Since segments ST and PQ are parallel, triangles SRT and PRQ are similar due to the AAA postulate. In general, the ratio between the corresponding sides of two similar triangles is constant; therefore,

$\frac{SR}{PR}=\frac{RT}{RQ}=\frac{TS}{QP}$

Furthermore,

$PS=PR-RS$

Finding PR and RS,

$\begin{gathered} ST=4+PS \\ SR=12 \\ PQ=15 \end{gathered}$

Then,

$\frac{12}{PR}=\frac{TS}{15}$ $\begin{gathered} \Rightarrow\frac{12}{PR}=\frac{4+PS}{15} \\ \Rightarrow\frac{12}{PS+12}=\frac{4+PS}{15} \end{gathered}$

Solving for PS,

$\begin{gathered} 12\cdot15=(PS+12)(PS+4) \\ \Rightarrow180=PS^2+16PS+48 \end{gathered}$

Solve the quadratic equation in terms of PS, as shown below

$\begin{gathered} \Rightarrow PS^2+16PS-132=0 \\ \Rightarrow PS=\frac{-16\pm\sqrt[]{16^2-4(-132)}}{2}=\frac{-16\pm28}{2} \\ \Rightarrow PS=-22,6 \end{gathered}$

And PS is a segment; therefore, it has to be positive.

Hence, the answer is PS=6

3 0

9 months ago

For any integer x,x2-x will always produce an even value

Levart [38]

Consider expression $x^2-x.$ First, you can factor it:

$x^2-x=x(x-1).$

Since x is integer number, then you can see that x-1 is previous integer number (x-1 is 1 unit smaller than x).

Therefore, x-1 and x are two consecutive integers. When you have two consecutive integers, one of them is always even and one is always odd. Multiplying even integer number by odd integer number you always get even integer number.

Thus, $x^2-x$ is always even.

8 0

2 years ago