\(In\ the\ diagram,\ angle\ A\ is\ right,\ E\ is \ the\ midpoint\ of\ AB,\ D\ is\ the\ midpoint \\ of\ AC,\ AB = 16\ and\ AC = 12. \\ \\What\ is\ the\ area \ of\ AEFGD,\ where \ EF\ and\ DG\ are\ perpendicular\ to\ BC?\)

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I've labeled everything on paper then I have no idea what to do...

I solved for CB which is 20, but I can't get past that...

I would love it if you could help me.

tommarvoloriddle Aug 13, 2019

#1**0 **

Calculate the area of the LARGE triangle...then subtrat the two smaller triangles.

You can calculate the angles of ADE.....find the hypotenuse....you are given two sides

then you can calculate the angles FEB CDG etc

Example : 180 - AED - 90 = FEB to find the side lengths of those triangles and their areas......etc,,,

Or calculate the area AED and add the rectangle when you find the side lengths.....

ElectricPavlov Aug 13, 2019

#3

#5**+3 **

From (1) t = 10

So from (2) x + y = 10 ...(5)

Subtract (3) from (4) y^{2} - x^{2} = 28 or (y+x)(y-x) = 28 ...(6)

Use (5) in (6) to get y - x = 28/10 or y - x = 14/5 ...(7)

Add (5) and (7) to get 2y = 64/5 or y = 32/5

Subtract (7) from (5) to get 2x = 36/5 or x = 18/5

Can you take it from there?

Alan
Aug 14, 2019

#10**+3 **

If you have 2x = 36/5 you can divide both sides by 2 to get x on its own. Dividing 36/5 by 2 gives 18/5, because 2 goes into 36 18 times.

Alan
Aug 14, 2019

#14**+2 **

Note that triangle CAB is a 12 - 16 - 20 Pythagorean Right triangle where BC = 20

Triangles CGD and EFB are similar to triangle CAB

Since CD = 1/2 AC....then CD = 6

So

DG/ CD = AB/ CB

DG / 6 = 16/ 20

DG = 6* 16/20 = 6 * 4/5 = 24/5 = 4.8

And CG = sqrt (CD^2 - DG^2) = sqrt ( 6^2 - 4.8^2) = sqrt (12.96) = 3.6

Likewise

Since EB = 1/2 AB....then EB = 8

So

FB/ EB = AB/CB

FB/ 8 = 16/20

FB = 8*16/20 = 8*4/5 = 32/5 = 6.4

So

GF = BC - GC - FB = 20 - 3.6 - 6.4 = 10

So [CFDE] = DG * GF = 4.8 * 10 = 48

And [ DAE ] =(1/2)(1/2)(AC) (1/2)(AB) = 1/8 (12 )(16) = 24

So

[ AEFGD] = [CFDE] + [DAE] = 48 + 24 = 72

CPhill Aug 15, 2019