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.
4 answers
Where is F ?
BF is between BA
BA is shorter side of rectangle
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
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
7 answers