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Given, arc QR is congruent to arc LN and arc OP is congruent to arc VW.
And the expressions for each arc in the diagram also given as:
Arc QR = 2x - y, arc LN = 11 , arc OP= 10 and arc VW=5x+y.
Hence, we will get the system of equations as following:
Arc QR = Arc LN
2x - y = 11 ...(1)
Arc OP = Arc VW
5x + y = 10 ...(2)
We need to find the value of x. So, we can add the equations to eliminate y so that we can solve the equations for x. Therefore,
2x+5x = 11 + 10
7x = 21
Divide each sides by 7.
So, x= 3
(C) 6 + 3√3
<u>Explanation:</u>
Area of the square = 3
a X a = 3
a² = 3
a = √3
Therefore, QR, RS, SP, PQ = √3
ΔBAC ≅ ΔBQR
Therefore,
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