Rectangular trapezoid

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

Correct result:

S =  9.84 cm²

Solution:

$$\begin{gathered} \mathrm{\mid }AB\mathrm{\mid }=\sqrt{3^2+4^2}=5 \\ {S}_1=S_2 \\ 3\cdot 4\mathrm{/}2=5h\mathrm{/}2 \\ h=3\cdot 4\mathrm{/}5=2.4cm=\mathrm{\mid }AD\mathrm{\mid }=\mathrm{\mid }CX\mathrm{\mid } \\ \mathrm{\mid }BX{\mathrm{\mid }}^2=3^2-h^2 \\ \mathrm{\mid }BX\mathrm{\mid }=\sqrt{9-2.4^2}=1.8cm \\ \mathrm{\mid }CD\mathrm{\mid }=\mathrm{\mid }AB\mathrm{\mid }-\mathrm{\mid }BX\mathrm{\mid }=5-1.8=3.2cm \\ S=\mathrm{\mid }CD\mathrm{\mid }\cdot h+h\cdot \mathrm{\mid }BX\mathrm{\mid }\mathrm{/}2=9.84{\text{cm}}^2 \end{gathered}$$

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