Could anyone help me understand this problem?

"Find the perimeter of the figure, where AC=26, AD=BF, and D is the midpoint of AC. Round your answer to the nearest tenth."

Could anyone help me understand this problem Find the perimeter of the figure where AC26 ADBF and D is the midpoint of AC Round your answer to the nearest tenth class=

Respuesta :

It seems you will have to use a LOT of trigonometric calculations.
Line AC = 26 
Line AD = 13 = Line DC = 13
Line BF = 13

Angle ABC = Angle GBF = Angle FBH = 55 degrees each

cosine 40 = 13 (Line AD) / Line AE  
Line AE = 13 / 0.76604  = 16.97

tangent 40 = ED / AD = ED / 13
Line ED = 0.8391 * 13 = 10.908

Line DC =13
 
Angle ABC = 55 degrees
tangent(55) = 26 / AB
AB = 26 / 1.4281  = 18.206

Well, that's about half the calculations.
(As I said it's a lot of work).


The perimeter of the figure is given by the sum of the lengths of the

external segments.

Response:

  • The perimeter of the figure is approximately 117.1

Which methods can be used to find the perimeter of the figure?

The given parameters ate;

AC = 26

AD = BF

D = The midpoint of AC

Required:

The perimeter of the figure

Solution:

AD = CD = [tex]\dfrac{26}{2}[/tex] = 13

[tex]AE = \mathbf{\dfrac{AD}{sin(50^{\circ})} }= \dfrac{13}{sin(50^{\circ})} \approx17[/tex]

[tex]ED = \dfrac{AD}{tan(50^{\circ})} = \mathbf{\dfrac{13}{tan(50^{\circ})}} \approx 10.9[/tex]

AB = AC × tan(35°) = 26 × tan(35°) ≈ 18.2

[tex]\dfrac{18.2}{26} = \dfrac{13}{GF}[/tex], [tex]GF = 13 \times \dfrac{26}{18.2} \approx 18.6[/tex]

GF = HF = 18.6 by similar triangles

According to Pythagorean theorem, we have;

BH = √(18.6² + 13²) ≈ 22.7

BG = BH ≈ 22.7

BC = √(26² + 18.2²) ≈ 31.7

GC ≈ BC - BG, which gives;

GC = 31.7 - 22.7 ≈ 9

  • The perimeter, P = AE + AB + BH + HF + GF + GC + CD

Which gives;

P ≈ 17 + 18.2 + 22.7 + 18.6 + 18.6 + 9 + 13 = 117.1

  • The perimeter of the figure, P ≈ 117.1

Learn more about similar triangles here:

https://brainly.com/question/24031436

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