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professor190 [17]
3 years ago
8

Which statement accurately explains whether a reflection over the y-axis and a 270° counterclockwise rotation would map figure A

CB onto itself? a coordinate plane with figure ACB with point A at 1, 1, C at 3, 4 and B at 5, 1 Yes, A″C″B″ is located at A″(1, 1), C″(4, 3), and B″(1, 5) Yes, A″C″B′ is located at A″(1, 1), C″(3, 4), and B″(5, 1) No, A″C″B″ is located at A″(1, 1), C″(4, 3), and B″(1, 5) No, A″C″B″ is located at A″(1, 1), C″(3, 4), and B″(5, 1)

Mathematics
1 answer:
wel3 years ago
5 0

Answer:

No, A″C″B″ is located at A″(1, 1), C″(4, 3), and B″(1, 5)

Step-by-step explanation:

The triangle has vertices A(1,1), B(5,1), C(3,4).

First transformation is reflection over y-axis with the rule

(x,y)→(-x,y)

so,

  • A(1,1)→A'(-1,1)
  • B(5,1)→B'(-5,1)
  • C(3,4)→C'(-3,4)

Second transformation is rotation by 270° counterclockwise with the rule

(x,y)→(y,-x)

so

  • A'(-1,1)→A''(1,1)
  • B'(-5,1)→B''(1,5)
  • C'(-3,4)→C''(4,3)

No, A″C″B″ is located at A″(1, 1), C″(4, 3), and B″(1, 5)

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In triangle ABC, AB = 7, AC = 15, and the length of median AM is 10. Find the area of triangle ABC.
steposvetlana [31]

Answer:

  42

Step-by-step explanation:

We can make use of Heron's formula for the area of the triangle. Let x represent half the length of BC:

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and the value used in Heron's formula, the semi-perimeter is ...

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The square of the area of ΔAMB is then ...

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Similarly, the square of the area of ΔABC is ...

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We want these two triangle areas to be the same, so we can solve for the value of x that makes it so.

  A1² - A2² = 0 = (-1/16)((x^4 -298x^2 +2601) -(x^4 -650x^2 +15625))

  352x^2 -13024 = 0 . . . . multiply by -16 and collect terms

  x^2 -37 = 0 . . . . . . . . . . . divide by 352

  x^2 = 37 . . . . . this is as far as we need to take it in order to find the area.

Substituting the value for x^2 into the expression for A1², we get ...

  A1² = (-1/16)((x^2 -298)x^2 +2601) = (-1/16)((37 -298)37 +2601) = 441

So, the area of ΔAMB is √441 = 21 square units. The area of ΔAMC is the same, so the area of ΔABC is ...

  area(ΔABC) = 2×area(ΔAMB) = 2×21 units²

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_____

<em>Addendum to the answer</em>

After thinking about this a little more, I realized the area is that of half a parallelogram with AB and AC as two sides and AM as half the diagonal. In other words, the area is the same as that of a triangle with sides 7, 15, and 2×10=20. A single straightforward application of Heron's formula gives the area as ...

  A = √(21(21-7)(21-15)(21-20)) = √(21(14)(6)) = 42

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