Don't get this question

Let WX = a, XY = b and YZ = c,

then

a + b = 9 and b + c = 7 ................... (1)

Let the height of the triangle be h,

then

from triangle AWX, ah/2 = 30 so a = 60/h,

and

from triangle AYZ, ch/2 = 20 so c = 40/h.

Substitute those into equations (1), and subtract one from the other to get 20/h = 2, so h = 10.

From that, a = 6, c = 4 and b = 3.

Triangle AXY will have an area bh/2 = 15.

In the diagram below, WY = 9, XZ = 7. The area of AWX = 30 and the area of AYZ = 20. Find the area of AXY

I will try and see if I can get it right!.

WY - XZ = 9 - 7 =2 =YZ

XZ - YZ =7 - 2 =5=XY

WY - XY =9 - 5 =4 =WX

Area of AWX =[Base x Height]/2 =30 =[4 x Height]/2 = 30

Height(AX) =15

Area of AXY =[5 x 15]/2 =37.5

Moderators: Please verify this. Thanks.

Good work both of you    

I get the same as guest 2   That is I get 15 u^2     (shouldn't forget the units)

Guest 2 I really like you logic :)     

Guest 1, it is really good that you had a go   

We are using the fact that triangles under the same height are to each other as their bases....that is

Area of Δ AYZ / Area of Δ AWX   =   YZ /WX

20 / 30  =  YZ / WX

2 / 3 = YZ/ WX

(2/3)WX  = YZ

And we know that

WX + XY  = 9

YZ + XY  = 7      making a substitution, we have that

WX        + XY   = 9

(2/3)WX + XY  = 7       subtract the second equation  from the first

(1/3) WX   = 2    →   WX  = 6    →  XY  = 3

And we can solve for the height as follows

Area  of Δ AWX  = 30

(1/2)WX * h   =

  (1/2)* 6 * h   =  30

3h  = 30

h = 10

So...the area of Δ AXY  = 

(1/2) XY * 10  =

(1/2) * 3 * 10   =

(1/2) * 30  =

15    ......just as Guest 2 found !!!