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Geometry Ch11.3 #38-41

Hannah asked:

I am doing Harold Jacobs Geometry,  I am on chapter 11 lesson 3.  I got to problems 39-41 and got stuck.  

Dr Callahan Answered:

Let us start at #38.

We can see that <ABC = 120 so <ADB = 30. Therefore we know AB = BD based on Theorem 10 page 159.

#39. 

AB = x  because given

BD = x  by substitution since AB  = BD

<BDC = 30 (since a right triangle with one angle 60)

2BC = BD (Them 51 on page 442)

2BC = x by substitution

BC = x/2 

#40

same logic using thrm 51 again  — of course starting from what we know from #39
CD = BC*SQRT(3)  Thrm 51 on page 442

BC = x/2 from #39

CD = x *SQRT(3) / 2

#41

We are to show how we get CD = 7/8 AB

AB = x

CD = x *SQRT(3) /2

We are to show CD/AB = 7/8 so we can divide what we have as AB and CD to get

CD/AB = (x *SQRT(3) /2 ) / x

CD/AB = SQRT(3) /2

Now you have to calculate SQRT(3) /2 to get about 0.866 and then do the math on 7/8 to get 0.875

They are not equal – but approximately equal.