Isosceles trapezium

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

Result

S =  28 cm2

Solution:

a=10 cm b=5 cm c=4 cm  x=(a−c)/2=(10−4)/2=3 cm h=b2−x2=52−32=4 cm  S=a+c2⋅ h=10+42⋅ 4=28 cm2a=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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