Question

question is in the image do I need to make another angle thats labeled A'B'C"? Considering that n =11/6

Answer 1

Considering that the dilation factor is 11/6 we have to multiply the sides of triangle ABC to calculate A'B'C'.

When a dilation is happening, you have to multiply the dilation factor times each coordinate of the original triangle.

0. We have to calculate the coordinates of the original triangle:

A(3, 0)

B(3, 6)

C(9, 0)

2. Now, we have to multiply the dilation factor to get the coordinates of A'B'C':

• A'

$A^{\prime}(3\cdot\frac{11}{6},0\cdot\frac{11}{6})$$A^{\prime}(\frac{11}{2},0)$

The coordinate of A' is (11/2, 0).

• B'

$B^{\prime}(3\cdot\frac{11}{6},6\cdot\frac{11}{6})$$B^{\prime}(\frac{11}{2},11)$

The coordinate of B' is (11/2, 11).

• C'

$C^{\prime}(9\cdot\frac{11}{6},0\cdot\frac{11}{6})$$C^{\prime}(\frac{33}{2},0)$

Then, the coordinates of the new triangle are A'(11/2, 0), B'(11/2, 11), and C'(33/2, 0).

Thus, the triangle A'B'C' is:

Then, the lengths of each segment of A'B'C'are:

• C'A' is 11 units.

As we have 0 in the y position, we just have to subtract x position in C' minus x position in A':

$\frac{33}{2}-\frac{11}{2}=\frac{22}{2}=11$

• A'B' is also 11 units.

In this case, we consider the position in y and the x position in the same.

$11-0=11$

Finally, as it is a right triangle we can use the Pythagorean Theorem to solve this triangle, were

$BC^{\prime2}=AB^{^{\prime}2}+AC^{^{\prime}2}$

Replacing the values:

$BC^{\prime}=\sqrt[]{11^2+11^2}$

Simplifying and solving:

$BC^{\prime}=\sqrt[]{11^2+11^2}\approx15.556$

Finally, from the graph we can obtain the segments of triangle DEF using the same procedure as above:

• DE = 11

$21-10=11$

• FD = 11

$14-3=11$

• EF =15.556

$EF^2=DE^{^{}2}+FD^2$$EF=\sqrt[]{11^2+11^2}\approx15.556$

Then, triangle A'B'C' is equal in sidelength with triangle DEF.

Answer:

• A'(11/2, 0)

,

• B'(11/2, 11)

,

• C'(33/2, 0).

,

• C'A' and DE = 11

,

• A'B' and FD = 11

,

• B'C' and EF = 15.56