For trapezoid JKLM, A and B are midpoints of the legs. Find AB.
1 answer:
Answer:
AB = 29
Explanation:
For a better understanding, we must work this problem in a graphic way. In the attached image we can see the solution.
First, we draw a vertical dotted lines from the point J & K to the line ML, then we can see two new portions with the same length. Then with this simple analysis:
2x = 39 - 19
x = 10
Then we know that x = 10, another important data to find the answer is that the AB line is located in the midpoints of the legs. We also can see the right triangle MJ and the dotted line.
Now for every single right triangle, no matter its size and relationship between the vertical and the horizontal lengths, if some point is located in the hypotenuse (leg) at the middle of its length. This will be proportional to the vertical and the horizontal cathetus, therefore we will have the middle point on those two lines.
So, the AB line will be the sum of JK plus two times 5
AB = 19 + 5 + 5 = 29
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Answer:
(a) the fundamental frequency of this string is 65 Hz
(b) the harmonics of the given frequencies are third and fourth respectively.
(c) the length of the string is 2.74 m
Explanation:
Given;
mass density of the string, μ = 3 x 10⁻³ kg/m
tension of the string, T = 380 N
resonating frequencies, 195 Hz and 260 N
For the given resonant frequencies;
(c) From any of the equations, solve for Length of the string (L);
(a) the fundamental frequency is calculated as;
(b) harmonics of the given frequencies;
the first harmonic (n = 1) = f₀ = 65 Hz
the second harmonic (n = 2) = 2f₀ = 130 Hz
the third harmonic (n = 3) = 3f₀ = 195 Hz
the fourth harmonic (n = 4) = 4f₀ = 260 Hz
Thus, the harmonics of the given frequencies are third and fourth respectively.