Session 6: 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 a² + b² = c² 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 a² + b² = c².

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

Next > Session 7: Symmetry

Session 6: Index | Notes | Solutions | Video