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(11)/(6),0\cdot(11)/(6))`
`A^(\prime)((11)/(2),0)`

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

• B'

`B^(\prime)(3\cdot(11)/(6),6\cdot(11)/(6))`
`B^(\prime)((11)/(2),11)`

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

• C'

`C^(\prime)(9\cdot(11)/(6),0\cdot(11)/(6))`
`C^(\prime)((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':

`(33)/(2)-(11)/(2)=(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