The area of a triangle is given by the formula (base x height)/2. If we have similar triangles, their sides are proportional with a ratio given by a number called the scale factor. The same scale factor also applies to other lines in the similar triangles – like their height, or to a combination of those lines like the perimeter. The area of similar triangles is also proportional, but here the scale factor will be squared – because we are multiplying two line segments (base and height) which are both proportional, and have the same scale factor.
The area of ΔBEC is (base x height)/2, and both the base and the height are given (CE is the height because CE is perpendicular to EB).
The hint in the problem is that AB||CD, so we can use our knowledge of angles created between parallel lines and an intersecting transversal line to show that the two triangles are similar.
∠DAE ≅ ∠CBE, as alternating interior angles between two parallel lines, and as m∠AED=m∠BEC=90 °, ΔAED∼ ΔBEC. The scale factor is BE/AE=6/3=2
We can now either use the scale factor to find the length of ED (ED/AE=CE/EB or ED=8/6 x 3= 4) and then use the formula for an area of a triangle (base x height)/2 to find the area, or find the area of triangle ΔBEC , whose base and height are given, and then use the scale factor squared to find the area of ΔAED.
(1) CD⊥AB //Given
(2) m∠AED=m∠BEC=90° //(1), defintion of perpendicular lines
(3) AB||CD //Given
(4) ∠DAE ≅ ∠CBE //Alternating interior angles between two parallel lines
(5) ΔAED∼ ΔBEC //Angle-Angle
(6) ED/AE=CE/EB //corresponding sides in similar triangles
(7) ED=8/6 x 3= 4 //CE=8, EB=6, AE=3, given
(8) AreaAED=ED x AE /2 //Formula for area of a triangle
(9) AreaAED=4 x 3 /2 = 6
(8) AreaBEC = CE x EB / 2 //Formula for area of a triangle
(9) AreaBEC= 8x 6 /2 = 24
(10) AreaAED = AreaBEC/4 //Area of similar triangle is proportional to scale factor squared