Answer:
AB = 8, HB = 6, Area of ∆ABC = 8[tex]\sqrt{3}[/tex], Perimeter of ∆ABC = 12 + 4[tex]\sqrt{3}[/tex]
Step-by-step explanation:
To find AB:
∆ABC is an 30,60,90∆
Using the theorem, you can find AB = 2AC = 2*4 = 8
AB = 8
To find HB:
You need to find AH to subtract from AB
Construct CH, a perpendicular bisector to side AB
From before you can put together that m∠CAB = 60°
∆ACH is an 30,60,90∆
Using this method again, AH = AC/2 = 4/2 = 2
Then you subtract AH from AB = 8-2 = 6
HB = 6
To find the area of ∆ABC:
You use the (base*height)/2 method
base = AB = 8
to find the height, CH, you need to use the Pytha Theorem
and get [tex]AH^{2}+CH^{2}=AC^{2}[/tex]
then substitute, and get [tex]2^{2} + CH^{2} = 4^{2}[/tex]
calculate and get CH = 2[tex]\sqrt{3}[/tex]
then the height = CH = 2[tex]\sqrt{3}[/tex]
solve the area and get
Area of ∆ABC = 8[tex]\sqrt{3}[/tex]
(optional perimeter)
to find perimeter of ∆ABC:
you add AC + CB + AB
you find CB by using opposite to 30°
CB = CH*2 = 2[tex]\sqrt{3}[/tex]*2 = 4[tex]\sqrt3}[/tex]
so AC + CB + AB = 4 + 4[tex]\sqrt3}[/tex] + 8
Perimeter of ∆ABC = 12 + 4[tex]\sqrt3}[/tex]
Hope this helps!!
