Suppose that ABCD is a trapezoid in which AD||BC.
Given AC is perpendicular to CD, AC bisects angle BAD,
and the area of ABCD is 42,
then compute the area of triangle ACD.
\(\text{Let $AB=a$ } \\ \text{Let $\angle BAC = \angle CAD = \alpha$ } \)
\(\text{Let $\angle ACB = \angle BAC = \alpha \qquad ( AD\parallel BC) $ } \)
sin - rule
\(\begin{array}{|rcll|} \hline \dfrac{ \sin(\angle BAC) } {BC} &=& \dfrac{ \sin(\angle ACB) } {AB} \\ \dfrac{ \sin(\alpha) } {BC} &=& \dfrac{ \sin(\alpha) } {AB} \\ BC &=& AB \quad & | \quad AB=a \\ & \boxed{BC = a} \\ \hline \end{array} \)
\(\text{Let $\angle ABC = 180^{\circ}-2\alpha $}\)
cos - rule
\(\begin{array}{|rcll|} \hline (AC)^2 &=& (BC)^2+(AB)^2-2\cdot BC \cdot AB \cdot \cos(180^{\circ}-2\alpha) \\ (AC)^2 &=& 2a^2+2a^2\cos( 2\alpha) \\ (AC)^2 &=& 2a^2(1+\cos(2\alpha)) \quad | \quad \cos( 2\alpha)=2\cos^2(\alpha)-1 \\ (AC)^2 &=& 2a^2\cdot ( 1+2\cos^2(\alpha)-1 ) \\ (AC)^2 &=& 2a^2\cdot 2\cos^2(\alpha) \\ (AC)^2 &=& 4a^2\cdot \cos^2(\alpha) \\ & & \boxed{ AC= 2a\cos(\alpha)} \quad | \quad \cos( \alpha)= \dfrac{AC}{AD} \\ AC &=& 2a\dfrac{AC}{AD} \\ 1 &=& \dfrac{2a}{AD} \\ & & \boxed{ AD= 2a } \\ \hline \end{array}\)
\(\mathbf{A_{\triangle} = \ ?} \)
\(\begin{array}{|rcll|} \hline A_{ABCD} &=& \left( \dfrac{AD+BC}{2} \right) \cdot h \quad | \quad AD= 2a,\quad BC = a \\ A_{ABCD} &=& \left( \dfrac{2a+a}{2} \right) \cdot h \\ A_{ABCD} &=& \dfrac{3}{2} a \cdot h \quad | \quad A_{ABCD}=42 \\ 42&=& \dfrac{3}{2} a \cdot h \\ a \cdot h &=& \mathbf{ \dfrac{2}{3} \cdot 42 } \\ \mathbf{a \cdot h} &\mathbf{=}& \mathbf{ 28 } \\\\ A_{\triangle} &=& \dfrac{AD\cdot h}{2} \quad | \quad AD= 2a \\ A_{\triangle} &=& \dfrac{2a\cdot h}{2} \\ A_{\triangle} &=& a\cdot h \quad | \quad a\cdot h = 28 \\ \mathbf{A_{\triangle}} &\mathbf{=}& \mathbf{ 28} \\ \hline \end{array}\)
The area of triangle ACD is 28.
+118690 
