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zhenek [66]
3 years ago
5

Find the sum of a finite geometric series.

Mathematics
2 answers:
NNADVOKAT [17]3 years ago
3 0

Answer: 17 in.

Step-by-step explanation:

The sides of an equilateral triangle measure 16 inches. The midpoints of the sides of the triangle are joined to form another equilateral triangle with sides that are half the length of the outer triangle. This process is continued until three triangles are inscribed in the first triangle. The sum of the perimeters of all four triangles is 17 in.

Bumek [7]3 years ago
3 0

Answer with explanation:

⇒Side of largest equilateral triangle in which three equilateral triangles are inscribed = 16 inches

Perimeter of a triangle = Sum of three sides of triangle

Perimeter of equilateral triangle having side length 16 inches = 16 +16+16=48 inches

⇒→Second equilateral triangle which is inscribed in this equilateral triangle having side length half of that equilateral triangle in which it is inscribed

 =\frac{16}{2}\\\\=8 inches

Perimeter of equilateral triangle having side length 8 inches = 8 +8+8=24 inches

⇒→Third equilateral triangle which is inscribed in this equilateral triangle having side length half of that equilateral triangle in which it is inscribed

 =\frac{8}{2}\\\\=4 inches

Perimeter of equilateral triangle having side length 4 inches =4+4+4=12 inches

⇒→Fourth equilateral triangle which is inscribed in this equilateral triangle having side length half of that equilateral triangle in which it is inscribed

 =\frac{4}{2}\\\\=2 inches

Perimeter of equilateral triangle having side length 2 inches =2+2+2=6 inches

→≡Total Perimeter of all four Equilateral Triangle

  =48 +24+12+6

= 90 inches

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Answer:

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Step-by-step explanation:

We have to maximize the area of the window, subject to a constraint in the perimeter of the window.

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A=A_r+A_c/2=a\cdot b+\dfrac{\pi r^2}{2}=ab+\dfrac{\pi}{2}\left (\dfrac{a}{2}\right)^2=ab+\dfrac{\pi a^2}{8}

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We can express b in function of a as:

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Then, the area become:

A=ab+\dfrac{\pi a^2}{8}=a(6-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a)+\dfrac{\pi a^2}{8}\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a^2+\dfrac{\pi a^2}{8}\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{4}-\dfrac{\pi}{8}\right)a^2\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{8}\right)a^2

To maximize the area, we derive and equal to zero:

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b=6-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a\\\\\\b=6-0.393*3.36=6-1.32\\\\b=4.68

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