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15 Nov 2021, 08:15
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C
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Difficulty:
35%
(medium)
Question Stats:
78% (02:58) correct
22%
(03:01)
wrong
based on 32
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In the figure, the areas of parallelograms EBFD and AECF are 3 and 2 respectively. What is the area of rectangle ABCD?
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15 Nov 2021, 08:30
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Area of quadrilateral EBFD = DF * Vertical Height (say h)
Area of quadrilateral AECF = FC * Vertical Height (say h)
DF * h = 3 --- (1)
CF * h = 2 --- (2)
Adding both the equations
(DF+CF) * h = 5
DF + CF is the base of the rectangle and height is BC (h)
IMO - C
Area of quadrilateral AECF = FC * Vertical Height (say h)
DF * h = 3 --- (1)
CF * h = 2 --- (2)
Adding both the equations
(DF+CF) * h = 5
DF + CF is the base of the rectangle and height is BC (h)
IMO - C
yashikaaggarwal
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15 Nov 2021, 08:37
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Given: the areas of parallelograms EBFD and AECF are 3 and 2 respectively.
Then,
The area of EGFH = 3-2 = 1
Making an area of AGE + Area of CFH = 1
Since it's a parallelogram.
Area of AGE = Area of CFH = 0.5
Also, area of DFG = area of EHB = 1
Now,
The area of a triangle whose one side lies on the opposite side of the parallelogram is equal to half of the area of a parallelogram.
Area of DEC = 2.5 (1+1+.5)
Area of parallelogram ABCD = 2*2.5 = 5
Answer is C
Posted from my mobile device
Then,
The area of EGFH = 3-2 = 1
Making an area of AGE + Area of CFH = 1
Since it's a parallelogram.
Area of AGE = Area of CFH = 0.5
Also, area of DFG = area of EHB = 1
Now,
The area of a triangle whose one side lies on the opposite side of the parallelogram is equal to half of the area of a parallelogram.
Area of DEC = 2.5 (1+1+.5)
Area of parallelogram ABCD = 2*2.5 = 5
Answer is C
Posted from my mobile device
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IMG_20211115_210051.jpg [ 52.96 KiB | Viewed 1628 times ]
