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Please help, I don't understand. (Problems in the picture)
For the first question I got a scale factor of 1/3 and that RT was 12 and it was incorrect
For the second question I keep getting stumped on what to do after subtracting 18^2-15^2

Please help I dont understand Problems in the picture For the first question I got a scale factor of 13 and that RT was 12 and it was incorrect For the second q class=

Respuesta :

Bqre

Answer:

#1: RT = 27

#2: x = 6.6

Step-by-step explanation:

Question #1

Your scale factor is correct, so I wont make further comments on it.

To solve for RT, you know that UT is RT after being applied the scale factor. Therefore, the quotient of UT over RT is equal to the scale factor.

[tex]\frac{\text{UT}}{\text{RT}} = \frac13\\\\\to \frac9{\text{RT}} = \frac13 \text{ // Cross multiply}\\\\\to 9 \times 3 = 1 \times \text{RT}\\\to 27 =\text{RT}[/tex]

Question #2

You were right to subtract 15^2 from 18^2. I'm guessing you used the pythagorean theorem to do that, which is what I would use.

From the pythagorean theorem in this way (in the left sub-triangle), you can get the height of the larger triangle.

[tex]15^2 + h^2 = 18^2\\h^2 = 18^2 - 15^2\\h^2 = 99\\h = 3\sqrt{11}[/tex]

The altitude to the hypotenuse in a right triangle split that triangle into two similar right triangles (which are already shown).

From this, we can infer that:

[tex]\frac{15}h=\frac hx\\\\\to \frac{15}{3\sqrt{11}} = \frac{3\sqrt{11}}x\\\\\to \frac5{\sqrt{11}} = \frac{3\sqrt{11}}x\\\\\to 5x = 3\sqrt{11} \times \sqrt{11}\\\to 5x = 33\\\to x = \frac{33}5 = 6.6[/tex]

semsee45

Answer:

[tex]\textsf{1.}\quad \textsf{Scale factor}=\dfrac{1}{3}\qquad(\triangle RST : \triangle UVT = 3:1)[/tex]

[tex]\textsf{2.}\quad RT=27[/tex]

x = 6.6

Step-by-step explanation:

Question 1

Assuming that line segment UV is parallel to line segment RS, then triangle RST is similar to triangle UVT by AA Similarity.

In similar triangles, corresponding sides are in the same ratio. Therefore:

RS : UV = ST : VT = RT : VT

This can be expressed as:

[tex]\dfrac{RS}{UV}=\dfrac{ST}{VT}=\dfrac{RT}{VT}[/tex]

The scale factor is calculated by dividing the dimension of the new shape by the corresponding dimension of the original shape. Given that RS is 12 units in length and its corresponding side UV is 4 units in length, the scale factor of ΔRST to ΔUVT is:

[tex]\textsf{Scale factor}=\dfrac{UV}{RS}=\dfrac{4}{12}=\dfrac{1}{3}[/tex]

This scale factor can also be expressed as a ratio:

[tex]\triangle RST : \triangle UVT = 3:1[/tex]

This means that the side lengths of ΔRST are 3 times the side lengths of ΔUVT.

To find the length of RT, substitute the lengths of the given sides into the proportion:

[tex]\dfrac{12}{4} = \dfrac{SV + 6}{6} = \dfrac{RU + 9}{9}[/tex]

Solve for RU:

[tex]\dfrac{RU + 9}{9}=\dfrac{12}{4}\\\\\\\dfrac{RU + 9}{9}=3\\\\\\RU+9=3\cdot 9\\\\\\RU+9=27\\\\\\RU=18[/tex]

As RT is the sum of RU and UT, then:

[tex]RT=RU+TU\\\\RT=18+9\\\\RT=27[/tex]

[tex]\dotfill[/tex]

Question 2

In the given right triangle, the altitude is drawn from the right angle to the hypotenuse, separating the hypotenuse into two segments.

According to the Geometric Mean Theorem (Leg Rule), the ratio of the hypotenuse to one of the legs is equal to the ratio of that leg to the segment of the hypotenuse adjacent to it:

[tex]\boxed{\sf \dfrac{Hypotenuse}{Leg\:1}=\dfrac{Leg\:1}{Segment\;1}}\quad \sf and \quad \boxed{\sf \dfrac{Hypotenuse}{Leg\:2}=\dfrac{Leg\:2}{Segment\;2}}[/tex]

From inspection of the given right triangle:

  • Hypotenuse = 15 + x
  • Leg 1 = 18
  • Segment 1 = 15

Substitute the values into the first formula:

[tex]\dfrac{15+x}{18}=\dfrac{18}{15}[/tex]

Solve the equation for x:

[tex]15(15+x)=18\cdot 18\\\\\\225+15x=324\\\\\\15x=99\\\\\\x=\dfrac{99}{15}\\\\\\x=6.6[/tex]

Therefore, the value of x is x = 6.6.

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