, where the lengths of the sides satisfy the relationship a^{2} + b^{2} = c^{2}.

a. |
Construct a right triangle with legs a and b (so there is a right angle between the sides with lengths a and b). |

b. |
What is the length of the hypotenuse of your new triangle? Why? |

c. |
Your new triangle and triangle ABC must be congruent. Why? |

d. |
Triangle ABC must be a right triangle. Why? |

**Problem H4**

Show that the altitude to the base (non-equal side) of an isosceles triangle bisects the base.

**Concurrencies**

Recall that in Session 1, it seemed (through experiments in folding paper) that the three perpendicular bisectors of any triangle were concurrent — they met at the same point, called the circumcenter of the triangle. We now have the tools to provide an explanation for this surprising fact.

All the points on the perpendicular bisector of AB are the same distance from A and B.

All the points on the perpendicular bisector of AC are the same distance from A and C.

**Problem H5**

Use the drawings and explanations above to describe point P, where the two perpendicular bisectors meet. Why must the perpendicular bisector of BC also go through point P?

All Homework problems (H1-H5) and text adapted from Connected Geometry, developed by Educational Development Center, Inc. pp. 65, 200, 201, 205, 206. © 2000 Glencoe/McGraw-Hill. Used with permission. www.glencoe.com/sec/math

### Solutions

**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 a^{2} = (a / 2)^{2} + h^{2}.

So h^{2} = a^{2} – (a / 2)^{2} = a^{2} – a^{2}/4 = 3 a^{2}/4.

.

Applying this formula, we have the following:

a. |
√3 / 2 |

b. |
√3 |

c. |
5 √3 |

d. |
n √3 / 2 |

**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. |

**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 a^{2} + b^{2} = c^{2}, so the length of the hypotenuse is √c^{2} = c. |

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. |

**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 h^{2} = a^{2} + x^{2} and h^{2} = a^{2} + y^{2}. So we must have a^{2} + x^{2} = a^{2} + y^{2}, or x^{2} = y^{2}, 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.

**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.