Question

In triangle $ABC$, if $AC=8,BC=7$ and $D$ lies between $A$ and $B$ such that $AD=2,BD=4$, then the length $CD$ equals

• A. $\sqrt{46}$
• B. $\sqrt{48}$
• C. $\sqrt{51}$
• D. $\sqrt{75}$

Answer 1

The correct option is C $\sqrt{51}$

By using cosine law

$\begin{array}{c}\Rightarrow {\left(CD\right)}^2=A{C}^2+A{D}^2-2AC\cdot AD\cos \angle CAB\\ \Rightarrow {\left(CD\right)}^2=A{C}^2+A{D}^2=2AC\cdot AD\left|\frac{A{C}^2+A{B}^2-B{C}^2}{2AC\cdot AB}\right|\\ \Rightarrow C{D}^2=8^2+2^2-\left(\frac{8^2+6^2-7^2}{6}\right)=64+4-2\times \frac{51}{6}=51\\ CD=\sqrt{51}\end{array}$