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Big Ideas Math Integrated I, 2016 View details

1. Angles of Triangles

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Exercise 57 Page 594

We know that HK ≅ BC.

Practice makes perfect

From the two figures, we see that HK ≅ BC. This means we can equate the expressions of these segments. HK ≅ BC: 8z-9=3z+6Since the equation only contains z as a variable, it can be solved.

8z-9=3z+6

SubEqn

LHS-3z=RHS-3z

5z-9=6

AddEqn

LHS+9=RHS+9

5z=15

DivEqn

.LHS /5.=.RHS /5.

z=3

When we know that z= 3 we can substitute this into the expression for BC: BC: 3( 3)+6=15