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ABCD is a rectangle with AB=16 units and...

`ABCD` is a rectangle with `AB=16` units and `BC=12` units. `F` is a point on `AB` and `E` is a point on `CD` such that `AFCE` is a rhombus. Find the length of `EF.`

A

` "15 units"`

B

` "16 units"`

C

` "17 units"`

D

` "18 units"`

Text Solution

Verified by Experts

The correct Answer is:
A
Let `FB` `x` units
`:. AF=16-x` units
Since AFCE is a rhombus.
`:. FC = AF =16-x`
Now, by Pythagoras theorem,
`CF^(2)=FB^(2)+BC^(2)`
`rArr (16-x)^(2)=x^(2)+144`
`rArr 256+x^(2)-32x=x^(2)+144`
`rArr 32 x=112`
`rArr x=(112)/(32)=(7)/(2) ` units
`:. ` Side of rhombus `=16-x=16-(7)/(2)=(25)/(2)` units
By Phythagoras theorem,
`aC^(2)=AB^(2)+BC^(2)`
`rArr AC^(2)=(16)^(2)+(12)^(2)=256+144=400 rArr AC = 20`
`:. AO =(1)/(2)xxAC=10 ` units
`:. AE^(2) = OE^(2)+AO^(2) " "` (by Phythagoras theorem)
`((25)/(2))^(2) = OE^(2)+(10)^(2)`
`rArr OE^(2)=156.25-100=56.25 rArr OE=7.5`
`:. EF=2xx OE = 2xx7.5 = 15` units
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