Don't understand the "the width is split in two: 36.41m and 46.41m" part.
Do you mean the width to the inside of the track is 36.41 m and the width to the outside of the track is 46.41 m ?
I will assume that.
The semicircles at the ends are alike, so combined they would make up whole circles.
Area of the combined outer circle : radius = 46.41/2 m or 23.205 m
Area = π(23.205)^2 m^2
similarly, the area of the inner circle = π(18.205)^2
The difference in these two areas would be the area of the circular parts of the track. I am sure you can find the width of the two straight tracks, and you know the length is 85 m
Add up the two straights and the difference in the circles and you got it.
Let me know what you got.
there is a diagram.
the running track in the diagram consists of two parallel sections with semicircular sections at each end. determine the area of the track.
The rectangle in the running track has only the measurements. the measurements for the rectangle: length is 85m and the width is split in two: 36.41m and 46.41m
2 answers
the diagram looks like this:
h ttp://oalevelsolutions.com/CIE_GCE_AS_Maths_P1_13_Nov_11_Q_8_files/image001.png
when I meant the width is split in two: 36.41m and 46.41m, I actually meant the diameter is split into two radius: 36.41m and 46.41m
h ttp://oalevelsolutions.com/CIE_GCE_AS_Maths_P1_13_Nov_11_Q_8_files/image001.png
when I meant the width is split in two: 36.41m and 46.41m, I actually meant the diameter is split into two radius: 36.41m and 46.41m