Trisection points divide the base into three equal segments, and both triangles share the same vertex altitude. Area equals (1/2) x base x height, so their areas are identical. Straight dissection along altitudes enables rearrangement.
Show that the areas of the shaded blue triangle and the shaded red triangle are equal. Find a way of cutting up the blue triangle into some number of pieces and rearranging the pieces to cover the red triangle.
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Let the top vertex be V and the opposite base be divided into three equal segments of length x by trisection points.
Both the blue triangle and the red triangle stand on a base of length x.
Because they share the common apex V, their perpendicular height h from V to the base line is exactly the same.
Area of blue triangle = (1/2) x x x h = (1/2)xh.
Area of red triangle = (1/2) x x x h = (1/2)xh.
Hence, their areas are equal.
By the Wallace-Bolyai-Gerwien theorem, two polygons of equal area can always be dissected into finitely many polygonal pieces using straight cuts and rearranged into each other. Dropping perpendicular cuts from corresponding vertices converts both into congruent sets of right-angled triangles and trapezia that match upon rearrangement.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/