Logan bought a baseball bat on a trip. The baseball bat is 29 inches long Logan thinks he can bring the baseball bat home in his suitcase. Is he correct? Use your understanding of the Pythagorean theorem to explain your answer

Step-by-step explanation:

Kindly find attached a description of Logan's suit case

given data

dimension of suitcase

adj, length=24in

opp, width=18in

dimension of base ball bat= 29in

we want to know if the bat can fit in the case, a way out is to solve for the diagonal or hypotenuse of the case using Pythagoras theorem.

we know that

hyp^2= opp^2+adj^2

hyp=√opp^2+adj^2

hyp=√18^2+24^2

hyp=√324+576

hyp=√900

hpy=30in

this means that the diagonal of the case is 30 in, seeing that the bat is 29 in, the bat cannot fit into the case

  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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