Let OA = OB = r. Semicircle on diameter AB has radius r/√2 and area (1/4)πr². Subtracting circular segment AFB from both the semicircle and the quarter circle OAB leaves equal areas for both shaded regions.
In the figure we see two shaded regions formed by a quarter circle, a semicircle, and a triangle. Show that the areas of the two shaded regions are equal.
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Let the radius of the large circle with centre O be r, so OA = OB = r.
Region AOB is a quarter circle of radius r:
Area(quarter circle AOB) = (1/4) x π x r² = (π/4)r².
In right-angled triangle AOB, the hypotenuse is AB = √(r² + r²) = r√2.
The semicircle AEB is drawn on diameter AB = r√2, so its radius is R = (r√2) / 2 = r / √2.
Area(semicircle AEB) = (1/2) x π x R² = (1/2) x π x (r² / 2) = (π/4)r².
Notice that:
Area(semicircle AEB) = Area(quarter circle AOB) = (π/4)r².
Let the unshaded circular segment between chord AB and arc AFB be S.
The upper shaded region (lune AEBFA) = Area(semicircle AEB) – S = (π/4)r² – S.
The lower shaded region (triangle AOB) = Area(quarter circle AOB) – S = (π/4)r² – S.
Since both expressions equal (π/4)r² – S, their areas are equal.
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/