# 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:

$b_{1}=10 \ \text{cm} \ \\ b_{2}=5 \ \text{cm} \ \\ \ \\ a_{1}^2 + a_{1}^2=b_{1}^2 \ \\ a_{2}^2 + a_{2}^2=b_{2}^2 \ \\ \ \\ a_{1}=b_{1} / \sqrt{ 2 }=10 / \sqrt{ 2 } \doteq 5 \ \sqrt{ 2 } \ \text{cm} \doteq 7.0711 \ \text{cm} \ \\ \ \\ a_{2}=b_{2} / \sqrt{ 2 }=5 / \sqrt{ 2 } \doteq 3.5355 \ \text{cm} \ \\ \ \\ A_{1}=\dfrac{ a_{1}^2 }{ 2 }=\dfrac{ 7.0711^2 }{ 2 }=25 \ \text{cm}^2 \ \\ A_{2}=\dfrac{ a_{2}^2 }{ 2 }=\dfrac{ 3.5355^2 }{ 2 }=\dfrac{ 25 }{ 4 }=6.25 \ \text{cm}^2 \ \\ \ \\ A=A_{1}-A_{2}=25-6.25=\dfrac{ 75 }{ 4 }=18.75 \ \text{cm}^2$

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