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

What is the length of BF?

40 m

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30 m

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50 m

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20 m

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Solution

The correct option is A
40 m

Consider the $△ A D F$

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

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

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

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

AF = 30 m.

BF = AB - AF = 70 - 30 = 40 m.

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What is the length of BF?

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What is the length of BF, if ABCD is a parallelogram?
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$\begin{matrix} M a r k s & 0.10 & 10 - 20 & 20 - 30 & 30 - 40 & 40 - 50 & 50 - 60 & 60 - 70 & 70 - 80 N o . o f s t u d e n t s (F r e q u e n c y) & 4 & 10 & 16 & 22 & 20 & 18 & 8 & 2 \end{matrix}$

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Marks	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$
Number of students	$7$	a	$8$	$10$	$5$

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30 40 50 35 55 60 40 30 60 75

75 60 30 40 60 55 35 50 40 30

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In the image, if you mark the points like this: What is the length of BF?

The correct option is A 40 m Consider the △ADF Since ∠AFD=90∘, AD is the hypotenuse. AF2+FD2=AD2 AF2+402=502 AF2=2500−1600=900 AF = 30 m. BF = AB - AF = 70 - 30 = 40 m.

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

What is the length of BF?

The correct option is A
40 m

Consider the $△ A D F$

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

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

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

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

AF = 30 m.

BF = AB - AF = 70 - 30 = 40 m.