What are the answer options?
Answer:
(√2)/2
Step-by-step explanation:
The ratio of the radius of the circle to the side of the inscribed square is the same regardless of the size of the objects.
The radius of the circle is half the length of the diagonal of the square. For simplicity, we can call the side of the square 1, so its diagonal is √(1²+1²) = √2 by the Pythagorean theorem. The radius is half that value, so is (√2)/2. The desired ratio is this value divided by 1.
Scaling up our unit square to one with a side length of 3 inches, we have ...
radius/side = ((3√2)/2) / 3 = (√2)/2
_____
A square with a side length of 3 inches will have an area of (3 in)² = 9 in².
All work shown, I hope that this works!
It’s 12:35, if you add 40+25 it equals 65, 60 minutes is an hour, subtract 65-60 and you’ll get 5. The hour brings you to 12:30 and then you add the remaining time (5 minutes), which will give you 12:35.
Answer:
Option (4). Rhombus
Step-by-step explanation:
From the figure attached,
Distance AB = ![\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%28x_%7B2%7D-x_%7B1%7D%29%5E2%2B%28y_%7B2%7D-y_%7B1%7D%29%5E2%7D)
= ![\sqrt{(1-4)^2+(-5+3)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%281-4%29%5E2%2B%28-5%2B3%29%5E2%7D)
= ![\sqrt{(-3)^2+(-2)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%28-3%29%5E2%2B%28-2%29%5E2%7D)
= ![\sqrt{13}](https://tex.z-dn.net/?f=%5Csqrt%7B13%7D)
Distance BC = ![\sqrt{(4-1)^2+(-3+1)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%284-1%29%5E2%2B%28-3%2B1%29%5E2%7D)
= ![\sqrt{9+4}](https://tex.z-dn.net/?f=%5Csqrt%7B9%2B4%7D)
= ![\sqrt{13}](https://tex.z-dn.net/?f=%5Csqrt%7B13%7D)
Distance CD = ![\sqrt{(-2-1)^2+(-3+1)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%28-2-1%29%5E2%2B%28-3%2B1%29%5E2%7D)
= ![\sqrt{9+4}](https://tex.z-dn.net/?f=%5Csqrt%7B9%2B4%7D)
= ![\sqrt{13}](https://tex.z-dn.net/?f=%5Csqrt%7B13%7D)
Distance AD = ![\sqrt{(1+2)^2+(-5+3)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%281%2B2%29%5E2%2B%28-5%2B3%29%5E2%7D)
= ![\sqrt{9+4}](https://tex.z-dn.net/?f=%5Csqrt%7B9%2B4%7D)
= ![\sqrt{13}](https://tex.z-dn.net/?f=%5Csqrt%7B13%7D)
Slope of AB (
) = ![\frac{y_{2}-y_{1}}{x_{2}-x_{1}}](https://tex.z-dn.net/?f=%5Cfrac%7By_%7B2%7D-y_%7B1%7D%7D%7Bx_%7B2%7D-x_%7B1%7D%7D)
= ![\frac{4-1}{-3+5}](https://tex.z-dn.net/?f=%5Cfrac%7B4-1%7D%7B-3%2B5%7D)
= ![\frac{3}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B2%7D)
Slope of BC (
) = ![\frac{4-1}{-3+1}](https://tex.z-dn.net/?f=%5Cfrac%7B4-1%7D%7B-3%2B1%7D)
= ![-\frac{3}{2}](https://tex.z-dn.net/?f=-%5Cfrac%7B3%7D%7B2%7D)
If AB and BC are perpendicular then,
![m_{1}\times m_{2}=-1](https://tex.z-dn.net/?f=m_%7B1%7D%5Ctimes%20m_%7B2%7D%3D-1)
But it's not true.
[
= -
]
It shows that the consecutive sides of the quadrilateral are not perpendicular.
Therefore, ABCD is neither square nor a rectangle.
Slope of diagonal BD =
= Not defined (parallel to y-axis)
Slope of diagonal AC =
= 0 [parallel to x-axis]
Therefore, both the diagonals AC and BD will be perpendicular.
And the quadrilateral formed by the given points will be a rhombus.