geometry

In quadrilateral ABCD, AB = 5, BC = 8 and CD = 20 units.

Angle B and angle C are both right angles.
What is the length of segment AD?

\(\text{Let $AD = x$}\)

\(\begin{array}{|rcll|} \hline AC^2 &=& 5^2+8^2 \\ AC^2 &=& 25 + 64 \\ \mathbf{AC^2} &=& \mathbf{89} \\ \hline \end{array} \begin{array}{|rcll|} \hline BD^2 &=& 20^2+8^2 \\ BD^2 &=& 400 + 64 \\ \mathbf{BD^2} &=& \mathbf{464} \\ \hline \end{array}\)

cos-rule:

\(\begin{array}{|rcll|} \hline AC^2 &=& 20^2+x^2-2*20x\cos(D) \\ AC^2 &=& 400+x^2-40x\cos(D) \\ 40x\cos(D) &=& 400+x^2 - AC^2 \qquad (1) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline BD^2 &=& 5^2+x^2-2*5x\cos(A) \quad | \quad A+D=180^\circ \quad \text{or} \quad A=180^\circ-D \\ BD^2 &=& 25+x^2-10x\cos(180^\circ-D) \quad | \quad \cos(180^\circ-D)=-\cos(D) \\ BD^2 &=& 25+x^2+10x\cos(D) \\ 10x\cos(D) &=& BD^2 - 25-x^2 \qquad (2) \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline \dfrac{(1)}{(2)}: & \dfrac{40x\cos(D)}{10x\cos(D)} &=& \dfrac{ 400+x^2 - AC^2}{BD^2 - 25-x^2 } \\\\ & 4 &=& \dfrac{ 400+x^2 - AC^2}{BD^2 - 25-x^2 } \\\\ & 4(BD^2 - 25-x^2) &=& 400+x^2 - AC^2 \quad | \quad BD^2=464,\ AC^2=89 \\\\ & 4(464 - 25-x^2) &=& 400+x^2 - 89 \\\\ & 4(439-x^2) &=& 311+x^2 \\\\ & 1756-4x^2 &=& 311+x^2 \\ & 1756-311 &=& 5x^2 \\ & 1445 &=& 5x^2 \\ & 5x^2 &=& 1445 \quad | \quad :5 \\ & x^2 &=& 289 \\ & \mathbf{ x } &=& \mathbf{ 17} \\ \hline \end{array}\)

The length of segment AD is 17