Sum of Perimeters of Triangles in Square

CAT 1995 Slot 1 · QA · Medium · Geometry

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ABCDABCD is a square of area 44, which is divided into four non-overlapping triangles by its diagonals. What is the sum of the perimeters of the four triangles?

  1. A.

    8(2+2)8(2 + \sqrt{2})

  2. B.

    8(1+2)8(1 + \sqrt{2})

  3. C.

    4(1+2)4(1 + \sqrt{2})

  4. D.

    4(2+2)4(2 + \sqrt{2})

Answer

B

Explanation

Area of square =4  ⟹  = 4 \implies side =2= 2. Perimeter of square =4×2=8= 4 \times 2 = 8. Diagonal length =22+22=22= \sqrt{2^2 + 2^2} = 2\sqrt{2}. Each triangle contains one outer side of the square and two half-diagonals. Each interior segment (half-diagonal) is shared between two triangles, so each of the 44 half-diagonals gets counted twice in the total perimeter calculation. Sum of perimeters =Perimeter of square+2×(Sum of 2 diagonals)=8+2×(2×22)=8+82=8(1+2)= \text{Perimeter of square} + 2 \times (\text{Sum of 2 diagonals}) = 8 + 2 \times (2 \times 2\sqrt{2}) = 8 + 8\sqrt{2} = 8(1 + \sqrt{2}).

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