A rectangular plank √10 m wide, is placed symmetrically along the diagonal of a square of side 10 m as shown in the figure. The area of the plank is:
Explanation:
Consider the figure given in the question.
The two small triangles formed in the region between the vertex of the square and the width of the plank (top left corner) will be congruent to each other. Also, because they are congruent and because the diagonal of the square is the angle bisector of the angle at the vertex, each of these two triangles will be a 45-45-90 triangle.
∴ Height of small triangle = width of small triangle/2 = √10/2
By symmetry, this will also be the height of the other small triangle (bottom right corner).
Diagonal of the square = 10√2 m
∴ Length of plank = 10√2 − (√10/2) − (√10/2) = 10√2 − √10 = √10(√20 − 1) m
Width of plank = √10 m
∴ Area of plank = √10 × √10(√20 − 1) = 10(√20 − 1) sq.m
Hence, option (a).
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