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605-655 (Medium)|   Geometry|      
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What is the area of the figure above? 21 Aug 2019, 08:34
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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45% (medium)

Question Stats:

71% (02:15) correct 29% (02:19) wrong based on 105 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


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C
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What is the area of the figure above? 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.
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What is the area of the figure above? 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
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What is the area of the figure above? Updated on: 21 Aug 2019, 09:11
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.
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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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What is the area of the figure above? 21 Aug 2019, 09:09
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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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What is the area of the figure above? 21 Aug 2019, 09:32
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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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What is the area of the figure above? 21 Aug 2019, 09:32
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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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What is the area of the figure above? 21 Aug 2019, 10:01
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This is how I did. Hope it helps.

Posted from my mobile device
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What is the area of the figure above? 21 Aug 2019, 10:08
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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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FDAB4625-FA4B-4D52-B481-160642219A45.jpeg
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What is the area of the figure above? 21 Aug 2019, 10:47
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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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What is the area of the figure above.jpg
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What is the area of the figure above? 21 Aug 2019, 11:34
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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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What is the area of the figure above? 21 Aug 2019, 11:39
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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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What is the area of the figure above? 21 Aug 2019, 11:45
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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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What is the area of the figure above? 21 Aug 2019, 12:45
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Quote:
What is the area of the figure above?

A. 40√2
B. 64
C. 68
D. 81
E. 92
Show more

shaded: area of square - area of triangle = 10*10 - (8*8/2) = 68
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What is the area of the figure above? 21 Aug 2019, 13:00
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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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What is the area of the figure above? 21 Aug 2019, 13:50
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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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What is the area of the figure above? 12 Sep 2023, 02:55
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