Triangle sequence

Geometry Level 2

A sequence of equilateral triangles is drawn. The altitude of each is $\sqrt{3}$ times the altitude of the preceeding triangle. The difference between the areas of the first triangle and the sixth triangle is $968\sqrt{3}$ sq.units. The perimeter of the first triangle is

60 48 12 24

1 solution

Amogha Pokkulandra
Jun 28, 2018

Let us visualize the sequence and initalize the first triangle's area as $x$ . If between one dimension the value is multiplied by $\sqrt{3}$ , then two dimensions (or area) should be multiplied by 3 . Let the top row of the table signify the $n$ th triangle while the bottom row represents the area of the triangle in terms of $x$ .

1	2	3	4	5	6
$x$	3 $x$	9 $x$	27 $x$	81 $x$	243 $x$

The difference between the 6th triangle and 1st triangle in terms of $x$ is 242 $x$ . 968 $\sqrt{3}$ and 242 $x$ divided by 242 on both sides yields $x$ = 4 $\sqrt{3}$ . Then use trig functions to derive the side as 4 and multiply by 3 to get the perimeter as $\boxed{12}$ .

Triangle sequence

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