Question

In the figure above, the vertices of square

Answer 1

(C) 6 + 3√3

Explanation:

Area of the square = 3

a X a = 3

a² = 3

a = √3

Therefore, QR, RS, SP, PQ = √3

ΔBAC ≅ ΔBQR

Therefore,

$\frac{BQ}{BA} = \frac{QR}{AC}$

$\frac{BQ}{BA} = \frac{\sqrt{3} }{BA}$

In ΔBAC, BA = AC = BC because the triangle is equilateral

So,

BQ = √3

So, BQ, QR, BR = √3 (equilateral triangle)

Let AP and SC be a

So, AQ and RC will be 2a

In ΔAPQ,

(AP)² + (QP)² = (AQ)²

(a)² + (√3)² = (2a)²

a² + 3 = 4a²

3 = 3a²

a = 1

Similarly, in ΔRSC

(SC)² + (RS)² = (RC)²

(a)² + (√3)² = (2a)²

a² + 3 = 4a²

3 = 3a²

a = 1

So, AP and SC = 1

and AQ and RC = 2 X 1 = 2

Therefore, perimeter of the triangle = BQ + QA + AP + PS + SC + RC + BR

Perimeter = √3 + 2 + 1 + √3 + 1 + 2 + √3

Perimeter = 6 + 3√3

Therefore, the perimeter of the triangle is 6 + 3√3