Question

in the given figure abcd is a trapizium in which AB =7cM AD=BC=5cm ,DC=Xcm and distance between ab and dc is 4 cm find the valu of x and area of a trapezium ABCD

Answer 1

Pic have the solution

Answer 2

Answer:

$\red {Area \: of \: the \: trapezium}\green {=40\:cm }$

Step-by-step explanation:

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

$Area \: of \: the \: trapezium\\ =\frac{1}{2} [(sum \:of \: the \: lengths \: of \: parallel \:sides )\times distance \: between \: them ]$

$= \frac{(AB+DC)h}{2}\\= \frac{(7+13)\times 4}{2}\\= \frac{ 20 \times 4}{2}\\= 10\times 4 \\= 40\:cm$

Therefore.,

$\red {Area \: of \: the \: trapezium}\green {=40\:cm }$

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