Basic Math - How to find the area of this composite shape?

[Calcolat]

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:

$$\begin{array}{l} A = \frac{1}{2}h(a + b) + 2*\frac{\theta }{{360}} \times \,\pi {r^2}\\ = \frac{1}{2}*\,10\,*\,(10 + 20) + 2\,*\,\frac{{60}}{{360}}*\,\pi *{10^2}\\ \approx 254.7c{m^2} \end{array}$$

However according to my maths book the correct answer for this is $150.8c{m^2}$ 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.

 

[Dick]

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:

$$\begin{array}{l} A = \frac{1}{2}h(a + b) + 2*\frac{\theta }{{360}} \times \,\pi {r^2}\\ = \frac{1}{2}*\,10\,*\,(10 + 20) + 2\,*\,\frac{{60}}{{360}}*\,\pi *{10^2}\\ \approx 254.7c{m^2} \end{array}$$

However according to my maths book the correct answer for this is $150.8c{m^2}$ 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.

The only thing I can see wrong is that I don't think h=10. But I also think the books answer isn't correct.

 

[Mentallic]

The value of h in a trapezium is the perpendicular distance (shortest distance) between the two parallel sides. The value of 10 in the picture is slanted. You need to use trigonometry to find the actual value of h.

 

[SammyS]

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:

$$\begin{array}{l} A = \frac{1}{2}h(a + b) + 2*\frac{\theta }{{360}} \times \,\pi {r^2}\\ = \frac{1}{2}*\,10\,*\,(10 + 20) + 2\,*\,\frac{{60}}{{360}}*\,\pi *{10^2}\\ \approx 254.7c{m^2} \end{array}$$

However according to my maths book the correct answer for this is $150.8c{m^2}$ 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.

The height of the trapezoid is not 10.

Use trig to find the trapezoid's height.

 

[Calcolat]

Ohhh I see... I incorrectly assumed that as the radius was 10cm it would make the height of the trapezium 10cm as well. Thanks for clarifying how that works.

Well I've now tried to have a go calculating the value of h using trig however I'm still way off the books answer of 150.8cm^2 :(

I made a right-angle triangle off one of the circle sectors and used (90-60 = 30 degrees) as the angle, the adjacent a as the unknown and hypotenuse as 10cm. Then...

$$\begin{array}{l} \cos {30^ \circ } = \frac{a}{{10}}\\ a = 10*\cos {30^ \circ }\\ a \approx 8.66cm \end{array}$$

However if I re-calculate my original answer using h = 8.66cm I still end up with only $234.6c{m^2}$

Am I still doing something wrong or is it possible that the books answer is incorrect as Dick mentioned?

 

[Dick]

Ohhh I see... I incorrectly assumed that as the radius was 10cm it would make the height of the trapezium 10cm as well. Thanks for clarifying how that works.

Well I've now tried to have a go calculating the value of h using trig however I'm still way off the books answer of 150.8cm^2 :(

I made a right-angle triangle off one of the circle sectors and used (90-60 = 30 degrees) as the angle, the adjacent a as the unknown and hypotenuse as 10cm. Then...

$$\begin{array}{l} \cos {30^ \circ } = \frac{a}{{10}}\\ a = 10*\cos {30^ \circ }\\ a \approx 8.66cm \end{array}$$

However if I re-calculate my original answer using h = 8.66cm I still end up with only $234.6c{m^2}$

Am I still doing something wrong or is it possible that the books answer is incorrect as Dick mentioned?

I agree with your value of h and the value of the area. Still think the book is wrong.

 

[SammyS]

I agree with your value of h and the value of the area. Still think the book is wrong.

I agree with Dick on all three accounts.

 

[Mentallic]

However if I re-calculate my original answer using h = 8.66cm I still end up with only $234.6c{m^2}$

Am I still doing something wrong or is it possible that the books answer is incorrect as Dick mentioned?

If we cut off the circular segment such that we're left with a rectangle with length 20cm and height 8.66cm (which is an area obviously smaller than before the cut was made) then we get an area of about 170cm² which is already more than the book's answer. So clearly the book is wrong

 

[Calcolat]

Wow thank you so much to all 3 of you as that was a huge help!

I spent over an hour on this last night re-reading the maths book and trying similar questions. Even tho I was okay with the other questions I still couldn't figure this one out and now I know why.

Thank you for not only helping correct the books wrong answer but for pointing out my mistake as well. If this forum had a "thumbs up" option I'd give all 3 of you a thumbs up