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Go for d!!
It has been proven that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
<h3>How to prove a Line Segment?</h3>
We know that in a triangle if one angle is 90 degrees, then the other angles have to be acute.
Let us take a line l and from point P as shown in the attached file, that is, not on line l, draw two line segments PN and PM. Let PN be perpendicular to line l and PM is drawn at some other angle.
In ΔPNM, ∠N = 90°
∠P + ∠N + ∠M = 180° (Angle sum property of a triangle)
∠P + ∠M = 90°
Clearly, ∠M is an acute angle.
Thus; ∠M < ∠N
PN < PM (The side opposite to the smaller angle is smaller)
Similarly, by drawing different line segments from P to l, it can be proved that PN is smaller in comparison to all of them. Therefore, it can be observed that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Read more about Line segment at; brainly.com/question/2437195
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Answer:
<h2>Your answer would be <em><u>0.34</u></em></h2>
Step-by-step explanation:
first we add them altogether:
0.25+0.17+0.21+0.03 = 0.66
than we subtact the 0.66 from the whole (1) =
<u>1-0.66= 0.34</u>
1/10 = .10
2/5 = .40
1/4 = .25
1/3 = .33
Therefore; 3.40>3.33>3.25>3.10