Geometry Help

[daniellagirl89]

Hello,
I've been struggling with geometry a lot lately. I'm completely confused on the areas with geometry especially with square roots included. I've posted 3 questions below I am most suck on. It would be helpful to see steps of solutions and answers this way I can figure it out and not just see the answer. Thank you to everyone who helps in advance. I can't even tell you how stuck I am on these.

Find the area of the parallelogram ABCD, if the measure of Angle A = 30 degrees, AB = 8, and AD = 6 square root(2).


Find the area of the parallelogram ABCD, if the measure of Angle A = 60 degrees, AB = 8, and AD = 6.

Find the area of the triangle, if the base is 4 square root (2) and the hypotenuse is 8

 

[soroban]

Hello, daniellagirl89!

Find the area of the parallelogram ABCD
if \(\displaystyle \angle A\,=\,30^o,\:AB\,=\,8,\:AD\,=\,6\sqrt{2}\)

                B * - - - - - - - - * C
                / :               /
              /   :             /
           8/     :h          /
          /       :         /
        / 30°     :       /
    A * - - - - - - - - * D
                6√2

We already know the base of the parallelogram: \(\displaystyle \,b = 6\sqrt{2}\)

At the left, we have a "30-60" right triangle.
. . Hence: \(\displaystyle \,h\,=\,4\)

Therefore, the area is: \(\displaystyle A\:=\:b\cdot h \:=\6\sqrt{2})(4) \:=\:24\sqrt{2}\)

Find the area of the parallelogram ABCD,
if \(\displaystyle \angle A\,=\,60^o,\:AB\,=\,8,\:AD\,=\,6\)

This is similar to the above problem.

Find the area of the right triangle
if the base is \(\displaystyle 4\sqrt{2}\) and the hypotenuse is \(\displaystyle 8\)

      *
      |  *
      |     *
     h|        * 8
      |           *
      |              *
      * - - - - - - - - *
              4√2

We know the area of a triangle: \(\displaystyle \:A \:=\:\frac{1}{2}bh\)

We know the base: \(\displaystyle \,b\,=\,4\sqrt{2}\)

From Pythagorus, we have: \(\displaystyle \,h^2\,+\,(4\sqrt{2})^2\:=\:8^2\;\;\Rightarrow\;\;h^2\,+\,32\:=\:64\)

. . \(\displaystyle h^2\:=\:32\;\;\Rightarrow\;\;h\:=\:\sqrt{32}\:=\:4\sqrt{2}\)

Therefore: \(\displaystyle \:A \:=\:\frac{1}{2}(4\sqrt{2})(4\sqrt{2}) \:=\:\frac{1}{2}(32) \:=\:16\)

 

[daniellagirl89]

Dear Soroban,
Thank you so much for all of the help!
I just have two questions

what does "30-60" triangle mean? and how did you get 4 for the height?

and for my second question
would i begin it the same way as the first or how?

 

[Denis]

Daniella, in case you're confused with the 90-60-30 right triangle:
this is a "nice" special case where the side opposite the 30 degrees
angle is always equal to half the hypotenuse; hence, as Soroban
shows, the hypotenuse = 8, so the side = 4; CLEAR?

 

[daniellagirl89]

Daniella, in case you're confused with the 90-60-30 right triangle:
this is a "nice" special case where the side opposite the 30 degrees
angle is always equal to half the hypotenuse; hence, as Soroban
shows, the hypotenuse = 8, so the side = 4; CLEAR?

Ohhhhh, sorry I didn't understand that before. But thanks for clearing it up! I get it now