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Math Expert V
Joined: 02 Sep 2009
Posts: 59017
What is the area of the figure above?  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 68% (01:59) correct 32% (01:53) wrong based on 64 sessions

### HideShow timer Statistics 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 1040 times ]

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Senior Manager  G
Joined: 25 Jul 2018
Posts: 338
Re: What is the area of the figure above?  [#permalink]

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2

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 right-angled isosceles triangle.
—> the area of ABC =(8*8)/2=32

—> the area of the figure = 100–32=68

Posted from my mobile device
Attachments FDAB4625-FA4B-4D52-B481-160642219A45.jpeg [ 869.06 KiB | Viewed 804 times ]

Director  P
Joined: 04 Sep 2015
Posts: 663
Location: India
WE: Information Technology (Computer Software)
Re: What is the area of the figure above?  [#permalink]

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1
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.
Attachments triangle.jpg [ 9.75 KiB | Viewed 917 times ]

SVP  P
Joined: 03 Jun 2019
Posts: 1837
Location: India
Re: What is the area of the figure above?  [#permalink]

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1
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?  [#permalink]

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IMO-C

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
Attachments WhatsApp Image 2019-08-21 at 9.29.01 PM.jpeg [ 76.63 KiB | Viewed 889 times ]

Originally posted by MayankSingh on 21 Aug 2019, 09:04.
Last edited by MayankSingh on 21 Aug 2019, 09:11, edited 1 time in total.
Senior Manager  P
Joined: 31 May 2018
Posts: 451
Location: United States
Concentration: Finance, Marketing
Re: What is the area of the figure above?  [#permalink]

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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
Attachments ps.png [ 7.88 KiB | Viewed 899 times ]

GMAT Club Legend  D
Joined: 18 Aug 2017
Posts: 5261
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: What is the area of the figure above?  [#permalink]

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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
Manager  G
Joined: 15 Jun 2019
Posts: 209
Re: What is the area of the figure above?  [#permalink]

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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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Senior Manager  P
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Re: What is the area of the figure above?  [#permalink]

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This is how I did. Hope it helps.

Posted from my mobile device
Attachments IMG_20190821_223035.jpg [ 2.83 MiB | Viewed 829 times ]

Senior Manager  G
Joined: 07 Mar 2019
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Location: India
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Re: What is the area of the figure above?  [#permalink]

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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.

Attachments

File comment: What is the area of the figure above What is the area of the figure above.jpg [ 1.08 MiB | Viewed 764 times ]

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GMAT Tutor S
Joined: 17 Sep 2014
Posts: 217
Location: United States
GMAT 1: 780 Q51 V45 GRE 1: Q170 V167 Re: What is the area of the figure above?  [#permalink]

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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.

Attachments completed image.png [ 35.47 KiB | Viewed 723 times ]

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Intern  B
Joined: 24 Jul 2019
Posts: 2
Re: What is the area of the figure above?  [#permalink]

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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
Senior Manager  P
Joined: 18 May 2019
Posts: 436
Re: What is the area of the figure above?  [#permalink]

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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 = EB-EF-GB = 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
Attachments image.jpg [ 1.58 MiB | Viewed 666 times ]

Director  P
Joined: 24 Nov 2016
Posts: 756
Location: United States
Re: What is the area of the figure above?  [#permalink]

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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
Director  P
Joined: 18 Dec 2017
Posts: 652
Location: United States (KS)
Re: What is the area of the figure above?  [#permalink]

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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.
100-32 is 68.
IMO C

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Senior Manager  G
Joined: 29 Jun 2019
Posts: 466
Re: What is the area of the figure above?  [#permalink]

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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,100-32=68
Option C

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Always waiting Re: What is the area of the figure above?   [#permalink] 21 Aug 2019, 13:50
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