Area of iso-trap

Find the area of an isosceles trapezoid if the lengths of its bases are 16 cm and 30 cm, and the diagonals are perpendicular to each other.

Correct result:

A =  529 cm²

Solution:

$$\begin{gathered} c=16\text{ cm } \\ a=30\text{ cm } \\ {a}^2=x^2+x^2 \\ x=a\mathrm{/}\sqrt{2}=30\mathrm{/}\sqrt{2}=15\sqrt{2}\text{ cm}\doteq 21.2132\text{ cm } \\ {c}^2=y^2+y^2 \\ y=c\mathrm{/}\sqrt{2}=16\mathrm{/}\sqrt{2}=8\sqrt{2}\text{ cm}\doteq 11.3137\text{ cm } \\ d=x+y=21.2132+11.3137=23\sqrt{2}\text{ cm}\doteq 32.5269\text{ cm } \\ {h}_1=\sqrt{x^2-(a\mathrm{/}2{)}^2}=\sqrt{21.2132^2-(30\mathrm{/}2{)}^2}=15\text{ cm } \\ {h}_2=\sqrt{y^2-(c\mathrm{/}2{)}^2}=\sqrt{11.3137^2-(16\mathrm{/}2{)}^2}=8\text{ cm } \\ h=h_1+h_2=15+8=23\text{ cm } \\ {A}_1={\displaystyle \frac{a+c}{2}}\cdot h={\displaystyle \frac{30+16}{2}}\cdot 23=529{\text{cm}}^2 \\ A={\displaystyle \frac{d\cdot d}{2}}={\displaystyle \frac{32.5269\cdot 32.5269}{2}}=529{\text{cm}}^2 \end{gathered}$$

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