- #1
Calcolat
- 3
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Hello all,
I've posted an image below of a very basic maths question I can't seem to solve. Even after looking at the correct answer I cannot seem to figure out how they got it no matter what I try.
The question is:
http://img684.imageshack.us/img684/6179/mr2q19.jpg
From what I can see it can be solved by breaking the composite shape into 2 circle sectors and a trapezium. Therefore I've tried to solve it by doing this:
[tex]\begin{array}{l}
A = \frac{1}{2}h(a + b) + 2*\frac{\theta }{{360}} \times \,\pi {r^2}\\
= \frac{1}{2}*\,10\,*\,(10 + 20) + 2\,*\,\frac{{60}}{{360}}*\,\pi *{10^2}\\
\approx 254.7c{m^2}
\end{array}[/tex]
However according to my maths book the correct answer for this is [itex]150.8c{m^2}[/itex] and no matter what I try I can't seem to come close to this answer. I've tried to solve it several different ways but the above answer is the closest I can get to the 150.8cm^2 answer the book gives me.
If someone could please let me know where I'm going wrong that would be much appreciated.
Thanks in advance.
I've posted an image below of a very basic maths question I can't seem to solve. Even after looking at the correct answer I cannot seem to figure out how they got it no matter what I try.
The question is:
http://img684.imageshack.us/img684/6179/mr2q19.jpg
From what I can see it can be solved by breaking the composite shape into 2 circle sectors and a trapezium. Therefore I've tried to solve it by doing this:
[tex]\begin{array}{l}
A = \frac{1}{2}h(a + b) + 2*\frac{\theta }{{360}} \times \,\pi {r^2}\\
= \frac{1}{2}*\,10\,*\,(10 + 20) + 2\,*\,\frac{{60}}{{360}}*\,\pi *{10^2}\\
\approx 254.7c{m^2}
\end{array}[/tex]
However according to my maths book the correct answer for this is [itex]150.8c{m^2}[/itex] and no matter what I try I can't seem to come close to this answer. I've tried to solve it several different ways but the above answer is the closest I can get to the 150.8cm^2 answer the book gives me.
If someone could please let me know where I'm going wrong that would be much appreciated.
Thanks in advance.
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