I will try and explain this since I can't draw the picture that went with this question.
A circular glass overlay was placed on the top of a circular table. Find the area of the shaded region, rounding to the nearest tenth.
The shaded region is the part of the table outside of the glass. the glass is in the center of the circle.
the information given is that that outside area ( the shaded part is 1 m).
The glass overlay has a radius of 1/2 m.
so I first found the area of the glass overlay. A=pi squared
pi*1/2 ^2 this is pi*1/4=.8
next I added the 1/2m(inner circle radius) and the 1m because together they give the radius of the full table.
using the same formula above for Area I got 7.1m
now I subtracted 7.1-.8=6.3m for the area of the shaded area.
If this was not too confusing to understand, does it seem like I did this correctly?
thank you for checking my work
4 answers
I would not have rounded off intermediate answer, only round off your final answer.
I assume you used a calculator, get into the habit of using the memories of the calculator to store intermediate answers.
for area of glass overlay I got
A = .785398.. (I stored that )
area of table = 1.785398... (stored that)
let the radius of table be r
πr^2 = 1.785398..
r^2 = 1.785398/π
r = .7538..
I find a discrepancy in your question.
It said, "Find the area of the shaded region", but then you state that the shaded region has an area of 1 m^2
Anyway, we have all kinds of answers we can now play with
How did you get the radius of the table?
where did the 1.785398 come from?
I got the radius from adding the 1/2m from the inside circle and the 1m from the shaded in portion=1.5
put that in the equation getting 7.065
Can you please explain how you did you work? thank you
Round table, we don't know its radius, I eventually let it be r
Smaller circle lies on the table, the radius of the smaller table is 1/2 or .5 m
The area outside the circle is shaded and you said the area of that shaded region is 1 m^2
so I find the area of the circle to be
π(.5)^2 = .7853...
I added this to the shaded region to get the whole table area
= 1 + .7853 = 1.7853...
so now πr^2 = 1.7853
r^2 = 1.7853/π = .5686...
r = √.5686.. = .7539<---- radius of table