Follow the pic of Kayasari Ryuunosuke :)

\(\Rightarrow a=\sqrt{100}=10\left(cm\right)\)

\(b=\sqrt{36}=6\left(cm\right)\)

Pythagoras \(\Rightarrow\)\(c=\sqrt{10^2-6^2}=8\left(cm\right)\)

\(\Rightarrow\)\(S_C=8^2=64\left(cm^2\right)\)

Condittion: \(3x+1\ne0\)

\(\Rightarrow x\ne-\dfrac{1}{3}\)

\(\dfrac{x-1}{3x+1}=\dfrac{1}{2}\)

\(\Rightarrow2\left(x-1\right)=3x+1\)

\(\Leftrightarrow2x-2=3x+1\)

\(\Leftrightarrow2x-3x=1+2\)

\(\Leftrightarrow-x=3\)

\(\Leftrightarrow x=-3\) (satisfy)

\(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}+\dfrac{3-\sqrt{5}}{3+\sqrt{5}}\)

\(=\dfrac{\left(3+\sqrt{5}\right)^2}{9-5}+\dfrac{\left(3-\sqrt{5}\right)^2}{9-5}\)

\(=\dfrac{\left(3+\sqrt{5}+3-\sqrt{5}\right)\left(3+\sqrt{5}-3+\sqrt{5}\right)}{4}\)

\(=\dfrac{6\cdot2\sqrt{5}}{4}=\dfrac{12\sqrt{5}}{4}=3\sqrt{5}\)

\(\left(100+1\right)\cdot100:2=5050\)

Answer: A

Because \(\Delta ABC\) is a square triangle.

\(\Rightarrow\)\(AB^2+AC^2=BC^2\) (Pytago)

\(\Rightarrow BC=\sqrt{3^2+4^2}=5\)

We have:
\(AB\cdot AC=BC\cdot AH\)

\(\Rightarrow AH=\dfrac{AB\cdot AC}{BC}=\dfrac{3\cdot4}{5}=\dfrac{12}{5}=2.4\left(cm\right)\)

Answer: 2.4 cm

\(\Delta PQS\) has: \(PQ=PS\)

\(\Rightarrow\)​\(\Delta PQS\) is an ​ isosceles triangle.

\(\Rightarrow\)\(\widehat{PSQ}=\frac{180^0-\widehat{QPS}}{2}\)

\(\Rightarrow\)\(\widehat{PSQ}=84^0\)

\(\Rightarrow\)\(\widehat{PSR}=180^0-84^0=96^0\)

\(\Delta PSR\) has: \(SP=SR\)

\(\Rightarrow\)\(\Delta PSR\) is an isosceles triangle

\(\Rightarrow\)\(\widehat{SPR}=\frac{180^0-\widehat{PSR}}{2}\)

\(\Rightarrow\)\(\widehat{SPR}=42^0\)

\(\Rightarrow\)\(\widehat{QPR}=12^0+42^0=54^0\)

Answer: C.