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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