Triangle/Quadrilateral theory and examples

 

You can say that BD = 0 in right angle triangle.

Solution:

First try to solve it using a square of side 1. You will quickly figure out that area of all parts of square except AXY adds up to 5/8 so AXY's area will be 3/8. So p + q = 11.
Now try to do it using a ||gram.
Let's say sides of ABCD are a,b.
In triangle BCD, area is k/2 where k = [ABCD], i.e. area of ABCD.

Triangle CXY is similar to BCD with side ratio 1/2 so area ratio = 1/4.
[CXY] = k/8.

One important thing in ||gram is that adjacent angles are supplementary and opposite angles are equal. So sin(x) for each angle is same.

[CXY] = 1/2(a/2)(b/2)sin(x) = k/4

=> ab.sin(x) = 2k

[ADY] = ab/2.1/2.sin(x) = k/4 = [ABX]

So [AXY] = k - k/4 - k/4 - k/8 = 3k/8

p + q = 11.

Solution:
angle ACE is right angle since sides of ACE make a Pythagorean triplet.

[ABF] = 1/2*3*15*sin(A), sin(A) = 16/20

[ABF] = 18

Similarly [EFD] = 5*12*1/2*sin(E) = 18

[BCD] = 1/2*9*4 = 18

[ACE] - [BFD] = [ABF] + [EFD] + [BCD] = 54

1 - 1/sqrt(2)

Solution:
[BXY] = [XYCA] = [ABC]/ 2 = K
[BXY]/[ABC] = 1/2
Since area ratio is 1/2 and BXY, ABC are similar so side ratio is 1/(sqrt(2)).
AX/AB = (AB-BX)/AB = 1 - BX/AB = 1 - 1/sqrt(2).