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100^2+100^2=c^2
10,000+10,000=c^2
20,000=c^2
141=c
X-intercept; -.25
y-intercept; -3
Answer:
The distance between the posts is
feet.
Step-by-step explanation:
In the attached figure, let DB and AC be the posts and let AD be the wire attached to the top of the posts.
BC is the distance between the posts and we need to find BC.
Note that BC = DE
Also, EC = DB = 8m and AE = AC - EC = 17 - 8 = 9 m.
Now, in the right triangle ADE,
![AD^{2} =AE^{2} +DE^{2}](https://tex.z-dn.net/?f=AD%5E%7B2%7D%20%3DAE%5E%7B2%7D%20%2BDE%5E%7B2%7D)
![24^{2} =9^{2} +DE^{2}](https://tex.z-dn.net/?f=24%5E%7B2%7D%20%3D9%5E%7B2%7D%20%2BDE%5E%7B2%7D)
![DE^{2} =24^{2} -9^{2}](https://tex.z-dn.net/?f=DE%5E%7B2%7D%20%3D24%5E%7B2%7D%20-9%5E%7B2%7D)
= (24+9) (24-9)
= 33(15)
= (3 × 11) × (3 × 5)
![DE=3\sqrt{55}](https://tex.z-dn.net/?f=DE%3D3%5Csqrt%7B55%7D)
Since DE = BC,
![BC=3\sqrt{55}](https://tex.z-dn.net/?f=BC%3D3%5Csqrt%7B55%7D)
Hence, the distance between the posts is
feet.
Perimeter is all the way around and area is l x w
So 5 x 2 would be 10.....and 5 + 5 + 2 + 2 = 14
So the dimenstions are 5 x 2