31) Find overlapping area. #maths #matholympiad #geometry

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Another method:

Let the side of the rhombus be a.

In △DAE, DA=5, AE=AB-EB=12-a, and DE=a

Using Pythagoras theorem, DA²+AE²=DE²
i.e., 5²+(12-a)²=a²

Simplifying, we get: a=169/24

Hence the area of the cm²

PS-mhts

After establishing the congruity of the outer 4 triangles, and assigning their hypotenuse to h, and the unknown leg to l, I solved the 2 equations in 2 unknowns:
h + l = 12, and
h² - l² = 5²
This gives the unknown leg as 119/24
The area of the inner rhombus may be calculated from the main rectangle (5 x 12) minus the two side triangles (5 x (119/24))/2.
So, 60 - 595/48 - 595/48 = 60 - 585/24 =
845/24 = 35 and 5/24, or approximately 35.2
Thanks!

skwest

My eyes tell me the yellow area is exactly half of either given rectangle

ZephyrusTheReal

31) Find overlapping area. #maths #matholympiad #geometry

31) Find overlapping area. #maths #matholympiad #geometry

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