nktdotgupta wrote:

An equilateral Triangle has a sqaure inscribed in it, side of sqaure =12 inch.

What is the perimeter of the triangle?

A. 36 inch

B.36 root3 inch

C.36 root3 inch + 24 inch

D.72 inch

E.24 root3 + 36 inch

Figure attached

ABC is equilateral traingle and DEFG is the square inscribed in it.

Since DE || FG so DE || BC so, <ADE = <B = 60 and <AED = <C = 60

So, Traingle ADE is equialteral. Similarly we can prove that Traingle DBG and Triangle ECF are equialteral.

So, AD = DE = 12inch

AB = AD*2= 24inch.

PErimeter of the traingle = 3*AB = 3*24= 72inch.

So, answer will be D

Hope it helps!

Triangle ADE is an equilateral triangle, but the other two triangles are not(however both are similar Tirangles).

Solving for BDG:

angle DBG=60(as per Q)

angle BG=90

hence, angle BDG=30.

the raito of sides for a 30,60,90 triangle >> 1:root3:2( refer to Bunuel's post

math-triangles-87197.html)

since DG=12

BG = 12/root3

hence : using P&T :>> DB = 24*root3

Same applies for the triangle EFC

therefore, Perimeter =DA+DB+BG+GF+FC+CE+EA

=12+24/root3+12/root3+12+12/root3+24/root3 +12

=36+72/root3

=36+(24*root3*root3)/root3

=36+24*root3

The Answer is E

hope the u dont mind the poor formatting

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