Answer:
The area of the kite is ![56\ cm^{2}](https://tex.z-dn.net/?f=56%5C%20cm%5E%7B2%7D)
Step-by-step explanation:
we know that
The area of the kite is equal to
![A=\frac{1}{2}(D1*D2)](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B1%7D%7B2%7D%28D1%2AD2%29)
where
D1 and D2 are the diagonals of the kite
we have
![D1=4+4=8\ cm](https://tex.z-dn.net/?f=D1%3D4%2B4%3D8%5C%20cm)
![D2=10+4=14\ cm](https://tex.z-dn.net/?f=D2%3D10%2B4%3D14%5C%20cm)
Find the area
![A=\frac{1}{2}(8*14)=56\ cm^{2}](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B1%7D%7B2%7D%288%2A14%29%3D56%5C%20cm%5E%7B2%7D)
Step - by - step - explanation
So first find out the pattern. As you can see it just minuses 2 each time. So first
18, 16, 14, 12 (then minus another 2) equals 10, 8, 6, 4, 2.
So 2 is the ninth of the pattern, so 2 will be your answer.
Hope this helps.
Answer:
The correct answers on edg are 2) AE and CE are always equal 3) BE and DE are always equal 4) Segment AC and segment BD always bisect each other
Step-by-step explanation:
Answer:
The answer is option 1.
Step-by-step explanation:
Given that the formula for tangent is tanθ = opposite/adjacent :
![\tan(θ) = \frac{oppo.}{adj.}](https://tex.z-dn.net/?f=%20%5Ctan%28%CE%B8%29%20%20%3D%20%20%5Cfrac%7Boppo.%7D%7Badj.%7D%20)
Let oppo. = BC = 16,
Let adj. = AB = 12,
![\tan(θ) = \frac{16}{12}](https://tex.z-dn.net/?f=%20%5Ctan%28%CE%B8%29%20%20%3D%20%20%5Cfrac%7B16%7D%7B12%7D%20)