In a figure, AB and CD are vertical.
AE = 24 cm, BE = 6 cm, CE = 18 cm, DE = 8 cm.
EJ =24 - 6 =18cm. EL =18 - 8 = 10cm.
The area of rectangle EJKL is 18 × 10 =180cm2.
If equivalence transfer of the part of a shadow area is carried out as shown in Fig. 1, the area of a shadow area and the area of a red portion will become equal.
Therefore, since the area of a shadow area is 1/2 of the remaining area after the area of the rectangle EJKL being subtracted from the area of a circle, the area of a shadow area is (785 - 180) × 12 = 302.5cm2.
AE = 24 cm, BE = 6 cm, CE = 18 cm, DE = 8 cm.
Answer
302.5 cm2
Solution
As shown in Fig. 1, the auxiliary line of HI and GF is drawn to be that AB = HI and CD = GF. EJ =24 - 6 =18cm. EL =18 - 8 = 10cm.
The area of rectangle EJKL is 18 × 10 =180cm2.
If equivalence transfer of the part of a shadow area is carried out as shown in Fig. 1, the area of a shadow area and the area of a red portion will become equal.
Therefore, since the area of a shadow area is 1/2 of the remaining area after the area of the rectangle EJKL being subtracted from the area of a circle, the area of a shadow area is (785 - 180) × 12 = 302.5cm2.
