The upper half of this figure is a triangle with height 3 and base 6. If we divide this vertically we get two congruent triangles of height 3 and base 3. Using the Pythagorean Theorem we find the length of the diagonal of one of these small triangles: (diagonal)^2 = 3^2 + 3^2, or (diagonal)^2 = 2*3^2.

Therefore the diagonal length is (diagonal) = 3√2, and thus the total length of the uppermost two sides of this figure is 6√2.

The lower half of the figure has the shape of a trapezoid. Its base is 4. Both to the left and to the right of the vertical centerline of this trapezoid is a triangle of base 1 and height 3; we need to find the length of the diagonal of one such triangle. Using the Pythagorean Theorem, we get

(diagonal)^2 = 1^2 + 3^2, or 1 + 9, or 10. Thus, the length of each diagonal is √10, and so two diagonals comes to 2√10.

Then the perimeter consists of the sum 2√10 + 4 + 6√2.

which, when done on a calculator, comes to 9.48. We must round this off to the nearest whole number, obtaining the final result 9.

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3 years ago