[tex]25)\text{ }x\text{ = 4 }\sqrt[]{2}[/tex]
Explanation:
25) The triangle is a right angled traingle
To get the missing side (solve for x), we will apply pythagoras theorem:
Hypotenuse² = opposite² + adjacent²
hypotenuse = longest side or highest value = 5√2
let adjacent = x, opposite = 3√2
substitute the values into the formula above:
[tex]\begin{gathered} (5\sqrt[]{2})^2\text{ = (3}\sqrt[]{2})^2+x^2 \\ 25(2)=9(2)+x^2 \\ 50=18+x^2 \\ 50-18=x^2 \\ 32=x^2 \end{gathered}[/tex][tex]\begin{gathered} \text{square root both sides:} \\ \sqrt[]{32}=\sqrt[]{x^2\text{ }} \\ x\text{ = }\sqrt[]{32}\text{ = }\sqrt[]{16\times2} \\ x\text{ = 4}\sqrt[]{2} \end{gathered}[/tex]
