What?? Im lost...Special triangles...

[ToOtSiE_PoP]

Alright..Im dealing with special triangles...45-45-90 and 30-60-90...and Ive got several ?s on my homework that are confusing me...
This is hard to explain..so if you dont understand my description...dont bother..lol..Ill try to figure it out...
...Find the area....
but I have a triangle, and its equiangular...and it is bisected...so there is a dashed line done the middle..both bottom <s are 60 degrees...then there is a right angle box on the dashed line, and the heighth of that dashed line is 8 root 3 cm..I figured out that the base of the triangle is 8 cm and the hypotenuse is 16...So I took half of 8 times 8 root 3..and got 32 root 3...checked my answer in the back of the book, and the answer is 64 root 3...How? Since your finding the area of a triangle..isnt it A=1/2(b) *h?? Im lost...

? #2...Find the area...I am given a parallelogram...the sides are all 4 yds...then a bottom angle is 45 degrees...I dont even know where to start...any hints...? Do I make a diagonal..to create a triangle..but how will that work> Im confused!
Any help is appreciated!!

 

[stapel]

1) Your method is spot on, but if you're finding the area of the original triangle (before you worked with the half-triangles formed by the dotted line), is the base length "8", or something else...?

Correct that value, and you'll get the book's answer.

2) Cut off a right triangle, a 45-45-90 triangle, from one of the corners. The hypotenuse is "4", right? Now think about your basic reference 45-45-90 triangle. What would you divide by, to go from the hypotenuse length (maybe sqrt[2]?) to the side length (of 1?). Divide by the same value to get the height of the parallelogram.

Then recall that the area of a parallelogram is found by multiplying the height (which you've just found) by the base (which is one of the known sides).

I know what you mean about this stuff being kind of hard to explain, if somebody doesn't already have a pretty good idea of what's going on. If my reference triangle (especially for the 45-45-90) uses different values from yours, or you otherwise get stuck, please reply with corrections or clarifications. I'm pretty sure you're on the right track, but it's easy to get bogged down on these.

Thank you.

Eliz.

 

[soroban]

Hello, ToOtSiE_PoP!

Eliz is absolutely correct . . . Did you take her hints?

             *
            /|\
           / | \
          /  |  \16
         /   |  _\
        /    |8√3 \
       /60°  |     \
      *------+------*
                8

See? . . . The base of the triangle is \(\displaystyle 16\).

           *-------- *
          /|        /
       4 / |       /
        /  |h     /
       /45°|     /
      *----+----*
            4

The area of a parallelgram is "Base times Height".
. . We know the base is 4.
Can you find the height \(\displaystyle h\) in that right triangle?