Question

The parallel sides of a trapezium are 25 cm and 13 cm ,its non parallel sides are equal,each being 10 cm.find the area of the trapezium.

Answer 1

Area of trapezium = 1/2*(sum of parallel sides )*height

To get the full answer see the image I hope I helped u

Answer 2

Answer:

Area of ABCD trapezium = 152 cm²

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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