The parallel side of a Trapezium are 25 and 13 cm its non parallel sides are equal each being begin 10 cm find the area of trapezium
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Step-by-step explanation:
Given:
ABCD is a isosceles trapezium.
Two parallel bases AB=25cm and CD = 13 cm
Equal sides = AD= BC = 10cm
Let Distance between two parallel sides = DE = CF = h cm
Let AE = FB = x cm
AB = 25cm
=> AE+EF+FB=25cm
=> x + DC + x = 25cm
=> 2x + 13cm = 25 cm
=> 2x = 25 cm - 13 cm
=> 2x = 12 cm
=> x = 12/2
=> x = 6 cm
Now ,
i) In ∆AED ,
<AED = 90°
AD = 10cm ,
BD = h cm,
AE = x cm = 6 cm
By Pythagoras theorem:
DE² = AD²-AE²
=> h² = 10²-x²
=> h² = 10² - 6²
=> h² = 100 - 36
=> h² = 64
=> h = √64
=> h = 8 cm
We \: know \: that \\</p><p>\boxed {Area \: of \: a \: trapezium(A) = \frac{(AB+CD)\times h }{2}}
\implies A = \frac{(25cm+13cm)\times 8cm}{2}\\=38cm \times 4cm\\=152 \: cm^{2}
Therefore,
Area of ABCD trapezium = 152 cm²
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