In figure, $A B C D$ is a trapezium in which $A B = 7 c m, A D = B C = 5 c m, D C = x c m$ , and distance between $A B$ and $D C$ is $4 c m$ . Find the value of $x$ and area of trapezium $A B C D$ .

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Solution

Given: $A B C D$ is a trapezium where $A B = 7 c m, A D = B C = 5 c m, D C = x c m$ and distance between $A B$ and $D C = 4 c m$

Since $A L$ and $B M$ are perpendiculars on $D C$ , then $A L = B M = 4 c m$ and $L M = 7 c m$ .

In $Δ B M C$ ,

Using Pythagoras theorem,

$(B C)^{2} = (B M)^{2} + (M C)^{2}$

$(5)^{2} = (4)^{2} + (M C)^{2}$

$(M C)^{2} = 25 - 16 = 9$

$M C = 3 c m$

Similarly, $D L = 3 c m$

$∴ x = D C = D L + L M + M C = 3 + 7 + 3$

$\Rightarrow x = 13 c m$

Now, Area of trapezium $A B C D = \frac{1}{2} (A B + C D) \times A L$

$= \frac{1}{2} (7 + 13) \times 4$

$= 40$

Hence, area of trapezium $A B C D$ is $40 c m^{2}$ .

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