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Use rules of transformations to answer each of the items below. Be sure to answer in complete sentences, and when necessary, include your calculations.

1.) Create your own set of coordinates for △BLU and graph it on a coordinate plane. Label each of the triangle's vertices.

2.) Write a scenario involving the dilation of △BLU onto its image. In the scenario, include the scale factor between the similar triangles and the coordinates for the image. Graph the image on the same coordinate plane as its pre-image.

3.) Use the distance formula or Pythagorean Theorem to prove mathematically that △BLU is similar to its image. Include all necessary calculations.

Respuesta :

merlynthewhizz
We choose the coordinates (4,4), (4,10), and (8,4) as the vertices of right-angle triangle ABC

We want to dilate triangle ABC by scale factor of 2 at the center (0,2)

The image is shown below as triangle A'B'C' with coordinates (8.6), (8,18), and (18,6). 

Triangle ABC has a base of 4 units and a height of 6 units
Triangle A'B'C' has a base of 8 units and a height of 12 units

Applying Pythagoras theorem to triangle ABC to find the hypotenuse
[tex] BC^{2}= AB^{2}+ AC^{2} [/tex]
[tex] BC^{2}= 4^{2}+ 6^{2} [/tex]
[tex] BC^{2} =52[/tex]
[tex]BC=2 \sqrt{13} [/tex]

Now, we calculate the hypotenuse of the image
[tex] B'C'^{2}= A'B'^{2}+ A'C'^{2} [/tex]
[tex] B'C'^{2}= 8^{2}+ 12^{2} [/tex]
[tex] B'C'^{2}=208 [/tex]
[tex] B'C'= \sqrt{208} =4 \sqrt{13} [/tex]

The hypotenuse of the image is twice the hypotenuse of triangle ABC which proof the two shapes are similar shape


Ver imagen merlynthewhizz
yousufmajidjanabi

We choose the coordinates (4,4), (4,10), and (8,4) as the vertices of right-angle triangle ABC

We want to dilate triangle ABC by scale factor of 2 at the center (0,2)

The image is shown below as triangle A'B'C' with coordinates (8.6), (8,18), and (18,6).  

Triangle ABC has a base of 4 units and a height of 6 units

Triangle A'B'C' has a base of 8 units and a height of 12 units

Applying Pythagoras theorem to triangle ABC to find the hypotenuse

BC^2 = AB^2 + AC^2

BC^2 = 4^2 + 6^2

BC^2 = 52

BC = 2/13

Now, we calculate the hypotenuse of the image

B'C'^2 = A'B'^2 + A'C'^2

B'C'^2 = 8^2 + 12^2

B'C'^2 = 208

B'C' = /208 = 4/13

The hypotenuse of the image is twice the hypotenuse of triangle ABC which proof the two shapes are similar shape

Ur welcome

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