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A thin piece of wire 40 meters long is cut into two pieces. One piece is used to form a circle with radius r, and the other is used to form a square. No wire is left over. Which of the following represents the total area, in square meters, of the circular and the square regions in terms of r?
A. \(\pi*r^2\)
B. \(\pi*r^2 + 10\)
C. \(\pi*r^2 + \frac{1}{4}*\pi^2*r^2\)
D. \(\pi*r^2 + (40 - 2\pi*r)^2\)
E. \(\pi*r^2 + (10 - \frac{1}{2}\pi*r)^2\)
A. \(\pi*r^2\)
B. \(\pi*r^2 + 10\)
C. \(\pi*r^2 + \frac{1}{4}*\pi^2*r^2\)
D. \(\pi*r^2 + (40 - 2\pi*r)^2\)
E. \(\pi*r^2 + (10 - \frac{1}{2}\pi*r)^2\)
21 Dec 2010, 08:20
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A thin piece of wire 40 meters long is cut into two pieces. One piece is used to form a circle with radius r, and the other is used to form a square. No wire is left over. Which of the following represents the total area, in square meters, of the circular and the square regions in terms of r?
A. \(\pi*r^2\)
B. \(\pi*r^2 + 10\)
C. \(\pi*r^2 + \frac{1}{4}*\pi^2*r^2\)
D. \(\pi*r^2 + (40 - 2\pi*r)^2\)
E. \(\pi*r^2 + (10 - \frac{1}{2}\pi*r)^2\)
The area of a circle will be - \(\pi{r^2}\) and \(2\pi{r}\) meters of wire will be used;
There will be \(40-2\pi{r}\) meters of wire left for a square. Side of this square will be \(\frac{40-2\pi{r}}{4}=10-\frac{\pi{r}}{2}\), hence the area of the square will be \((10-\frac{\pi{r}}{2})^2\).
The total area will be - \(\pi{r^2}+(10-\frac{\pi{r}}{2})^2\).
Answer: E.
A. \(\pi*r^2\)
B. \(\pi*r^2 + 10\)
C. \(\pi*r^2 + \frac{1}{4}*\pi^2*r^2\)
D. \(\pi*r^2 + (40 - 2\pi*r)^2\)
E. \(\pi*r^2 + (10 - \frac{1}{2}\pi*r)^2\)
The area of a circle will be - \(\pi{r^2}\) and \(2\pi{r}\) meters of wire will be used;
There will be \(40-2\pi{r}\) meters of wire left for a square. Side of this square will be \(\frac{40-2\pi{r}}{4}=10-\frac{\pi{r}}{2}\), hence the area of the square will be \((10-\frac{\pi{r}}{2})^2\).
The total area will be - \(\pi{r^2}+(10-\frac{\pi{r}}{2})^2\).
Answer: E.
10 Nov 2014, 21:41
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In this problem, its not at all mentioned that "wire 40 meters long is cut into two equal pieces"
We have to take a variable for distribution of 40 meters in square & circle.
Refer diagram below:
sqa.png [ 3.51 KiB | Viewed 195812 times ]
Let the perimeter of the square = d
then circumference of the circle = 40-d
\(2\pi r = 40-d\)
\(d = 40 - 2\pi r\) ..................... (1)
Area of Circle \(= \pi r^2\) .............. (2)
Each side of square \(= \frac{d}{4} = \frac{40 - 2\pi r}{4} = 10 - \frac{\pi r}{2}\) .................... From (1)
Area of square \(= (10 - \frac{\pi r}{2})^2\)
Total Area \(= \pi r^2 + (10 - \frac{\pi r}{2})^2\)
Answer = E
We have to take a variable for distribution of 40 meters in square & circle.
Refer diagram below:
Attachment:
sqa.png [ 3.51 KiB | Viewed 195812 times ]
Let the perimeter of the square = d
then circumference of the circle = 40-d
\(2\pi r = 40-d\)
\(d = 40 - 2\pi r\) ..................... (1)
Area of Circle \(= \pi r^2\) .............. (2)
Each side of square \(= \frac{d}{4} = \frac{40 - 2\pi r}{4} = 10 - \frac{\pi r}{2}\) .................... From (1)
Area of square \(= (10 - \frac{\pi r}{2})^2\)
Total Area \(= \pi r^2 + (10 - \frac{\pi r}{2})^2\)
Answer = E
