Question

Geometry math question

Answer 1

Look at the picture.

Let |AE| = |AD| = b

We have a proportion:

`(x)/(13){y}=(19)/(y+a)=(x)/(y+2a)`

Solve for y from first proportion

`(13)/(y)=(19)/(y+a)\ \ \ |\text{cross multiply}\\\\19y=13(y+a)\\\\19y=13y+13a\ \ \ \ |-13y\\\\6y=13a\ \ \ |:6\\\\y=(13)/(6)a`

Substitute to the second proportion

`(19)/(y+a)=(x)/(y+2a)\\\downarrow\\(19)/((13)/(6)a+a)=(x)/((13)/(6)a+2a)\\\\(19)/((13)/(6)a+(6)/(6)a)=(x)/((13)/(6)a+(12)/(6)a)\\\\(19)/((19)/(6)a)=(x)/((25)/(6)a)\ \ \ \ |\dot a\\eq0\\\\(19)/((19)/(6))=(x)/((25)/(6))\\\\19\cdot(6)/(19)=x\cdot(6)/(25)\\\\6=(6x)/(25)\ \ \ |\cdot25\\\\6x=6\cdot25\ \ \ |:6\\\\x=25`

Answer: B. 25