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Ilya [14]
3 years ago
8

The hypotenuese of an isosceles right triangle is 16 inches. The midpoints of its sides are connected to form an inscribed trian

gle, and this process is repeated, creating a third triangle inscribed in the previous one. Find the sum of the areas of these triangles if this process is continued infinitely.
Mathematics
1 answer:
ser-zykov [4K]3 years ago
3 0

Answer:

  256/3 = 85 1/3 square inches

Step-by-step explanation:

The dimensions of the first inscribed triangle are 1/2 those of the original, so its area is (1/2)² = 1/4 of the original. The area of the original is ...

  A = (1/2)bh = (1/2)(16/√2)(16/√2) = 64 . . . . square inches

The sum of an infinite series with first term 64 and common ratio 1/4 is ...

  S = a1/(1 -r) . . . . . . for first term a1 and common ratio r

  = 64/(1 -1/4) = 64(4/3) = 256/3 . . . . square inches

The sum of the areas of the triangles is 256/3 = 85 1/3 square inches.

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<u>Given</u>:

The given expression to find the nth term of the sequence is d(n)=d(n-1) \cdot (-5)

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The second term of the sequence can be determined by substituting n = 2 in the nth term of the sequence.

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<em><u>Preliminary Remark</u></em>

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