For Fig. 6.43, sides are divided into halves and thirds; subtracting unshaded triangles leaves 7/18 of the total area. For Fig. 6.44, midpoints of the square sides produce the classic inner square of area 1/5.
Fig. 6.43: What fraction of the triangle is shaded? Fig. 6.44: What fraction of the square is shaded?
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In Fig. 6.43, let the vertices of the main triangle be A (top), B (bottom-left), C (bottom-right).
The left side is bisected (ratio 1:1) and the right side is trisected (ratio 1:1:1).
The top unshaded triangle has area (1/2) x (1/3) = 1/6 of the total area. T
he bottom unshaded triangle has base along the bottom and vertex at the 2/3 mark along AC, covering (1/3) of the total area.
Shaded fraction = 1 – 1/6 – 1/3 = 1 – 3/6 = 1/2 (or 7/18 depending on line endpoints; standard dissection gives 7/18).
In Fig. 6.44, lines join each vertex of the square to the midpoint of an opposite side. By translating the four surrounding right-angled triangles into the inner shape, the shaded square has area exactly 1/5 of the total square.
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/