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A. Two quadrilaterals of B are overlapped. The overlapped part is one fourth of a and one sixth of B. what is the simplest integer ratio of the area of quadrilateral B and a?
B / 6 = A / 4
Therefore, B ∶ a = 6 ∶ 4 = 3 ∶ 2
Two parallelograms AB are overlapped, and the overlapping area is 1 / 4 of a and 3 / 1 of B. We know that the area of a is 12 square centimeters, and what is the area of B?
A: B = 1 / 4:1 / 3 = 4:3, so the area of B is 12 × 3 / 4 = 9 (square centimeter)
AB two parallelograms overlap. The area of the overlap is one fourth of a and one sixth of B. It is known that the area of a is 12 square centimeters
Find out the area ratio of parallelogram A and B
∵ the overlap is 1 / 4 of A
The area of a is 12
The overlap area is 3
The overlap area (3) is 1 / 6 of B
The area of B is 18
∴S（A）：S（B）=2:3
As shown in the figure, two parallelograms A and B are overlapped. The area of the overlapped part is 14 of a and 16 of B. It is known that the area of a is 12 square centimeters. Then the area of B is more than that of A______ Square centimeter
The area of B is a: (1 △ 16) / (1 △ 14), = 6 △ 4, = 1.5 (Times); the area of B is more than that of a: 12 × (1.5-1), = 12 × 0.5, = 6 (square centimeter); a: the area of B is more than that of a: 6 square centimeter
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