Question

Caluculate the length of AC to 1 decimal place in the trapezium below.

Answer 1

Check the picture below.

usign the pythagorean theorem let's find the side CD, then let's get the side AC using the same pythagorean threorem.

`\textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies √(c^2 - b^2)=a \qquad \begin{cases} c=\stackrel{hypotenuse}{16}\\ a=\stackrel{adjacent}{CD}\\ b=\stackrel{opposite}{7}\\ \end{cases} \\\\\\ √(16^2 - 7^2)=CD\implies √(207)=CD \\\\[-0.35em] ~\dotfill`

`c^2=a^2+b^2\implies c=√(a^2 + b^2) \qquad \begin{cases} c=\stackrel{hypotenuse}{AC}\\ a=\stackrel{adjacent}{CD}\\ b=\stackrel{opposite}{11}\\ \end{cases} \\\\\\ AC=\sqrt{(√(207))^2~~ + ~~11^2}\implies AC=√(207 + 121)\implies \boxed{AC\approx 18.1}`