Given:
[tex]ST=4y,TU=2y+6[/tex]
To find:
The value of y, ST and TU.
Solution:
In triangle STV and UTV,
[tex]m\angle TSV\cong n\angle TUV[/tex] (Given right angles)
[tex]m\angle SVT\cong n\angle UVT[/tex] (Given)
[tex]TV\cong TV[/tex] (Common side)
The corresponding two angles and a non included sides in both triangles are congruent. So, the triangles are congruent by using AAS congruence postulate.
[tex]\Delta STV\cong \Delta UTV[/tex] (AAS congruence postulate)
Corresponding parts of congruent triangles are congruent. So,
[tex]ST\cong UT[/tex] (CPCTC)
[tex]4y=2y+6[/tex]
[tex]4y-2y=6[/tex]
[tex]2y=6[/tex]
[tex]y=3[/tex]
Now,
[tex]ST=4y[/tex]
[tex]ST=4(3)[/tex]
[tex]ST=12[/tex]
And,
[tex]TU=2y+6[/tex]
[tex]TU=2(3)+6[/tex]
[tex]TU=6+6[/tex]
[tex]TU=12[/tex]
Therefore, [tex]y=3,\ ST=12\ units,\ TU=12\ units[/tex].
