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One side of an equilateral triangle is 24cm. The mid points of its sides are joined to form another triangle. Again , joining the mid points of this triangle forms another triangle and the process is continued infinitely. Determine the sum of the perimeters of all such triangles.
hi, for first triangle perimeter = 3 *24 = 72 for second , ( its sides would be half the first one ) perimeter = 3 * 24 /2 = 72 /2 for third , perimeter = 3 * (24/2 )/2 = 72 / 4 and so on so the sum of perimeter will be = 72 + 72/2 +72 /4 +72/ 8 + ....................... = 72 ( 1 + 1/2 +1/4 + 1/8 +........................) = 72 { 1/( 1-1/2) } { sum of ∞ g.p. = a/ 1-r ) = 72 *2 = 144 cm
hi,
for first triangle perimeter = 3 *24 = 72
for second , ( its sides would be half the first one )
perimeter = 3 * 24 /2 = 72 /2
for third , perimeter = 3 * (24/2 )/2 = 72 / 4
and so on
so the sum of perimeter will be =
72 + 72/2 +72 /4 +72/ 8 + .......................
= 72 ( 1 + 1/2 +1/4 + 1/8 +........................)
= 72 { 1/( 1-1/2) } { sum of ∞ g.p. = a/ 1-r )
= 72 *2
= 144 cm
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