Provide the missing reasons for the proof of part of the triangle midsegment theorem.
2 answers:
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<h2>Question #22 Answer</h2>
B. 2 in.
<h3>Explanation:</h3>Cross out the common factor
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<h2>Question #23 Answer (Picture attached)</h2>D. proportional, equal
<h3>Explanation:</h3> Δ
Δ ×
Answer:
h = 1.4
c = 2.8
Step-by-step explanation:
For each problem, remember the special triangle side ratios then use a proportion. To solve, isolate the variable.
For the triangle with the variable h:
Since two of the angles are 45, this is an isosceles triangle. All isosceles triangles have two equal sides that are not the hypotenuse.
In a right isosceles triangle, the ratio for regular side to hypotenuse is 1 to √2.
h = √2
h ≈ 1.4
For the triangle with the variable c:
The is an equilateral triangle cut in half because the angles are 30 and 60.
The side ratio of altitude to hypotenuse is √3 to 2.
c = 2√2
c ≈ 2.8
Answer:
b ≈ 101.52
Step-by-step explanation:
Given two sides and the angle between, the Law of Cosines is useful.
b^2 = a^2 +c^2 -2ac·cos(B)
b^2 = 105^2 +9^2 -2·105·9·cos(65°) ≈ 10307.251
b ≈ √10307.251
b ≈ 101.52
