saiesta wrote:
Would anyone please be so kind to help me with this question? I have no idea how to tackle it. I literally spent hours trying to solve the question.
The rhombus (AFCE) is inscribed in rectangle (ABCD). The length of BF = DE.
If the rectangle has a width of 20 yards and a length of 25 yards, what would be the total length (the perimeter) of a fence along the sides defined by AFCE?
A. 80 yards
B. 82 yards
C. 84 yards
D. 85 yards
E. 90 yards
Hi,
the method explained above is the correct method..
But another method which you could use is the POE..
now the perimeter of rectangle is 20+25+20+25=90..
so perimeter of rhombus has to be less than 90.. elimiate E..
Also perimeter of the rhombus has to be greater than 20+20+20+20=80, as the side of rhombus is the hypotenuse , which has to be more than the side of the same triangle, 20
eliminate A..
lets see the remaining choices..
A rhombus has all equal sides, so each side = perimeter/4
B. 82 yards... side= 20.5
C. 84 yards..side=21
D. 85 yards..side=21.25
now this side = hypotenuse osf a right angle triangle whose one side is 20..
lets see if hyp=20.5 and one side = 20, what is the second side...
\(\sqrt{20.5^2-20^2}\)=\(\sqrt{(20.5-20)(20.5+20)}\)
...\(\sqrt{.5*40.5}\)=\(\sqrt{.5*.5*81}\)=4.5..
this exactly fits in as other side of rhombus..
25=4.5=20.5..
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