The diagram above shows two congruent right triangles 12-16-20 overlapping to form a dark blue region.
Find the area of this dark blue region.
+128475 Let the bottom let vertex of the blue triangle =A
Let the bottom right vertex of the blue triangle = B
Let the remaining vertex of the blue triangle = C
Let the bottom right vertex of the figure = D
Let DE be the edge that is perpendicular to AD
And traingles
ABC and ADE are similar triangles
So
BC / AB = DE / AD
BC / 12 = 12 / 16
16* BC = 12 * 12
16* BC = 144
BC = 144/16 = 9
So the dark area is the area of triangle ABC =
(1/2) (AB) ( BC) =
(1/2)(12) (9) =
(1/2) 108 =
54 units^2
