Rectangular trapezoid

Calculate the area of a rectangular trapezoid with a right angle at point A and if |AC| = 4 cm, |BC| = 3 cm, and the diagonal AC is perpendicular to the side BC.

Final Answer:

S =  9.84 cm²

Step-by-step explanation:

$$\begin{gathered} AC=4\text{cm} \\ BC=3\text{cm} \\ AB=\sqrt{B{C}^2+A{C}^2}=\sqrt{3^2+4^2}=5\text{cm} \\ {S}_1=S_2 \\ 3\cdot 4/2=5h/2 \\ h=3\cdot \frac{4}{5}=\frac{3\cdot 4}{5}=\frac{12}{5}=2.4\text{cm} \\ h=|AD|=|CX| \\ B{X}^2=B{C}^2-h^2 \\ BX=\sqrt{B{C}^2-h^2}=\sqrt{3^2-2.4^2}=\frac{9}{5}=1.8\text{cm} \\ CD=AB-BX=5-\frac{9}{5}=\frac{5\cdot 5}{5}-\frac{9}{5}=\frac{25}{5}-\frac{9}{5}=\frac{25-9}{5}=\frac{16}{5}=3.2\text{cm} \\ S=CD\cdot h+\frac{h\cdot BX}{2}=\frac{16}{5}\cdot \frac{12}{5}+\frac{\frac{12}{5}\cdot \frac{9}{5}}{2}=\frac{246}{25}{\text{cm}}^2=9.84{\text{cm}}^2 \end{gathered}$$

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