In the given figure,AB = 12 , cm,CD = 8 , cm, | vedclass

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(A) Let $\angle BAE = 	heta$. Since $	riangle ABE$ is a right-angled triangle at $B$,$\angle AEB = 90^{\circ} - 	heta$.Given $\angle AEC = 90^{\cir

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$x$ has two possible values whose difference is $4$.

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(A) Let $\angle BAE = 	heta$. Since $	riangle ABE$ is a right-angled triangle at $B$,$\angle AEB = 90^{\circ} - 	heta$.Given $\angle AEC = 90^{\circ}$,we have $\angle CED = 180^{\circ} - (90^{\circ} - 	heta) - 90^{\circ} = 	heta$.In $	riangle CDE$,$\angle CED = 	heta$ and $\angle CDE = 90^{\circ}$,so $\angle ECD = 90^{\circ} - 	heta$.Thus,$	riangle ABE \sim 	riangle ECD$ by $AA$ similarity.Therefore,$\frac{AB}{BE} = \frac{ED}{CD}$.Given $AB = 12$,$CD = 8$,$BD = 20$,and $BE = x$,we have $ED = 20 - x$.Substituting these values: $\frac{12}{x} = \frac{20 - x}{8}$.$96 = 20x - x^2$.$x^2 - 20x + 96 = 0$.$(x - 12)(x - 8) = 0$.So,$x = 8$ or $x = 12$.The difference between the two values is $|12 - 8| = 4$.

In the given figure,$AB = 12 \, cm$,$CD = 8 \, cm$,$BD = 20 \, cm$,and $\angle ABD = \angle AEC = \angle EDC = 90^{\circ}$. If $BE = x$,then:

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