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Deffense [45]
3 years ago
14

Triangle ABC is isosceles with AB=CB. Circle M is inscribed in Triangle ABC such that it is tangent at points D, E, and F. If th

e length of BF is twice the length of CF and the perimeter of Triangle ABC is 32 inches, then determine the length of side BC in inches.
Mathematics
1 answer:
Ierofanga [76]3 years ago
8 0

Answer:

The answer is "12 inches".

Step-by-step explanation:

Please find the image file of the graph.

We know AE = CF \ and \ EB = FB so because the triangle is isosceles.

EB = 2x, FB = 2x,\ and \ CF = x when AE is called by x.

We know that AE = AD = x since E and D are tangent points to a circle.

They have CD = CF = x since D and F are tangent points only to circle.

Thus, if the perimeter is 32, the following is the result:

\to AE + EB + BF + FC + CD + DA = 32\\\\\to x + 2x + 2x + x + x + x = 32\\\\\to 8x = 32\\\\\to x = 4

Therefore the length of BC = BF + FC = 2x + x = 3x = 12\  inches

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Answer:

  (2.27, 0.96)

Step-by-step explanation:

The orthocenter is the point of intersection of altitudes of the triangle. The equation for an altitude is the equation of a line through a vertex that is perpendicular to the opposite side.

For example, the line perpendicular to side AC can be found as ...

  (∆x, ∆y) = C-A = (3, -1) -(-4, 2) = (7, -3)

  ∆x(x -h) +∆y(y -k) = 0 . . . . perpendicular through point (h, k)

For side AC, we want the point to be B(4, 5), so the equation is ...

  7(x -4) -3(y -5) = 0

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Similarly, the altitude to side AB can be written as ...

  (∆x, ∆y) = B -A = (4, 5) -(-4, 2) = (8, 3)

  8(x -3) +3(y +1) = 0

  8x +3y -21 = 0

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By the "cross multiplication method", the solution to these equations is ...

  x = (-3(-21) -(3(-13))/(7(3) -8(-3)) = (63+39)/(21+24) = 102/45 = 2 4/15 ≈ 2.27

  y = (-13(8) -(-21)(7))/45 = 43/45 ≈ 0.96

The orthocenter is near (2.27, 0.96).

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A graphing application confirms this result.

_____

<em>Additional information about the cross multiplication method</em>

For the general form equations ...

  • ax +by +c = 0
  • dx +ey +g = 0

The "cross multiplication method" has you write the array ...

  \begin{array}{cccc}a&b&c&a\\d&e&g&d\end{array}

and form the "cross products" in groups of four coefficients:

  D = ae -db, X = bg -ec, Y = cd -ga

Then the solution to the set of equations is ...

  1/D = x/X = y/Y   ⇒   x = X/D, y = Y/D

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<em>Additional note</em>

Videos of this method show you writing the array as bcab/egde and using the final equation x/X=y/Y=1/D. The above gets the same result in a more straightforward manner.

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