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What is the area of the figure above?
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21 Aug 2019, 07:34
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64% (01:56) correct 36% (01:59) wrong based on 72 sessions
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What is the area of the figure above? A. \(40\sqrt{2}\) B. 64 C. 68 D. 81 E. 92 Attachment:
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Re: What is the area of the figure above?
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21 Aug 2019, 07:56
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 454590 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.
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Re: What is the area of the figure above?
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21 Aug 2019, 07:59
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



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Re: What is the area of the figure above?
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Updated on: 21 Aug 2019, 08:11
IMOC Refer attached Image. Draw a line joining the vertices B & E This divides the figure into an isosceles triangle & a trapezium. From fig, Sides of parallel side of trapezium are 10 sqrt2 & (10 sqrt2 2 sqrt2 ) and distance between them sqrt2 Area= 1/2 x (10 x 10) + 1/2 x (10 sqrt2 + 8sqrt2 ) x sqrt2 = 68
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WhatsApp Image 20190821 at 9.29.01 PM.jpeg [ 76.63 KiB  Viewed 1000 times ]
Originally posted by MayankSingh on 21 Aug 2019, 08:04.
Last edited by MayankSingh on 21 Aug 2019, 08:11, edited 1 time in total.



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Re: What is the area of the figure above?
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21 Aug 2019, 08:09
area of the given figure = area of a triangle ABE + area of a trapezium BCDE = \(\frac{1}{2}\)*10*10 + \(\frac{1}{2}\)(10√2 +8√2 )√2 = 50 + 18 = 68 C is the answer
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Re: What is the area of the figure above?
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21 Aug 2019, 08:32
Image What is the area of the figure above?
A. 402‾√402 B. 64 C. 68 D. 81 E. 92
divide the figure we get 2 rectangles; 10*2 + 8*2 ; 36 and an isosceles ∆ 8:8:8√2 ; area ; 1/2 * 8*8 ; 32 total area ; 36+ 32 ; 68 IMO C



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Re: What is the area of the figure above?
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21 Aug 2019, 08:32
its area of square  area of right angled isoceles triangle of non hypotenuse side of 8
=10 ^2 1/2 *8*8
=100 32 =68



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Re: What is the area of the figure above?
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21 Aug 2019, 09:01
This is how I did. Hope it helps. Posted from my mobile device
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Re: What is the area of the figure above?
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21 Aug 2019, 09:08
Please, find the attached file. We can continue the small sides (the length of the side =2) to get the square( the length of the side of square =10. The area of the square =10*10=100 —> See the picture below. Triangle ABC is rightangled isosceles triangle. —> the area of ABC =(8*8)/2=32 —> the area of the figure = 100–32=68 The answer is C. Posted from my mobile device
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Re: What is the area of the figure above?
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21 Aug 2019, 09:47
What is the area of the figure above? A. 402√402 B. 64 C. 68 D. 81 E. 92 Refer attached snapshot. Alternative method is slightly longer. Answer(C)
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What is the area of the figure above.jpg [ 1.08 MiB  Viewed 874 times ]
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Re: What is the area of the figure above?
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21 Aug 2019, 10:34
See the image attached below, We can complete a square since we have 3 angles of 90 degrees, and two sides that are equal. Then we can see the area is a square with length of 10, minus an isosceles right triangle with a side of 8. The triangle is isosceles because both sides are 8. The added outer angle must be 90 because a 4 sided polynomial must have all interior angles adding up to 360. Therefore the area is 10*10  8*8/2 = 100  32 = 68. Answer: C
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Re: What is the area of the figure above?
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21 Aug 2019, 10:39
The answer is 68.
Area of the figure = Area of the triangle formed by extending the sides with value 10 and 10 to form one big isosceles triangle  area of 2 small isosceles triangle which have one side as 2.
side of the smaller triangle: one side will be 2 and the hypotenuse = 2*sqrt2
Area of the smaller triangle = (1/2)*2*2 = 2
Side of bigger triangle: Base and height = 10+2=12, 12 and hypotenuse = 12*sqrt2 Area of the bigger triangle = (1/2)*12*12 = 72
Area of the figure = 72  2*2 = 68



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Re: What is the area of the figure above?
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21 Aug 2019, 10:45
From the image below, triangle ABE is a right angled isosceles triangle. This means that EB=(10^2 + 10^2)^0.5 = 200^0.5 = 10(2)^0.5 Area of solid ABCDE = Area of Triangle ABE + Area of trapezoid EBCD. Area of triangle ABE = (10*10)/2 = 50. Angle AED (90) = angle AEB + angle DEF ....(1) Since we know that triangle ABE is a right angled isosceles triangle, angle AEB = angle ABE = 45. Hence angle DEF=45, implying triangle EFD is a right angled isosceles triangle. This also means that EF=FD=(2)^0.5 Likewise GB=CG=(2)^0.5 Hence DC = EBEFGB = 8(2)^0.5 So, Area of EBCD= {(10(2)^0.5 + 8(2)^0.5)*(2)^0.5}/2 ={(18)(2)/2} =18 So Area of ABCDE = 50 + 18 = 68 Posted from my mobile device
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Re: What is the area of the figure above?
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21 Aug 2019, 11:45
Quote: What is the area of the figure above?
A. 40√2 B. 64 C. 68 D. 81 E. 92 shaded: area of square  area of triangle = 10*10  (8*8/2) = 68



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Re: What is the area of the figure above?
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21 Aug 2019, 12:00
Just think of it as a square of 10*10 having area as 100. A right triangle whose 2 sides are 8 and 8 is cut. Area of that triangle is 32. 10032 is 68. IMO C Note : 1min 37sec. But could have solved way quicker if had i realized the shape
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Re: What is the area of the figure above?
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21 Aug 2019, 12:50
As we see in the figure, we could have a squire with sides 10.but the shape is not perfect. So the area of the shape is less than 100=10*10. In deed we got a triangle out of the shape whose area is 8*8/2=32 So,10032=68 Option C Posted from my mobile device
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Re: What is the area of the figure above?
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