Find the Area of the Blue Shaded Quadrilateral. Geometry video.

#geometry #pythagoras_theorem #quadrilateral
Solve this geometry problem with me. Step-by-step instruction.

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[santiagoarosam]

BCD es triángulo 3, 4, 5, y CKD y MKD son semejantes 》KD=(6/5)3=18/5 》KM=(3×18)/(5×4)=54/20 》ABKM=(6×8/2) - (18/5)(54/20)/2 = 1914/100 =19, 14
Saludos cordiales

[Nikioko]

OK, we have three similar rectangular triangles with a leg ratio of 1 : 0, 75.
The large and medium ones combined have an area of A = 8 · 6 / 2 = 0, 375 · 8² = 24.
The medium and small ones combined have an area of 0, 375 · 6² = 13, 5
If we subtract these from the rectangle, we would subtract the medium triangle twice. So we have to add it again. To find its long leg, we have apply Pythagoras' Theorem, Euclid's Leg Theorem and Euclid's Height Theorem:
c = √(a²+ b²) = √(8² + 6²) = 10.
p = a²/c = 6²/10 = 3, 6
q = b²/c = 8²/10 = 6, 4
h = √(p · q) = √(3, 6 · 6, 4) = 4, 8
A = 0, 375 · 4, 8² = 8, 64

And now we can calculate the blue shaded area:
A = 48 - 24 - 13, 5 + 8, 64 = 19, 14 SU.

[misterenter-izrz]

MD=6*(3/4)=9/2, area of MKD=(1/2)*9/2*(3/5)*9/2*(4/5)=243/50, therefore the area is 24-243/50=957/50.