Exercise 31 Page 350

Practice makes perfect

a We know that the triangular pennants are isosceles and congruent. Let's indicate congruent sides on the diagram.

There are several congruence statements we can write. Let's see a few of them!

Pairs of Congruent Segments	Reason
AB&≅CB DE&≅FE	The triangles are isosceles.
A B&≅D E C B&≅F E A C&≅D F	The triangles are congruent, so the corresponding sides are congruent.
AB&≅FE DE&≅CB	Transitive Property of Congruence

b We want to know how long the pennant string will need to be if the side length of a square roped off by the string is x feet. To do so, recall the formula for the area of a square. A= s^2 Here, s represents the side length of the square. In our case, the side length is equal to x. Then, the area is x^2 square feet. Now that we know the area needed, we can find the side length.

A=100

Substitute

A= x^2

x^2=100

SqrtEqn

sqrt(LHS)=sqrt(RHS)

x=sqrt(100)

CalcRoot

Calculate root

x=10

The side length of the roped off area is 10 feet. Since the region is a square, all four sides have this length. Length of pennant string:4 * 10=40ft

c We know the length of the pennant string in feet. The distance between pennants is given in inches. We can convert this distance to feet. Let's do it!

6in.=0.5ft We can now find the number of pennants on the string by dividing the length of the string by the distance between the pennants. Number of pennants:40/0.5=80ft

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