Answer:
yes
Step-by-step explanation:
is there supposed to be a picture? if so it isn’t there
Since, a circle is inscribed in a square, it means that the side of a square is equal to the diameter of a circle.
Since, the perimeter of a square is 32 inches.
A perimeter is a path that surrounds the two dimensional shape.
Perimeter of square is given by ![4 \times side](https://tex.z-dn.net/?f=%204%20%5Ctimes%20side%20)
![32 = 4 \times side](https://tex.z-dn.net/?f=%2032%20%3D%204%20%5Ctimes%20side%20)
side ![= \frac{32}{4}](https://tex.z-dn.net/?f=%20%3D%20%5Cfrac%7B32%7D%7B4%7D%20)
side = 8 inches.
Since, side of a square = diameter of a circle
Therefore, diameter of a circle = 8 inches.
radius of a circle =
, where d is the diameter
radius =
= 4 inches.
The perimeter of a circle is termed as Circumference.
Circumference of a circle = ![2\Pi r](https://tex.z-dn.net/?f=%202%5CPi%20r%20)
= ![2\Pi \times 4](https://tex.z-dn.net/?f=%202%5CPi%20%5Ctimes%204%20)
=
inches.
Answer:
85 square units
Step-by-step explanation:
The area of a kite is half of the product of the diagonals.
The left half of the horizontal diagonal is labeled 5, so the entire horizontal diagonal measures 10.
Now we need to find the length of the vertical diagonal.
The vertical diagonal is made up of two segments. Each segment is a leg in a right triangle. We can use the Pythagorean theorem twice to find the lengths of the two segments of the vertical diagonal.
Upper vertical segment:
a^2 + b^2 = c^2
5^2 + x^2 = [5sqrt(2)]^2
25 + x^2 = 50
x^2 = 25
x = 5
The upper segment of the vertical diagonal has length 5.
Lower vertical segment:
a^2 + b^2 = c^2
5^2 + x^2 = 13^2
25 + x^2 = 169
x^2 = 144
x = 12
The lower segment of the vertical diagonal has length 12.
The length of the vertical diagonal is the sum of the lengths of the two vertical segments:
5 + 12 = 17
The diagonals of the kite measure 1`0 and 17.
area = d1 * d2/2
area = 10 * 17/2 = 170/2 = 85
Answer:
![(5-7)^{2}](https://tex.z-dn.net/?f=%285-7%29%5E%7B2%7D)
Step-by-step explanation:
(g o h)(5) = g(h(5))
h(5) = 5-7
plug in h(5) where you have x.
g(h(5)) = ![(5-7)^{2}](https://tex.z-dn.net/?f=%285-7%29%5E%7B2%7D)