Let square side be s. The sum of heights to AB and CD equals s, so Area(ΔPAB) + Area(ΔPCD) = (1/2) x s x s = s²/2. Both regions equal half the square’s area, making their ratio 1:1.
Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD (Fig. 6.33). What is the ratio of the areas of the red region (ΔPAB and ΔPCD) and the green region (ΔPBC and ΔPDA)?
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Let the side length of square ABCD be s.
Draw a line through point P perpendicular to AB and CD.
Let the perpendicular distance from P to AB be h1 and to CD be h2.
Then h1 + h2 = s.
Area of red region = Area(ΔPAB) + Area(ΔPCD)
= (1/2) x s x h1 + (1/2) x s x h2 = (1/2) x s x (h1 + h2) = (1/2) x s².
Since total area of the square is s², the green region also equals s² – s²/2 = (1/2) x s².
Thus, the ratio of areas is 1:1.
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/