Question

perimeter of equilateral triangle with enclosed semi circle

Answer 1

Refer the figure above. The following concepts are important before we solve:

i) Since the triangle is Equilateral (side S = 6 cm), it’s Perpendicular Bisector (Altitude) = Median = Angle Bisectors. And all meet at the same point (O).

ii) In order to inscribe a triangle within a circle, the Centre of the Circle should be the Circum-centre (that is where the perpendicular bisectors meet). Thus, here O being the meeting point of the perpendicular bisectors, the centre of the circle.

iii) For the circle, radius is denoted by “a”

iv) Medians gets intersected in the ratio 2:1 at the point of intersection. Thus, longer segment is 2/3 of the median length and shorter segment in 1/3 of the median length

v) We know that for an equilateral triangle, Height (perpendicular) = (√3/2).(side)

Here, Height = (√3/2).S

From (iv), we now get that length of the line segment denoted by “a” = (2/3). Height =

T(2/3).(√3/2).S =(S/ √3) = Radius

So, Radius = (6/ √3) = 2√3 cm

Notes:

A) This problem can also be solved by Trigonometry. But in many competitive exams (especially GRE, GMAT etc.) Trigonometry is not a part of the syllabus. Hence thought of sharing the alternative

B) This problem can also be solved by the concept of 30–60–90 Triangles. But since the concept is not widely used, I have used the very basic concepts of Median, Height etc.

C) For any Equilateral Triangle

The radius of Incircle = (Side/ 2√3)

The radius of Circumcircle = (Side/ √3)