Question

the parallel sides of a trapezium are 25cm & 13cm & non parallel sides are 15cm each. Find area of trapezium.​

Answer 1

Answer:

see the attachment

Step-by-step explanation:

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

\begin{gathered} We \: know \: that \\ < /p > < p > \boxed {Area \: of \: a \: trapezium(A) = \frac{(AB+CD)\times h }{2}}\end{gathered}Weknowthat</p><p>Areaofatrapezium(A)=2(AB+CD)×h

\begin{gathered}\implies A = \frac{(25cm+13cm)\times 8cm}{2}\\=38cm \times 4cm\\=152 \: cm^{2}\end{gathered}⟹A=2(25cm+13cm)×8cm=38cm×4cm=152cm2

Therefore,

Area of ABCD trapezium = 152 cm²

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