Circumference 2 x (22/7) x r = 44 yields radius r = 7 cm. A quadrant has angle 90°, so its area is (1/4) x π x r² = (1/4) x (22/7) x 7² = 77/2 = 38.5 cm².
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Using sector area formula (θ / 360) x π x r² with r = 7 cm and θ = 60° gives (60 / 360) x (22/7) x 7² = (1/6) x 154 = 77/3 = 25.67 cm².
Since CQ || PD, triangles DPQ and DPC share base PD and parallel lines, giving Area(ΔDPQ) = Area(ΔDPC). Thus, Area(ΔBPQ) = Area(ΔBPD) + Area(ΔDPQ) = Area(ΔBPD) + Area(ΔDPC) = Area(ΔBDC) = (1/2) x Area(ΔABC).
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.
Median AD divides ΔABC into equal areas: Area(ΔABD) = Area(ΔACD). Similarly, in ΔPBC, median PD gives Area(ΔPBD) = Area(ΔPCD). Subtracting these two equations yields Area(ΔABP) = Area(ΔACP).