Question

right triangle test 1 k
simplify the radical. show your work for full credit. (4 pts each)

1. $6\sqrt{192} = $
2. $\sqrt{\frac{28}{121}} = $

find the missing sides. leave answers that are not integers in radical form. (5 pts. each)
3
jg = ______
x = ______

1. the bottom of a ladder is placed 12 m away from the bottom of a wall. the ladder is leaned against the wall so that the angle made by the wall and ladder is $30^\circ$. how far above the ground is the top of the ladder (where it meets the wall)? draw and label a picture for full credit. (6 pts)

find the missing sides. leave answers that are not integers in radical form. (5 pts. each)
6
ab =
ac =
7.
de =
fd =

Explanation:

Step1: Simplify $\sqrt{192}$

$192=64\times3$, so $\sqrt{192}=\sqrt{64\times3}=8\sqrt{3}$

Step2: Multiply by 6

$6\times8\sqrt{3}=48\sqrt{3}$

Step3: Simplify $\sqrt{\frac{28}{121}}$

$\sqrt{\frac{28}{121}}=\frac{\sqrt{28}}{\sqrt{121}}=\frac{2\sqrt{7}}{11}$

Step4: Use Pythagorean theorem for $\triangle DJG$

$JG=\sqrt{DG^2-DJ^2}=\sqrt{16^2-(\sqrt{6})^2}=\sqrt{256-6}=\sqrt{250}=5\sqrt{10}$

Step5: Let height = $h$, use geometric mean

For the triangle, $h=\sqrt{6\times12}=\sqrt{72}=6\sqrt{2}$. Then $x=\sqrt{12^2+(6\sqrt{2})^2}=\sqrt{144+72}=\sqrt{216}=6\sqrt{6}$

Step6: Use 30-60-90 triangle properties

Let height = $h$, ladder length = $2h$. By Pythagoras: $(2h)^2=h^2+12^2$
$4h^2-h^2=144$
$3h^2=144$
$h^2=48$
$h=4\sqrt{3}$

Step7: Solve $\triangle ABC$ (30-60-90)

$AB=12\sqrt{3}\times\tan60^\circ=12\sqrt{3}\times\sqrt{3}=36$
$AC=\frac{12\sqrt{3}}{\cos60^\circ}=\frac{12\sqrt{3}}{\frac{1}{2}}=24\sqrt{3}$

Step8: Solve $\triangle DEF$ (45-45-90)

$DE=FE=9$ (legs equal in isosceles right triangle)
$FD=9\sqrt{2}$ (hypotenuse = leg$\times\sqrt{2}$)

Answer:

1. $\boldsymbol{48\sqrt{3}}$
2. $\boldsymbol{\frac{2\sqrt{7}}{11}}$
3. $\boldsymbol{5\sqrt{10}}$
4. $\boldsymbol{6\sqrt{6}}$
5. $\boldsymbol{4\sqrt{3}}$ m
6. $AB=\boldsymbol{36}$, $AC=\boldsymbol{24\sqrt{3}}$
7. $DE=\boldsymbol{9}$, $FD=\boldsymbol{9\sqrt{2}}$