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.
Answers
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
\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
=152cm
2
Therefore,
Area of ABCD trapezium = 152 cm²
Step-by-step explanation:
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