Answer:
<u>Formulas</u>
![\textsf{Area of a circle}=\pi r^2 \quad \textsf{(where r is the radius)}](https://tex.z-dn.net/?f=%5Ctextsf%7BArea%20of%20a%20circle%7D%3D%5Cpi%20r%5E2%20%5Cquad%20%5Ctextsf%7B%28where%20r%20is%20the%20radius%29%7D)
![\textsf{Radius of a circle}=\dfrac{1}{2}d \quad \textsf{(where d is the diameter)}](https://tex.z-dn.net/?f=%5Ctextsf%7BRadius%20of%20a%20circle%7D%3D%5Cdfrac%7B1%7D%7B2%7Dd%20%5Cquad%20%5Ctextsf%7B%28where%20d%20is%20the%20diameter%29%7D)
![\textsf{Area of a square}=s^2 \quad \textsf{(where s is the side length)}](https://tex.z-dn.net/?f=%5Ctextsf%7BArea%20of%20a%20square%7D%3Ds%5E2%20%5Cquad%20%5Ctextsf%7B%28where%20s%20is%20the%20side%20length%29%7D)
![\textsf{Diagonal of a square}=s\sqrt{2} \quad \textsf{(where s is the side length)}](https://tex.z-dn.net/?f=%5Ctextsf%7BDiagonal%20of%20a%20square%7D%3Ds%5Csqrt%7B2%7D%20%5Cquad%20%5Ctextsf%7B%28where%20s%20is%20the%20side%20length%29%7D)
<u>Question 14</u>
If a circle is inscribed in a square, then the diameter of the circle is equal to the side length of the square. Therefore, as the radius of a circle is half the diameter, the radius of the circle is half the side length of the square.
Given:
Therefore, the areas of the square and circle are:
![\begin{aligned} \textsf{Area of the circle} & =\pi (6x)^2\\ & = 36 \pi x^2 \:\: \sf cm^2 \end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%20%5Ctextsf%7BArea%20of%20the%20circle%7D%20%26%20%3D%5Cpi%20%286x%29%5E2%5C%5C%20%26%20%3D%2036%20%5Cpi%20x%5E2%20%5C%3A%5C%3A%20%5Csf%20cm%5E2%20%5Cend%7Baligned%7D)
![\begin{aligned}\textsf{Area of the square}& =(12x)^2\\ & = 144x^2 \:\: \sf cm^2 \end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Ctextsf%7BArea%20of%20the%20square%7D%26%20%3D%2812x%29%5E2%5C%5C%20%26%20%3D%20144x%5E2%20%5C%3A%5C%3A%20%5Csf%20cm%5E2%20%5Cend%7Baligned%7D)
Therefore, the ratio of the circle to square is:
![\implies \sf circle : square](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20circle%20%3A%20square)
![\implies 36 \pi x^2:144x^2](https://tex.z-dn.net/?f=%5Cimplies%2036%20%5Cpi%20x%5E2%3A144x%5E2)
![\implies 36 \pi :144](https://tex.z-dn.net/?f=%5Cimplies%2036%20%5Cpi%20%3A144)
![\implies \pi : 4](https://tex.z-dn.net/?f=%5Cimplies%20%5Cpi%20%3A%204)
![\implies \dfrac{1}{4} \pi : 1](https://tex.z-dn.net/?f=%5Cimplies%20%5Cdfrac%7B1%7D%7B4%7D%20%5Cpi%20%3A%201)
So the circle is ¹/₄π the size of the square.
<u>Question 15</u>
If a square is inscribed in a circle, then the diagonal of the square is the diameter of the circle. Therefore, as the radius of a circle is half the diameter, the radius of the circle is half the diagonal of the square.
Given:
![\begin{aligned} \textsf{Area of the circle} & =\pi (5a^2)^2\\ & = 25 \pi a^4 \:\: \sf cm^2 \end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%20%5Ctextsf%7BArea%20of%20the%20circle%7D%20%26%20%3D%5Cpi%20%285a%5E2%29%5E2%5C%5C%20%26%20%3D%2025%20%5Cpi%20a%5E4%20%5C%3A%5C%3A%20%5Csf%20cm%5E2%20%5Cend%7Baligned%7D)
![\begin{aligned}\textsf{Diagonal of a square} & =s\sqrt{2}\\10a^2 & = s \sqrt{2}\\ s & =\dfrac{10a^2}{\sqrt{2}}\\ s & = 5\sqrt{2}a^2\:\: \sf cm^ \end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Ctextsf%7BDiagonal%20of%20a%20square%7D%20%26%20%3Ds%5Csqrt%7B2%7D%5C%5C10a%5E2%20%26%20%3D%20s%20%5Csqrt%7B2%7D%5C%5C%20s%20%26%20%3D%5Cdfrac%7B10a%5E2%7D%7B%5Csqrt%7B2%7D%7D%5C%5C%20s%20%26%20%3D%205%5Csqrt%7B2%7Da%5E2%5C%3A%5C%3A%20%5Csf%20cm%5E%20%5Cend%7Baligned%7D)
![\begin{aligned}\textsf{Area of the square}& =(5\sqrt{2}a^2)^2\\ & = 50a^4 \:\: \sf cm^2 \end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Ctextsf%7BArea%20of%20the%20square%7D%26%20%3D%285%5Csqrt%7B2%7Da%5E2%29%5E2%5C%5C%20%26%20%3D%2050a%5E4%20%5C%3A%5C%3A%20%5Csf%20cm%5E2%20%5Cend%7Baligned%7D)
Therefore, the ratio of the circle to square is:
![\implies \sf circle : square](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20circle%20%3A%20square)
![\implies 25 \pi a^4:50a^4](https://tex.z-dn.net/?f=%5Cimplies%2025%20%5Cpi%20a%5E4%3A50a%5E4)
![\implies 25 \pi :50](https://tex.z-dn.net/?f=%5Cimplies%2025%20%5Cpi%20%3A50)
![\implies \pi : 2](https://tex.z-dn.net/?f=%5Cimplies%20%5Cpi%20%3A%202)
![\implies \dfrac{1}{2} \pi:1](https://tex.z-dn.net/?f=%5Cimplies%20%5Cdfrac%7B1%7D%7B2%7D%20%5Cpi%3A1)
So the circle is ¹/₂π the size of the square.
Answer:
answer is 30
Step-by-step explanation:
x=5 6(5)=30
Anwer is the first choice. When you plug in the values, the only choice that works is the first one
That’s answer is A)-3,4) because you reflect as follows
Answer: 7/8
Step-by-step explanation: Hope this helps.