Question

What is the value of x? x= units

Answer 1

Step 1

In the right triangle RTQ

Find the value of $RQ^{2}$

Applying the Pythagorean Theorem

$RQ^{2} =RT^{2}+TQ^{2}$

we have

$RT=x\ units \\TQ=16\ units$

substitute the values

$RQ^{2} =x^{2}+16^{2}$

$RQ^{2} =x^{2}+256$

Step 2

In the right triangle RST

Find the value of $RS^{2}$

Applying the Pythagorean Theorem

$RS^{2} =RT^{2}+ST^{2}$

we have

$RT=x\ units \\ST=9\ units$

substitute the values

$RS^{2} =x^{2}+9^{2}$

$RS^{2}=x^{2}+81$

Step 3

In the right triangle RSQ

Find the value of x

Applying the Pythagorean Theorem

$SQ^{2} =RS^{2}+RQ^{2}$

we have

$RS^{2}=(x^{2}+81)\ units^2\\RQ^{2}=(x^{2}+256)\ units^2$

$SQ^{2}=(16+9)^2=625\ units^2$

substitute the values

$625=(x^{2}+81)+(x^{2}+256)$

$625=(2x^{2}+337)$

$2x^{2}=625-337$

$2x^{2}=288$

$x^{2}=144$

$x=12\ units$

therefore

the answer is

$x=12\ units$

Answer 2

X = square root ( 9*16)
x = square root (144)
x = 12