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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
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WhatsApp Image 2019-08-21 at 9.29.01 PM.jpeg
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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
C is the answer
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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
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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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This is how I did. Hope it helps.

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

—> the area of the figure = 100–32=68
The answer is C.

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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.

Answer(C)
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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.

Answer: C
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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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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

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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
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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

Note : 1min 37sec. But could have solved way quicker if had i realized the shape :)
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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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