Question

WILL GIVE TESTIMONIAL AND MEDAL. What is the length of AD in the parallelogram below?

Answer 1

Let the point on CB be M



By pythagoras:

(CD)^2 = (CM)^2 + (DM)^2

CD = AB = 16
DM = 12
(16)^2 = (CM)^2 + (12)^2
256-144 =(CM)^2
(CM) = sqrt 112

Area of Parallelogram = Area of Triangle DCM + Area of Trapezoid ADMB
Base x Height = (1/2 base height) + (1/2 x height x sum of parallel sides)
16 x 6 = (1/2 x sqrt112 x 12) + (1/2 x 12 x AD+BM)
96 = 6sqrt(112) + (6)(AD+BM)

BM = CB - CM = AD - CM
96 - 6sqrt(112) = 6 (AD + AD - CM)
96 - 6sqrt(112) = 6 (2AD - CM)
96 - 12sqrt(112) = 12AD - 6CM
96 - 12sqrt(112) = 12AD - 6sqrt112
96 - 6sqrt(112) = 12AD
8 - 0.5sqrt(112) = AD
AD = ?????

Answer 2

????????????????? @dauspex

Answer 3

Which don't you get

Answer 4

This is actually a surprisingly easy question if looked at the right way. There is a hard way to find it, but here is the easy way:

Angle A is equal to angle C, so the 2 triangles are similar. Plus, AB = CD = 16

AD/6 = 16/12 -> AD = (16)(6)/12 = 8

Answer 5

All good now, @TheRealAmiee ?

Answer 6

btw, the 2 triangles that are similar are ADE and CDF :

Answer 7

Yes, @tcarroll010 thank you thank you thank you!

Answer 8

uw! Good luck to you in all of your studies and thx for the recognition! @TheRealAmiee