# Isosceles trapezium

Calculate the area of an isosceles trapezium ABCD if a = 10cm, b = 5cm, c = 4cm.

Result

S =  28 cm2

#### Solution:

$a=10 \ \text{cm} \ \\ b=5 \ \text{cm} \ \\ c=4 \ \text{cm} \ \\ \ \\ x=(a-c)/2=(10-4)/2=3 \ \text{cm} \ \\ h=\sqrt{ b^2-x^2 }=\sqrt{ 5^2-3^2 }=4 \ \text{cm} \ \\ \ \\ S=\dfrac{ a+c }{ 2 } \cdot \ h=\dfrac{ 10+4 }{ 2 } \cdot \ 4=28 \ \text{cm}^2$

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