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21 Aug 2019, 08:34
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C
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Difficulty:
45%
(medium)
Question Stats:
71% (02:15) correct
29%
(02:19)
wrong
based on 105
sessions
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Time
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Not Attempted Yet
What is the area of the figure above?
A. \(40\sqrt{2}\)
B. 64
C. 68
D. 81
E. 92
Attachment:
image2264.jpg [ 3.38 KiB | Viewed 4943 times ]
21 Aug 2019, 08:56
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What is the area:
A. 402√402
B. 64
C. 68
D. 81
E. 92
draw line parallel from both ends and to the opposite side,
then we have a square at the top and two rectangles of dimentions 2*8
so the triangle we are left with 8,8 ,x this will form a 45-45-90 trianlge and rule x,x,xroot2
or we can also say that the area of the triangle =32
then the area of the two rectangles=2*8*2=32
and the square at the top=2*2=4
so total area=32+32+4=68
Option C is correct.
A. 402√402
B. 64
C. 68
D. 81
E. 92
draw line parallel from both ends and to the opposite side,
then we have a square at the top and two rectangles of dimentions 2*8
so the triangle we are left with 8,8 ,x this will form a 45-45-90 trianlge and rule x,x,xroot2
or we can also say that the area of the triangle =32
then the area of the two rectangles=2*8*2=32
and the square at the top=2*2=4
so total area=32+32+4=68
Option C is correct.
Attachments
triangle.jpg [ 9.75 KiB | Viewed 4560 times ]
21 Aug 2019, 08:59
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What is the area of the figure above?
A. 40√2
B. 64
C. 68
D. 81
E. 92
If we divide the figure as
A triangle of area = 8*8/2 = 32
A rectangle with area = 10*2 = 20
A rectangle with area = 8*2 = 16
Adding 3 areas = 32 + 20 + 16 = 68
IMO C
A. 40√2
B. 64
C. 68
D. 81
E. 92
If we divide the figure as
A triangle of area = 8*8/2 = 32
A rectangle with area = 10*2 = 20
A rectangle with area = 8*2 = 16
Adding 3 areas = 32 + 20 + 16 = 68
IMO C
