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pshichka [43]
3 years ago
7

How do you use parallel and perpendicular lines to solve real-world problems?

Mathematics
2 answers:
klasskru [66]3 years ago
6 0
If you work construction it’s in blue print salon with different angles on where you would cut wood or weld metal together.
ASHA 777 [7]3 years ago
4 0

Answer:

In engineering, construction, and technical fields

Step-by-step explanation:

The use of parallel lines has many wide applications

For example, the parallel lines are important in science in preventing parallax errors in measuring, say the meniscus of the water in a glass.

Another use is in construction. The use of the plumb line shows if the object being built is straight or not. The principle is that the force of the pull acts downwards. Another example is the spirit lever. This can be used by builders to check if the surface is even or not.

In engineering, the properties are important in building bridges and other high-tension structures.

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100 points! Mhanifa can you please help? Look at the picture attached. I will mark brainliest!
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Answer:

1 and 2.

Midpoints calculated, plotted and connected to make the triangle DEF, see the attached.

  • D= (-2, 2), E = (-1, -2), F = (-4, -1)

3.

As per definition, midsegment is parallel to a side.

Parallel lines have same slope.

<u>Find slopes of FD and CB and compare. </u>

  • m(FD) = (2 - (-1))/(-2 -(-4)) = 3/2
  • m(CB) = (1 - (-5))/(1 - (-3)) = 6/4 = 3/2
  • As we see the slopes are same

<u>Find the slopes of FE and AB and compare.</u>

  • m(FE) = (-2 - (- 1))/(-1 - (-4)) = -1/3
  • m(AB) = (1 - 3)/(1 - (-5)) = -2/6 = -1/3
  • Slopes are same

<u>Find the slopes of DE and AC and compare.</u>

  • m(DE) = (-2 - 2)/(-1 - (-2)) = -4/1 = -4
  • m(AC) = (-5 - 3)/(-3 - (-5)) = -8/2 = -4
  • Slopes are same

4.

As per definition, midsegment is half the parallel side.

<u>We'll show that FD = 1/2CB</u>

  • FD = \sqrt{(2+1)^2+(-2+4)^2} = \sqrt{3^2+2^2} = \sqrt{13}
  • CB = \sqrt{(1 + 5)^2+(1+3)^2} = \sqrt{6^2+4^2} = 2\sqrt{13}
  • As we see FD = 1/2CB

<u>FE = 1/2AB</u>

  • FE = \sqrt{(-4+1)^2+(-1+2)^2} = \sqrt{3^2+1^2} = \sqrt{10}
  • AB = \sqrt{(-5 -1)^2+(3-1)^2} = \sqrt{6^2+2^2} = 2\sqrt{10}
  • As we see FE = 1/2AB

<u>DE = 1/2AC</u>

  • DE = \sqrt{(-2+1)^2+(2+2)^2} = \sqrt{1^2+4^2} = \sqrt{17}
  • AC = \sqrt{(-5 +3)^2+(3+5)^2} = \sqrt{2^2+8^2} = 2\sqrt{17}
  • As we see DE = 1/2AC

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