# Find the area of the shaded region. Explain. Figure: http://www.flickr.com/photos/93084714@N07/8638417297/in/photostream

durbanville | Certified Educator

We need to find the area of the rectangle `A=l times b`  and

the area of the 2 triangles which is the same for both `A=1/2 b times h`

`therefore A_(rect)= 10 times 5sqrt2`  =`= 50sqrt2` and the base of the triangle is 5mm and the perpendicular height (that is the height reaching from the right angle) is also 5mm and we have 2 triangles:

`therefore` `A_(tri) = 2( 1/2) 5 times 5`  `= 25`

(care to use the correct dimensions for the triangles' bases and height - must be the perpendicular height)

Now deduct the area of the 2 triangles (25mm`^2` ) from the area of the rectangle (`50sqrt2` )

`therefore A _(rect - tri)= 50sqrt2 - 25`

`A = 45.71 mm^2`

`therefore` The area of the shaded region is 45,71mm`^2`

oldnick | Student

IT'S RIGHT SOLUTION!

The angle  inside the triangle is of 90°

the since the tirangle is isosceles  the  tu sides are equals:

then:    so l= 5

Now  the heigts realtive hypotenuse divides the basis of tirangle in two equals halfs of value . So from Euclide Theorem we can get heigts h:

so

The totala area is

To get the green Area onli have  to subtract from A the  area of two equal triagle isosceles  T:

so the greeen area G will be:

`G= A - 2T = 50sqrt(2) -25/2= (100sqrt(2) -25)/2=25/2(4sqrt(2) -1)`

Posted by oldnick on April 11, 2013 at 3:32 AM (Answer #1)

oldnick | Student

The angle  inside the triangle is of 90°

the since the tirangle is isosceles  the  tu sides are equals:

then:  `sqrt(l^2 + l^2) =lsqrt(2) = 5sqrt(2)`  so l= 5

Now  the heigts realtive hypotenuse divides the basis of tirangle in two equals halfs of value `5sqrt(2)/2` . So from Euclide Theorem we can get heigts h:

`5sqrt(2)/2 5sqrt(2)/2 = h^2`   so `h=5sqrt(2)/2`

The totala area is `A=50 sqrt(2)`

To get the green Area onli have  to subtract from A the  area of two equal triagle isosceles  T:

`T= 1/2 5 sqrt(2)/2 5sqrt(2)/2= 25/4`

so the greeen area G will be:

`G= A -T= 50sqrt(2)-25/4= [200sqrt(2) -25]/4= 25/4 (8sqrt(2)-1)`