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08 May 2019, 01:51
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B
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Difficulty:
25%
(medium)
Question Stats:
81% (02:17) correct
19%
(02:15)
wrong
based on 96
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In the figure above, DF has a length of \(5\sqrt{2}\) and ∠ACD measures 135°. If the area of parallelogram ACDF is 75, what is the area of rectangle ABDE?
A. \(50 \sqrt{2}\)
B. 100
C. \(75 \sqrt{2}\)
D. \(100 \sqrt{2}\)
E. 150
Attachment:
2019-05-08_1244.png [ 13.35 KiB | Viewed 7781 times ]
Archit3110
08 May 2019, 02:39
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Bunuel
side DE =FE = 5 ; as its 45:45:90 ∆
given area of llogram = 75 ; sum of base * height /1/2 ; we get CD+AF=30 + 10 ; 40 ; each side 20
area of rectangle ; 20*5 ; 100
IMO B
Rakesh1987
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08 May 2019, 03:24
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Bunuel
∠ACD=135°, Therefore ∠AFD=135° (opposite angles of a parallelogram)
∠EFD=45° (1-∠AFD which is 135°), And ∠EDF=45° {1-∠DEF(90°, it is given in question that ABDE is rectangle)-∠EFD(45°)}
Triangle DEF is a 90-45-45 right triangle and the measure of its sides is therefore \(\sqrt{2}\):1:1, As DF=\(5\sqrt{2}\), DE=EF=5.
Triangle ABC and Triangle DEF are congruent as all their corresponding angles are equal and DE=AB (opposite sides of a rectangle).
Therefore area of the two triangles ABC and DEF combined is= 2*(1/2*5*5)=25
Area of parallelogram ACDF =75 (given in question)
Therefore, Area of rectangle ABDE= Triangle ABC+ Triangle DEF+ Parallelogram ACDF=25+75 =100. Option B.
Please hit +1 Kudos if you liked the explanation.
