The trapezium

The trapezium is formed by cutting the top of the right-angled isosceles triangle. The base of the trapezium is 10 cm and the top is 5 cm. Find the area of trapezium.

Correct result:

A =  18.75 cm²

Solution:

$$\begin{gathered} b_1=10\text{ cm } \\ {b}_2=5\text{ cm } \\ {a}_1^2+a_1^2=b_1^2 \\ {a}_2^2+a_2^2=b_2^2 \\ {a}_1=b_1\mathrm{/}\sqrt{2}=10\mathrm{/}\sqrt{2}=5\sqrt{2}\text{ cm}\doteq 7.0711\text{ cm } \\ {a}_2=b_2\mathrm{/}\sqrt{2}=5\mathrm{/}\sqrt{2}\doteq 3.5355\text{ cm } \\ {A}_1={\displaystyle \frac{a_1^2}{2}}={\displaystyle \frac{7.0711^2}{2}}=25{\text{cm}}^2 \\ {A}_2={\displaystyle \frac{a_2^2}{2}}={\displaystyle \frac{3.5355^2}{2}}={\displaystyle \frac{25}{4}}=6.25{\text{cm}}^2 \\ A=A_1-A_2=25-6.25={\displaystyle \frac{75}{4}}={\displaystyle \frac{75}{4}}{\text{cm}}^2=18.75{\text{cm}}^2 \end{gathered}$$

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