Patricia has a lawn, ABCD. She has placed a watering hose, BD, as shown below. Part A: Patricia plans to put a fence along the length AD of her lawn. What is the length of the fence required? Part B: Using complete sentences, explain how you arrived at the answer for Part A.
@Mertsj can you help me?
@Robb828 , do you know pythagoras theorem? a^2 + b^2 = c^2
yes i do
Yes, apply that and get length AD
I mean first find BD and then find AD
That is take triangle BCD and apply pythagoras theorem and get BD. Then take triangle ABD and apply pythagoras theorem and find AD
so i do 30^2 + 16^2 = c^2
yes. so get the value of c^2. Now take triangle ABD and apply sin 60 = BD / AD
Sorry Initially I thought AD is given. But AD is not given and hence you cannot apply pythagoras theorem for ABD
so what i got was 1156 = c^2
Get c. Then\ sin (60) = c / AB
you lost me...
No I didn't c= sqrt(1156) Now sin(60) = BD/ AB Do you know trigonometry??
not that much
Then sin of any angle is sin(A) = opposite side to A / Hypotenuse of triangle
Now you get it ??
yea
so....
Now apply this, we get sin(60) = BD / AB we know that sin (60) = sqrt(3) / 2 BD = sqrt (1156) Apply this and and get your AB
i got 34
@Robb828 BD= 34 Now you know \[\sin(60) = {\sqrt{3} \over 2}\] \[{\sqrt{3} \over 2} = {34 \over AB}\] \[AB = 34 * { 2 \over \sqrt{3}}\]
@Robb828 , make sure you practice trigonometry.
ok thanks for you help and trust me i will lol
yw :)
@shivam_bhalla i forgot to ask do i multiply those last two or just leave them like that??
@Robb828 , If you have a calcultor at home, then simplify it, else leave it that way
*calculator
ok i do thank you
yw :)
\[BD=\sqrt{30^2+16^2}=34\]
\[AD=\frac{34}{\sqrt{3}}=\frac{34\sqrt{3}}{3}\]
Triangle BCD is a right triangle with legs 30 and 16. Using the Pythagorean theorem I found the hypotenuse BD of that triangle. That hypotenuse BD is the long leg of the 30-60-90 right triangle BDA. AD is the short leg of that triangle. The short leg of a 30-60-90 triangle can be found by dividing the long leg by the square root of three.
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