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lisathedork

Member

Registered: 2013-10-24

Posts: 36

Area of Circles and Sectors

May I get some assistance on these problems? Can you also give me some helpful notes also?

2. SQRT(3)
A = (PI)r^2
A = (PI)SQRT(3)^2
A = 9 PI

14. t^2(PI)
t2(PI) = (PI)r^2

15. t^2(PI)
t2(PI) = (PI)r^2

16. 26
A = (n/360)[^2]
A = (120/360)[(PI)62]
A = (3/3)[(PI)36]
A = 108/3(PI)
A = 36

20. 120
A = (n/360)[(PI)r^2]
A = (120/360)[(PI)62]
A = (3/3)[(PI)36]
A = 108/3(PI)
A = 36

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Area of Circles and Sectors

hi lisathedork

Have a look here for some excellent notes on this:

http://www.mathsisfun.com/geometry/circ … gment.html

2. SQRT(3)   Is this the radius?

A = (PI)r^2
A = (PI)SQRT(3)^2

root 3 squared is 3 not 9

A = 9 PI

14. t^2(PI)    ?? What is the question here?

t2(PI) = (PI)r^2

15. t^2(PI) and here?

t2(PI) = (PI)r^2

16. 26 26 is what?

In what follows it looks like the angle of the sector is 120 and the radius is 6.
A = (n/360)[^2]
A = (120/360)[(PI)6^2]  Good start.
A = (3/3)[(PI)36]But what happened here.  6 squared is 36 ... ok  ... but 120/360 simplifies to 1/3
A = 108/3(PI)
A = 36

20. 120  Now I'm confused.  This is the same as question 16.
A = (n/360)[(PI)r^2]
A = (120/360)[(PI)62]
A = (3/3)[(PI)36]
A = 108/3(PI)
A = 36

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

lisathedork

Member

Registered: 2013-10-24

Posts: 36

Re: Area of Circles and Sectors

What is the area of the circle of the radius is:
2.    SQRT(3)
A = (PI)r2
A = (PI)SQRT(3)2
A = 9 PI

What is the radius if the area of the circle is:
14. t^2(PI)
t2(PI) = (PI)r2
Need Help

15. t^2(PI)
t2(PI) = (PI)r2
Need help

What is the area of the sector if the radius is 6 and the degree measure is:

16.     26
A = (n/360)[(PI)r2]
A = (26/360)[(PI)62]
A = (13/180)[(PI)36]
A = 468/180(PI)
A = 2 3/5

20.    120
A = (n/360)[(PI)r2]
A = (120/360)[(PI)62]
A = (3/3)[(PI)36]
A = 108/3(PI)
A = 36

Sorry for the confusion!

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Area of Circles and Sectors

hi lisathedork,

Would you mind if I called you Lisa ?  I find it clashes with my optimistic nature to give in to your self assessment. 

What is the area of the circle of the radius is:
2.    SQRT(3)
A = (PI)r2
A = (PI)SQRT(3)2
  sqrt(3) is about 1.732...  When you square it you get 3 not 9.

A = 9 PI

What is the radius if the area of the circle is:
14. t^2(PI)
t2(PI) = (PI)r2
Need Help  The pi cancels leaving you with a radius of 't'.

15. t^2(PI)
t2(PI) = (PI)r2
Need help
This seems to be the same question again ???

What is the area of the sector if the radius is 6 and the degree measure is:

16.     26
A = (n/360)[(PI)r2]
A = (26/360)[(PI)62]
A = (13/180)[(PI)36]
A = 468/180(PI)
A = 2 3/5   You just need to put a pi in this answer.

Another thought occurs to me.  Is your teacher recognising that you mean two and three fifths ?  That looked like twenty three over five when I first looked at this question.

20.    120
A = (n/360)[(PI)r2]
A = (120/360)[(PI)62]
A = (3/3)[(PI)36]

Something funny has happened to the fraction 120/360.  That cancels down like this:

A = 108/3(PI)
A = 36

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

lisathedork

Member

Registered: 2013-10-24

Posts: 36

Re: Area of Circles and Sectors

Thank you so much for the help! I figure out how to do 14 & 15. Plus I made the minor changes in the other problems also. Yes, I wouldn't mind being called Lisa!

Thanks again!