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A side of an equilateral triangle is $20 \mathrm{~cm}$ long. A second equilateral triangle is inscribed in it by joining the midpoints of the sides of the first triangle. This process is continued. Find the perimeter of the sixth inscribed equilateral triangle.
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Verified Answer
By joining the mid-points of the first triangle, we get another equilateral triangle of side equal to half of the length of side of first triangle.
Continuing in this way, we get a G.P. where
Perimeter of first triangle $=20 \times 3=60 \mathrm{~cm}$
Perimeter of second triangle $=10 \times 3=30 \mathrm{~cm}$
Perimeter of third triangle $=5 \times 3=15 \mathrm{~cm}$
Now, the GP will be $60,30,15, \ldots$
Here, $a=60$
$$
\therefore r=\frac{30}{60}=\frac{1}{2}
$$
It is the sixth term of the series.
$\therefore$ perimeter of sixth inscribed triangle
$$
\begin{aligned}
&=a_6=a r^{6-1} \quad\left[\because a_n=a r^{n-1}\right] \\
&=60 \times\left(\frac{1}{2}\right)^5=\frac{60}{32}=\frac{15}{8} \mathrm{~cm}
\end{aligned}
$$
Continuing in this way, we get a G.P. where
Perimeter of first triangle $=20 \times 3=60 \mathrm{~cm}$
Perimeter of second triangle $=10 \times 3=30 \mathrm{~cm}$
Perimeter of third triangle $=5 \times 3=15 \mathrm{~cm}$
Now, the GP will be $60,30,15, \ldots$
Here, $a=60$
$$
\therefore r=\frac{30}{60}=\frac{1}{2}
$$
It is the sixth term of the series.
$\therefore$ perimeter of sixth inscribed triangle
$$
\begin{aligned}
&=a_6=a r^{6-1} \quad\left[\because a_n=a r^{n-1}\right] \\
&=60 \times\left(\frac{1}{2}\right)^5=\frac{60}{32}=\frac{15}{8} \mathrm{~cm}
\end{aligned}
$$
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