Question

Given that ABCD is a rhombus, what is the value of x?

Answer 1

Answer:

Diagonals of a rhombus intersect at right angles

Therefore, $\angle AOD=90$

$Now,in\: \bigtriangleup AOD$

$\angle OAD+\angle AOD+\angle ADO=180$

$x+90+3x+12=180$ $\leftarrow [combing \: like\: terms]$

$4x+102=180\\-102 \: -102$ $\leftarrow [subtracting \:102 ]$

$\frac{4x}{4} =\frac{78}{4} \leftarrow [dividing \: by\: 4]$

$x=19.5$ $\Longleftarrow$

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