Mathematics, 15.07.2020 01:01 lwaites18

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For number 5, we know that the triangles are not similar. Therefore, their corresponding sides are not proportional. I believe you could prove this by diving Triangle PQR’s side lengths with their corresponding side lengths in Triangle STU. I’ll do it below:
Side PQ corresponds to side TS. 9/7 is roughly 1.29, which would make that its scale factor. (I’m dividing because when side lengths are proportional, there is a specific value, the scale factor, that you can multiply or divide by to find the unknown side length of a corresponding side). Side QR corresponds to side UT. 15/10 is 1.5, thus making that the scale factor between these two sides. Finally, we see that side PR corresponds to side US. 12/8 is 1.5, making the scale factor between those two sides 1.5. Although two pairs of corresponding sides have the same scale factor, the same cannot be said for the third pair of corresponding sides. In order for the triangles to be similar, the scale factor should remain constant between each corresponding side, therefore, the triangles are not similar.
For number 6, since you are only given the side lengths, you would not be able to use the AA (Angle-Angle Similarity Theorem), or SAS (Side-Angle-Side Similarity Theorem). The calculations are shown below:
PQ~TS: 9/6= 1.5
QR~UT: 15/10= 1.5
PR~US: 12/8= 1.5

The scale factor for each pair of corresponding sides is congruent, proving that the triangles are proportional, making them similar.

Answer from: Quest

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.

Find the value of the variable and the length of each chord for #5 and #6.

Answer from: Quest

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