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.
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
To prove that the sum of the areas of triangles CBF and ADF is equal to 36 units squared, we can use the formula for the area of a triangle.
Let's start by determining the length of AF, the height of triangle ADF. Since the area of the rectangle ABCD is given as 72 units squared, we can find its length by dividing the area by the width. Let's assume the width of the rectangle, AB, is x units.
Area of Rectangle ABCD = Length * Width = 72
Length * x = 72
Length = 72 / x
Now, we are given that length BF is 4 units, and since BF is parallel to AD, it becomes the height of triangle CBF.
Area of Triangle CBF = (Base * Height) / 2 = (x * 4) / 2 = 2x
To find the area of triangle ADF, we use the formula ADF = (Base * Height) / 2. The base is AD, which is the same as the width of the rectangle.
Area of Triangle ADF = (x * AF) / 2
Since the length of the rectangle (AD) is equal to the length of BF plus AF, we can express it as:
Length = (BF + AF) = 4 + AF
AF = Length - BF = (72 / x) - 4
Now, we can substitute the value of AF into the formula for the area of triangle ADF:
Area of Triangle ADF = (x * ((72 / x) - 4)) / 2 = (72 - 4x) / 2 = 36 - 2x
To prove that the sum of the areas of triangles CBF and ADF is equal to 36 units squared, we can add their areas together:
Area of Triangle CBF + Area of Triangle ADF = 2x + (36 - 2x) = 36
Therefore, the sum of the areas of triangles CBF and ADF is indeed 36 units squared, algebraically proven.