Question #bd22c

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Answer 1 · 2017-04-06T00:40:06

This can be solved by using Pythagorean theorem

To find x you must find DC and to do this you can use Pythagorean theorem

$a^2 + b^2 = c^2$
$2.25^2 + b^2 = 3^2$
$5.0625+ b^2 = 9$
$5.0625 - 5.0625+ b^2 = 9-5.0625$
$b^2=3.9375$
$sqrtb^2=sqrt3.9375$
$b = 1.984$

1.984 is the value of Dc

Same thing for X

$a^2 + b^2 = c^2$
$1.984^2 + b^2 = 4^2$
$3.9375 + b^2 = 16$
$3.9375 - 3.9375 + b^2 = 16 -3.9375$
$b^2 = 12.064$
$sqrtb^2 = sqrt12.064$
$b = 3.473$

3.473 is the value of x

Answer 2 · 2017-04-06T00:40:21

Assuming I understand the intent of the diagram:
$color(white)("XXX")x=3$

If the diagram is intended to show that $/_ACD =/_DCB$
then by the Angle Bisector Theorem
$color(white)("XXX")abs(AD)/abs(AC)=abs(DB)/abs(CB)$

Replacing with values shown on the diagram:
$color(white)("XXX")(2.25)/3=x/4$

$color(white)("XXX")rarr 9/3=x$

$color(white)("XXX")rarr x=3$