If the given Quadrilateral is a rectangle, find the area of the shaded

Official Explanation

:D at least tried to solve it
we have two isosceles triangles (45 45 90) with two sides equal 1, so to find altitude of these triangles i need to to draw the line that bisects this tringle let height be x then x^2+(sqrt2)^2=1^2 ----> x^2= 1^2-(sqrt2)^2 ---> x^2 = -1 so altitude is 1. Area of triangle = base *height ---> 1*1 =1

i also need to find areas of two longer triangles.


Diaogonal of rectangle is sqroot W^2+l^2 ----> 5^2+8^2 = 89 i will round it to 90

so diagonal of rectangle is 90

Area of rhombus is length * width

Width is 1 and length is of each rhombus is 90/2 i.e. 45

so Area of Rhombus is 1*45/2 = 22.5

got a "bit" confused .... anyway will solve it to the end

so if area of one rhombus is 22.5 then area of two rhombus is 90

so area of shaded region is 90

hope my answer is correct

and why hell i was calculating area of triangle

IMO D.

Both triangles are similar. Find the area of one shaded region and multiply it by 2.

13/4 is area of one region and when multiplied by 2 gives 13/2 .

Area of the rectangle = 8*5 = 40
Diagonal AC divides it in half , 20 each.
DO divides area AOC into 2 parts. 10 each
KO divides AOD into further 2 parts 5 each.

Area of AOK = 5 when base is 2.5.
when base is 1 , area = 2

area of OGC = 5 when base is 4
when base is 1 , area = 5/4 = 1.25

Area of shaded region = 2(2+1.25) = 6.5

If we divide the rectangle into four equal parts and then analyse 1st part APOS
Area of this part = 4*5/2 = 10
There will be two unshaded triangles.
Area of 1st unshaded triangle = 1/2*3*5/2 = 15/4
Area of 2nd unshaded triangle = 1/2*3/2*3 = 9/4
So area of shaded part in the 1st part = 10-15/4-9/4 = 10-6 = 4
2nd part BPOQ has no shaded part
3rd part CQOR has shaded part and its area will also be =4
4th part DROS has no shaded part.

Total area of shaded part =4+4 = 8

Answer E

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