Lisa bought a tank for her hermit crab. the tank can hold 1,331 cubic inches of water and includes a top. the base fo the aquarium is a square. the height of the aquarium is 11 inches. a. what is the length of each side of the base of the tank

11 in

Step-by-step explanation:

All we care about here is volume, not area.  So the top is irrelevant.

The volume of a rectangular tank with a square bottom, b, and height, h, is

V = b²h = 1,331 in³                       1,331 in³

Then b² = area of the base  = = 121 in²

                                                    11 in

The square of the length of any one side of the base is 121 in².

Taking the square root of this yields 11 in.

The length of each side of the base is 11 in.

  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.

step-by-step explanation:

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