Exercise 143 Page 459

Practice makes perfect

a Let's consider each of the statements one at a time using the given p and q.

p =& Polygon is a parallelogram q =& area of the polygon & equals base times height. We can write the conditional statement p⇒ q in an if-then form. If a polygon is a parallelogram, then the area of the polygon equals its base times its height.

b The converse of a conditional statement q⇒ p reverses the hypothesis and the conclusion of the conditional statement.

If the area of a polygon equals its base times its height, then the polygon is a parallelogram. This statement is true. To see why, let's recall the formulas for areas of polygons that use bases and heights.

Polygon	Formula for Area
Triangle	1/2bh
Trapezoid	1/2(b_1+b_2)h
Parallelogram	bh

As we can see, only the area for the parallelogram is equal to the base times the height.

c We want to write a conjecture about triangles that is similar to the one in Part A. Since the statement involves the area of a triangle, let's recall that formula.

A = 1/2bh As we can see, the formula is similar to the one in Part A, but it also includes a factor of 12. The formula can be read as the area A is one-half its base b times its height h. Let's adjust the parts p and q from Part A so that the statement fits triangles.

p =& Polygon is a triangle q =& area of the polygon equals &one-half its base times its height. Let's write the conditional statement p ⇒ q in an if-then form. If a polygon is a triangle, then the area of the polygon equals one-half its base times its height. We also need to write it as an arrow diagram. Polygon is a triangle ⇓ area of the polygon equals one-half base times height

d We want to the converse of the conditional statement from Part C. Since our statement is p ⇒ q, the converse statement is q ⇒ p.

If the area of the polygon equals one-half its base times its height, then the polygon is a triangle. This statement is true. To see why, let's recall the formulas for areas of polygons that use bases and heights.