Question

refer the attachment carefully
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Answer 1

$ABCD \: be \: the \: trapezium. \\ AB||CD$

$Area \: of \: trapezium:$

$\: \: \: \: = \frac{1}{2 } \times \: \binom{sum \: of \: parallel}{sides} \times height$

$∴ ABCD= \frac{1}{2} ×(AB+CD)×AM \\ </p><p>33= \frac{1}{2} (x+2x+1)×(x-4)$

$66 = 3x²-12x+x-4$

$3x²-11x - 70=0$

$3x²-21x + 10x - 70=0$

$3x(x-7)+ 10(x-7)= 0$

$</p><p>∴(3x+10) (x-7)=0$

By using the property, if the proud of two numbers is zero, then at least one of them is zero, we get

$(3x + 10) = 0 or (x-7)=0$

$x = \frac{ - 10}{3} \: or \: x = 7$

$But, \: length \: is \: never \: negative.$

$x ≠ \: \frac{ - 10}{3}$

$x \: = 7$

$AB => x=7cm$

$CD=2x+1=2(7)+1=15 cm$

$AD = BC = x-2=7-2=5 cm$