. In the given figure, AD= BC = 5 cm, AB = 7 cm.
The parallel sides AB, DC are 4 cm apart. DC=r
cm. Find x and the area of the trapezium ABCD
Answers
Answer:
Given:
ABCD is a trapezium in which AB = 7 cm,
AD = BC = 5 cm ,
DC = x cm,
AB // DC ,
Distance between AB and DC is 4 cm
To find :
Value of x = ?
solution :
In ∆BFC , <BFC = 90°,
BC² = BF² + FC² ( Phythagorean theorem )
=> 5² = 4² + FC²
=> FC² = 5² - 4²
=> FC² = 25 - 16 = 9
=> FC = √9 = √3² = 3 cm
=> DC = DE + EF + FC
=> DC = 3 + 7 + 3 = 13 cm
\begin{gathered} Area \: of \: the \: trapezium\\ =\frac{1}{2} [(sum \:of \: the \: lengths \: of \: parallel \:sides )\times distance \: between \: them ]\end{gathered}
Areaofthetrapezium
=
2
1
[(sumofthelengthsofparallelsides)×distancebetweenthem]
\begin{gathered} = \frac{(AB+DC)h}{2}\\= \frac{(7+13)\times 4}{2}\\= \frac{ 20 \times 4}{2}\\= 10\times 4 \\= 40\:cm \end{gathered}
=
2
(AB+DC)h
=
2
(7+13)×4
=
2
20×4
=10×4
=40cm
Therefore.,
\red {Area \: of \: the \: trapezium}\green {=40\:cm }Areaofthetrapezium=40cm
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