In the image, if you mark the points like this:

What is the length of BF, if ABCD is a parallelogram?
(3 marks)

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Solution

Consider the $△ A D F$

Since $∠ A F D = 90^{\circ}$ , AD is the hypotenuse.

${A F}^{2} + {F D}^{2} = {A D}^{2}$

${A F}^{2} + 40^{2} = 50^{2}$

${A F}^{2} = 2500 - 1600 = 900$

$A F = 30 m .$
(2 marks)

$B F = A B - A F = 70 - 30 = 40 m .$
(1 mark)

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In the image, if you mark the points like this: What is the length of BF, if ABCD is a parallelogram? (3 marks)

Consider the △ADF Since ∠AFD=90∘, AD is the hypotenuse. AF2+FD2=AD2 AF2+402=502 AF2=2500−1600=900 AF=30m. (2 marks) BF=AB−AF=70−30=40m. (1 mark)

In the image, if you mark the points like this:

What is the length of BF, if ABCD is a parallelogram?
(3 marks)

Consider the $△ A D F$

Since $∠ A F D = 90^{\circ}$ , AD is the hypotenuse.

${A F}^{2} + {F D}^{2} = {A D}^{2}$

${A F}^{2} + 40^{2} = 50^{2}$

${A F}^{2} = 2500 - 1600 = 900$

$A F = 30 m .$
(2 marks)

$B F = A B - A F = 70 - 30 = 40 m .$
(1 mark)