Question

A sequence of equilateral triangles is drawn. The altitude of each is \sqrt 3 times the altitude of the preceding triangle, the difference between the area of the first triangle and the sixth triangle is 968 \sqrt root3 square unit. The perimeter of the first triangle is : ​

Answer 1

Given: Altitude = √3 (altitude of the preceding triangle), difference between the area is 968√3 square units.

To find: The perimeter of the first triangle?

Solution:

• Now we have given the relation between altitudes, so let the altitude of first triangle be x and let side of the triangle given is a.

• Then x = √3a/2

• Now for rest of the triangles, altitude will be:

             3a/2 , 3√3a/2 , 9a/2 , 9√3a/2 , 27a/2

• Now the side will be:

             2x/√3     ...............(for 1 st triangle)

             2x/27      ...............(for 6 th triangle)

• Now the difference in area is given as 968√3 square units.

• So:

             √3/4 (4x^2/3 - 4x^2/27^2) = 968√3

             x^2 (1/3 - 1/729) = 968

             x^2(242/729) = 968

             x^2 = 4 x 729

             x = 2 x 27

             x = 54

• So the side of first triangle will be:

             2(54)/√3

• Then the perimeter will be:

             4(2(54)/√3 = 432/√3

Answer:

           So the perimeter is 432/√3 units.