Question

What is the length of side TS?

Answer 1

We know that

 in the triangle TQS
applying the Pythagorean theorem
QS²=TS²+TQ²---------> TQ²=QS²-TS²--------> TQ²=18²-9x²-----> equation 1

in the triangle TRS
TS²=TR²+RS²--------------> TR²=TS²-RS²-------> TR²=9x²-144----> equation 2

in the triangle QTR
TQ²=TR²+36-----------> equation 3


I substitute 1 and 2 in 3
18²-9x²=9x²-144+36--------> 18x²-432=0------> x²=24-------> x=√24
x=2√6
TS=3*x------> 3*2√6-----> 6√6
TS=6√6 units

the answer is
TS=6√6 units 
6 is the square root of 6 units

Answer 2

We are given a triangle TQS.

And in triangle TQS, TR is a perpendicular to QS.

Therefore, in triangle TQS and TRS

<QTS = <TRS  :    Right angles .

<S = <S           : Common angles of triangles TQS and TRS.

Therefore  triangles TQS and TRS are similar.

Note: Similar triangle has sides in proportion.

Therefore,

QS/TS =TS/RS

(6+12)/3x = 3x/12.

18/3x = 3x/12

ON cross multiplication, we get

3x*3x = 18 * 12

$9x^2 = 216$.

Dividing both sides by 9, we get

$\frac{9x^2}{9} =\frac{216}{9}$

$x^2=24$

$x=\sqrt{24}$

Plugging x=$\sqrt{24}$ in 3x, we get

3x = 3($\sqrt{24}$) = $3(2\sqrt{6})$ =$6\sqrt{6}$.

Therefore, the length of side TS is $6\sqrt{6}$ units.