20 point if you answer all questions Riley has a piece of yarn that is 6 1/2 ft long. She wants to cut the yarn into sections that are each 1/4 ft long.

How many sections will the yarn be cut into?

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A carpenter has 16 1/2m of wood. He cuts the wood into pieces that are each 2 3/4m long.
Kevin has a box of cereal that contains 10 1/2 cups of cereal. A serving size of the cereal is 3/4 cup.

How many servings are in the box of cereal?

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How many pieces of wood does the carpenter cut?

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Jason draws a line in the sand at the beach. The line is 7/8 ft long. He wants to divide the line into sections that are each 1/8 ft long.

How many sections will the line be divided into?

Enter your answer in the box.
Divide.

2 1/9÷2/3

13/8 45/3/8  7 7/8

Step-by-step explanation:

  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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step-by-step explanation:

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