Question 1202110
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https://drive.google.com/file/d/1zsj-S2V9xnu6WYPC88f8xYPvpK_pVFyO/view?usp=sharing

What is the value of x?

Use the rules of special right triangles to find x.

The other person apparently has no clue how to do this problem. Plus, his RIDICULOUS answer is WRONG!!


In right ΔABC, &#8737CAB = <font color = red><font size = 4><b>45<sup>o</sup></font></font></b>, which makes this triangle a special 45-45-90 right-triangle. Its hypotenuse also measures {{{matrix(1,2, 6sqrt(2), units)}}}. 

In any 45-45-90 right Δ, the measure of each congruent leg is the right Δ's {{{Hypotenuse/sqrt(2)}}}. 
In this case, each congruent leg, AB and BC measures {{{matrix(1,6, Hypotenuse/sqrt(2), "=", 6sqrt(2)/sqrt(2), "=", 6, units)}}} 


In right ΔBCD, &#8737CBD = <font color = red><font size = 4><b>60<sup>o</sup></font></font></b>, which makes &#8737BCD = <font color = red><font size = 4><b>90 - 60 = 30<sup>o</sup></font></font></b>. This means that ΔBCD is a special 30-60-90 right-triangle.
It's also seen, from above, that its hypotenuse, BC, measures <font color = red><font size = 4><b>6 units</font></font></b>

Note that side BD is the shorter of ΔBCD's 2 legs, as it's OPPOSITE the smaller of the 2 acute angles (30<sup>o</sup> & 60<sup>o</sup>).

In any 30-60-90 right Δ, the length of the shorter leg, knowing hypotenuse, is the {{{Hypotenuse/2}}}.
In this case, x (or BD), the shorter leg of right ΔBCD measures {{{highlight_green(matrix(1,8, Hypotenuse/2, "=", BC/2, "=", 6/2, "=", highlight(3), units))}}}</pre>