Question

In the figure below, ABCD and ABEF are parallelograms:Which of the following are true?
1. Area(ABCD) = Area(ABEF)
2. Area(AYB) = 1/2 Area(ABCD)
3. Area(AYB) = 1/2 Area(ABEF)
4. Area(AXF) = 1/2 Area(ABCD)
5. Area(AXF) = 1/2 Area(ABEF)
6. Area(AYB) = Area(AXF)

Answer 1

True or false with proper reason given below:

          1. Area(ABCD) = Area(ABEF) : True

          Reason: As the theorem states that parallelograms on same base and between same parallels are equal in area.

          2. Area(AYB) = 1/2 Area(ABCD) : True

          Reason: As the  theorem states that if triangle and parallelogram are on same  base and between same parallels then the area of triangle is half of the parallelogram.

          3. Area(AYB) = 1/2 Area(ABEF) : True ..............(i)

          Reason: Now in 1, we proved that Area(ABCD) = Area(ABEF)

          So Area(AYB) = 1/2 Area(ABEF)  = 1/2 Area(ABCD)

          4. Area(AXF) = 1/2 Area(ABCD)  : True

          Reason: Now according to 1,  Area(ABCD) = Area(ABEF), so substituting ABCD to ABEF, then:

          Area(AXF) = 1/2 Area (ABEF): True  

          And the theorem states that if triangle and parallelogram are on same  base and between same parallels then the area of triangle is half of the parallelogram.

          5. Area(AXF) = 1/2 Area(ABEF) : True   ..................(ii)

          Reason: As the  theorem states that if triangle and parallelogram are on same  base and between same parallels then the area of triangle is half of the parallelogram.

          6. Area(AYB) = Area(AXF) : True

          Reason : from (i) and (ii), we can say that Area(AYB) = Area(AXF) =  1/2 Area (ABEF)

Answer 2

Answer:

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True or false with proper reason given below:

         1. Area(ABCD) = Area(ABEF) : True

         Reason: As the theorem states that parallelograms on same base and between same parallels are equal in area.

         2. Area(AYB) = 1/2 Area(ABCD) : True

         Reason: As the  theorem states that if triangle and parallelogram are on same  base and between same parallels then the area of triangle is half of the parallelogram.

         3. Area(AYB) = 1/2 Area(ABEF) : True …..(i)

         Reason: Now in 1, we proved that Area(ABCD) = Area(ABEF)

         So Area(AYB) = 1/2 Area(ABEF)  = 1/2 Area(ABCD)

         4. Area(AXF) = 1/2 Area(ABCD)  : True

         Reason: Now according to 1,  Area(ABCD) = Area(ABEF), so substituting ABCD to ABEF, then:

         Area(AXF) = 1/2 Area (ABEF): True  

         And the theorem states that if triangle and parallelogram are on same  base and between same parallels then the area of triangle is half of the parallelogram.

         5. Area(AXF) = 1/2 Area(ABEF) : True   …(ii)

         Reason: As the  theorem states that if triangle and parallelogram are on same  base and between same parallels then the area of triangle is half of the parallelogram.

         6. Area(AYB) = Area(AXF) : True

         Reason : from (i) and (ii), we can say that Area(AYB) = Area(AXF) =  1/2 Area (ABEF)