Noah built a fenced dog run that is 8 yards long and 5 yards wide. He placed posts at every corner and every yard along the length and width of the run.

The question is not complete, below is the completed question

Noah built a fenced dog run that is 8 yards long and 6 yards wide. He placed posts at every corner and every yard along the length and width of the run. How many posts did he use?

Answer

There are 28 posts in total

Step-by-step explanation:

Attached to this answer is a picture showing the fenced dog run and the position of the posts.

The number of posts is determined as follows:

Number of posts for the four corners = 4 posts

Number of posts for the 8-yard side = 7 posts

N:B There are 7 posts instead of 8 posts because one post has been counted as a corner post already, hence we will not count it twice.

∴ Total number of posts for both 8-yard sides = 7 × 2 = 14 posts

Number of posts for the 6-yard sides = 5 posts (same reason as above)

Ttal number of posts for both 6 yard sides = 5 × 2 = 10 posts.

Total number of posts = 4 + 14 + 10 = 28 posts

∴ There are 28 posts in total

  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.

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