Category: Geometry Chapter 11
Geometry Ch11.3 #38-41
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.
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
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
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.