Mathematics: In the given figure ABCD is a tripepizeum with AB||CD. If AO = x-1, CO = BO = x+1 and CD = x+4. Find the value of x.

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In the given figure ABCD is a tripepizeum with

angiecamachoac1728

30.10.2021

$\longmapsto$ The value of "x" is 5.

$\large\underline{\sf{Solution-}}$

Given that,

A trapezium ABCD in which AB || CD such that

AO = x - 1CO = BO = x + 1OD = x + 4

Now,

$\rm In \: \triangle \: AOB \: and \: \triangle \: COD$

$\rm \: \angle \: AOB \: and \: \angle \: COD \: \: \{vertically \: opposite \: angles \}$

$\rm \: \angle \:ABO \: and \: \angle \: CDO \: \: \{alternate \: interior \: angles \}$

$\bf \: \triangle \: AOB \: \sim \: \triangle \: COD \: \: \: \{AA \: similarity \}$

$\bf\longmapsto\:\dfrac{AO}{CO} = \dfrac{BO}{DO}$

$\rm \longmapsto\:\dfrac{x - 1}{x + 1} = \dfrac{x + 1}{x + 4}$

$\rm \longmapsto\:(x - 1)(x + 4) = {(x + 1)}^{2}$

$\rm \longmapsto\: {x}^{2} - x + 4x - 4 = {x}^{2} + 1 + 2x$

$\rm \longmapsto\: 3x - 4 = 1 + 2x$

$\rm \longmapsto\: 3x - 2x= 1 +4$

$\bf\longmapsto \:x = 5$

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