Exercise 36 Page 646

Let's begin with recalling the 45^(∘)-45^(∘)-90^(∘) Triangle Theorem. This theorem tells us that in an isosceles right triangle the legs l are congruent and the length of the hypotenuse is sqrt(2) times the length of a leg.

Let's look at the given picture. We will name the vertices with consecutive letters. First let's notice that △ ABC is a right isosceles triangle. This means that ∠ B and ∠ C have measures of 45^(∘).

Now we can see that △ ABD is also a 45^(∘)-45^(∘)-90^(∘) triangle. Therefore, the length of its hypotenuse, AB is sqrt(2) times the length of its leg, AD. AB=sqrt(2)AD Since we are given that the length of AB is 2 feet, we can solve the above equation for AD, which represents the length that we want to find.

AB=sqrt(2)AD

Substitute

AB= 2

2=sqrt(2)AD

▼

Solve for AD

DivEqn

.LHS /sqrt(2).=.RHS /sqrt(2).

2/sqrt(2)=AD

RearrangeEqn

Rearrange equation

AD=2/sqrt(2)

MultEqn

LHS * 1=RHS* 1

AD*1=2/sqrt(2)*1

OneToFrac

Rewrite 1 as sqrt(2)/sqrt(2)

AD*1=2/sqrt(2)*sqrt(2)/sqrt(2)

MultByOne

a * 1=a

AD=2/sqrt(2)*sqrt(2)/sqrt(2)

MultFrac

Multiply fractions

AD=2sqrt(2)/sqrt(2)*sqrt(2)

ProdToPowTwoFac

a* a=a^2

AD2sqrt(2)/(sqrt(2))^2

PowSqrt

( sqrt(a) )^2 = a

AD=2sqrt(2)/2

ReduceFrac

a/b=.a /2./.b /2.

AD=sqrt(2)/1

DivByOne

a/1=a

AD=sqrt(2)

RoundDec

Round to 2 decimal place(s)

AD≈ 1.41

Kei should make the support approximately 1.41 feet.