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Semmy [17]
3 years ago
10

In Triangle XYZ, A is the midpoint of XY, B is the midpoint of YZ, and C is the midpoint of XZ.

Mathematics
1 answer:
Roman55 [17]3 years ago
5 0

The perimeter of triangle ABC is 24 units

Step-by-step explanation:

If a segment joining the mid points of two sides of a triangle, then

this segment is:

  • Parallel to the third side
  • Its length is half the length of the third side

In The triangle XYZ

∵ A is the mid point of XY

∵ B is the mid point of YZ

∴ AB = \frac{1}{2} XZ

∵ XZ = 18 units

- Substitute the value of XZ in AB

∴ AB = \frac{1}{2} × 18 = 9 units

∵ B is the mid point of YZ

∵ C is the mid point of XZ

∴ BC = \frac{1}{2} XY

∵ AY = 7 units

∵ AY = \frac{1}{2} XY

∴ XY = 2 × AY

∴ XY = 2 × 7

∴ XY = 14 units

∴ BC = \frac{1}{2} × 14 = 7 units

∵ A is the mid point of XY

∵ C is the mid point of XZ

∴ AC = \frac{1}{2} YZ

∵ BZ = 8 units

∵ BZ = \frac{1}{2} YZ

∴ YZ = 2 × BZ

∴ YZ = 2 × 8

∴ YZ = 16 units

∴ AC = \frac{1}{2} × 16 = 8 units

∵ The perimeter of a triangle = the sum of the lengths of its sides

∴ Perimeter Δ ABC = AB + BC + AC

∴ Perimeter Δ ABC = 9 + 7 + 8 = 24 units

The perimeter of triangle ABC is 24 units

Learn more:

You can learn more about triangles in brainly.com/question/5924921

#LearnwithBrainly

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Step-by-step explanation:

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Jacob transformed quadrilateral FGHJ to F'G'H'J'.
Rudik [331]

Answer:

A. Reflection across the line x = 1

Step-by-step explanation:  

Please find the attachment.

We have been given that Jacob transformed quadrilateral FGHJ to F'G'H'J'.  We are asked to find which transformation Jacob used to reflect FGHJ to F'G'H'J'.

Since we know that while reflecting a figure, the line of reflection will lie between the original figure and reflected figure. Each point of the reflected figure will have the same distance from the line of reflection as the corresponding point of the original figure.              

Now let us see our given choices one by one.

A. Reflection across the line x = 1.

Upon looking at point G and G' we can see that both points are equidistant (2 units) from the line x=1. Other corresponding points of both quadrilaterals are also equidistant from line x=1, therefore, Jacob used the reflection across the line x=1 to transform quadrilateral FGHJ to F'G'H'J'.

B. Reflection across the line y = 1 .      

If Jacob had reflected quadrilateral across line y=1, the points of quadrilateral F'G'H'J' will lie in second and third quadrant. We can see from our graph that F'G'H'J' lies in 1st quadrant, therefore, option B is not a correct choice.

C. Reflection across the line y-axis.

If Jacob had reflected quadrilateral across y-axis, the coordinates of points of quadrilateral F'G'H'J' will be G'(1,4), H'(1,2), J'(4,0) and F'(2,5). Therefore, option B is not a correct choice.

D.  Reflection across the x -axis.

If Jacob had reflected quadrilateral across x-axis, the points of quadrilateral F'G'H'J' will lie in third quadrant. We can see from our graph that F'G'H'J' lies in 1st quadrant, therefore, option D is not a correct choice.

3 0
3 years ago
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