Find the area of a square with side lengths of 2/5 inch
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Answer:
√2:1
Step-by-step explanation:
First we need to know that the length of the side of the square is equal to the diameter of the inscribed circle i.e
L = di
Given the area of the square to be 9in², we can get the length of the square.
Area of a square = L²
L is the length of the square.
9 = L²
L = √9
L = 3in
Hence the length of one side of the square is 3in
This means that the diameter of the inscribed circle di is also 3in.
Circumference of a circle = π×diameter of the circle(di)
Circumference of inscribed circle = π×3
= 3π in
For the circumscribed circumscribed circle, diameter of the outer circle will be equivalent to the diagonal of the square.
To get the diagonal d0, we will apply the Pythagoras theorem.
d0² = L²+L²
d0² = 3²+3²
d0² = 9+9
d0² = 18
d0 = √18
d0 = √9×√2
d0 = 3√2 in
Hence the diameter of the circumscribed circle (d0) is 3√2 in
Circumference of the circumscribed circle = πd0
= π(3√2)
= 3√2 π in
Hence, ratio of the circumference of the circumscribed circle to the one of the inscribed will be 3√2 π/3π = √2:1
The solution is the point of intersection between the two equations.
Assuming you have a graphing calculator or a program to lets you graph equations (I use desmos) you simply put in the equetions and note down the coordinates of the point of intersection.
In the graph the first equation is in blue and the second in red.
The point of intersection = the solution = (-6 , -1)
If you dont have access to a graphing calculator you could draw the graphs by hand;
1) Draw a table of values for each equation; you do this by setting three or four values for x and calculating its image in y (you can use any values of x)
y = 0.5 x + 2 (Im writing 0.5 instead of 1/2 because I find its easier in this format)
x | y
-1 | 1.5 * y = 0.5 (-1) + 2 = 1.5
0 | 2 * y = 0.5 (0) + 2 = 2
1 | 2.5 * y = 0.5 (1) + 2 = 2.5
2 | 3 * y = 0.5 (2) + 2 = 3
y = x + 5
x | y
-1 | 4 * y = (-1) + 5 = 4
0 | 5 * y = (0) + 5 = 5
1 | 6 * y = (1) + 5 = 6
2 | 7 * y = (2) + 5 = 7
2) Plot these point on the graph
I suggest to use diffrent colored points or diffrent kinds of point markers (an x or a dot) to avoid confusion about which point belongs to which graph
3) Using a ruler draw a line connection all the dots of one graph and do the same for the other
4) The point of intersection is the solution
Assuming you have a graphing calculator or a program to lets you graph equations (I use desmos) you simply put in the equetions and note down the coordinates of the point of intersection.
In the graph the first equation is in blue and the second in red.
The point of intersection = the solution = (-6 , -1)
If you dont have access to a graphing calculator you could draw the graphs by hand;
1) Draw a table of values for each equation; you do this by setting three or four values for x and calculating its image in y (you can use any values of x)
y = 0.5 x + 2 (Im writing 0.5 instead of 1/2 because I find its easier in this format)
x | y
-1 | 1.5 * y = 0.5 (-1) + 2 = 1.5
0 | 2 * y = 0.5 (0) + 2 = 2
1 | 2.5 * y = 0.5 (1) + 2 = 2.5
2 | 3 * y = 0.5 (2) + 2 = 3
y = x + 5
x | y
-1 | 4 * y = (-1) + 5 = 4
0 | 5 * y = (0) + 5 = 5
1 | 6 * y = (1) + 5 = 6
2 | 7 * y = (2) + 5 = 7
2) Plot these point on the graph
I suggest to use diffrent colored points or diffrent kinds of point markers (an x or a dot) to avoid confusion about which point belongs to which graph
3) Using a ruler draw a line connection all the dots of one graph and do the same for the other
4) The point of intersection is the solution