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Home » A side of an equilateral triangle is 20cm long. A second equilateral triangle is inscribed in it by joining the mid points of the sides of the first triangle. The process is continued as shown in the accompanying diagram. Find the perimeter of the sixth inscribed equilateral triangle.
A side of an equilateral triangle is 20cm long. A second equilateral triangle is inscribed in it by joining the mid points of the sides of the first triangle. The process is continued as shown in the accompanying diagram. Find the perimeter of the sixth inscribed equilateral triangle.
CBSE | Class 11 | Maths | NCERT Exemplar | Sequences and Series
A side of an equilateral triangle is 20cm long. A second equilateral triangle is inscribed in it by joining the mid points of the sides of the first triangle. The process is continued as shown in the accompanying diagram. Find the perimeter of the sixth inscribed equilateral triangle.

Solution:

NCERT Exemplar Solutions for Class 11 Maths Chapter 9 - Image 8

Say ABC is a triangle with AB\text{ }=\text{ }BC\text{ }=\text{ }AC\text{ }=\text{ }20 cm

Let’s say that D, E and F are respectively the midpoints of AC, CB and AB  that are joined to form an equilateral triangle named as DEF

We now need to find the length of side of ΔDEF

Now let us consider ΔCDE

CD = CE = 10 cm … D is the midpoint of AC and E is the midpoint of CB

\therefore ΔCDE is isosceles

\Rightarrow ~\angle CDE\text{ }=~\angle CED… isosceles triangle’s base angle

But \angle DCE\text{ }=\text{ }60{}^\circ …∠ABC is equilateral

\therefore \angle CDE\text{ }=~\angle CED\text{ }=\text{ }60{}^\circ

Therefore, ΔCDE is equilateral

As a result, DE = 10 cm

In the similar way, it can be shown that GH = 5 cm

Therefore, the series so formed of the equilateral triangle’s sides will be 20, 10, 5 …

The series so formed is Geometric Progression with the first term a = 20 and common ratio r =\frac{1}{2}

To find the perimeter of 6th triangle inscribed we first have to find the side of 6th triangle that is the 6th term in the series

nth term in GP is given by {{t}_{n}}~=\text{ }a{{r}^{n-1}}

\Rightarrow ~{{t}_{6}}~=\text{ }\left( 20 \right)\text{ }{{\left( 1/2 \right)}^{6-1}}

\Rightarrow ~{{t}_{6}}~=\text{ }20/\text{ }{{2}^{5}}

=\text{ }20/\text{ }(4\text{ }\times \text{ }{{2}^{3}})

\Rightarrow ~{{t}_{6}}~=\text{ }5/8

As a result, the side of inscribed 6th equilateral triangle is 5/8 cm and therefore its perimeter would be thrice the length of its side since it’s an equilateral triangle

The perimeter of the inscribed 6th equilateral triangle is 3\text{ }\times \text{ }5/8\text{ }=\text{ }15/8 cm

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