Let the rectangle have dimensions X and Y. The vertical dividing line of height h gives C = (1/2)hx₂. Combining expressions A + C = (1/2)hX and B + C = (1/2)x₂Y yields the rectangle area XY = [2(A + C)(B + C)] / C.
In Fig. 6.54, we see three triangles within a rectangle. The areas of the triangles are A, B, C, as marked. Show that the area of the rectangle is [2(A + C)(B + C)] / C.
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Let the rectangle have horizontal length X and vertical height Y, so total Area = XY.
From Fig. 6.54, the common vertical segment shared by triangles A, B, and C has length h.
Let the horizontal distance from the left edge to this segment be x₁, and to the right edge be x₂,
so X = x₁ + x₂.
Triangles A and C together form a triangle of base h and altitude X:
A + C = (1/2) x h x X.
Triangles B and C together form a triangle on base h with altitude Y:
B + C = (1/2) x h x Y (or using horizontal base x₂ with total height Y: B + C = (1/2) x x₂ x Y).
Triangle C has base h and horizontal width x₂:
C = (1/2) x h x x₂.
Multiplying (A + C) and (B + C):
(A + C)(B + C) = [(1/2)hX] x [(1/2)x₂Y]
= (1/2) x [(1/2)hx₂] x XY = (1/2) x C x (Area of rectangle).
Rearranging:
Area of rectangle = [2(A + C)(B + C)] / C.
Hence proved.
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/