Posted by **Anthony** on Thursday, March 7, 2013 at 7:19am.

Rectangle ABCD has an area of 72 units squared. The length of BF is 4 units. Algebracially prove that the sum of the areas of triangles CBF and ADF are equal to 36 units squared.

- Algebra -
**Reiny**, Thursday, March 7, 2013 at 10:30am
Where is F ?

- Algebra -
**Anonymous**, Thursday, March 7, 2013 at 11:58am
BF is between BA

- Algebra -
**Anthony**, Thursday, March 7, 2013 at 12:46pm
BA is shorter side of rectangle

- Algebra - REINY -
**Anthony**, Thursday, March 7, 2013 at 11:02pm
Hi - Do you think this is correct?

BC = AD

Area of Rectangle = (FB+AF).BC

Area of Rectangle = 72 units squared

Area = 1/2 bh

Area 1 = 1/2 FB.BC

Area 2 = 1/2 AF.AD

1/2 = AF.BC

A1 + A2 = 36 units squared

1/2 FB.BC + 1/2 AF.BC

1/2 [FB.BC + AF.BC]

1/2 [BC (FB + AF)

1/2.72 units squared - 36 units squared

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