1. ## Help me solve this exercise, please? Parallelogram Area

Hello ^^

Please help me solve this. This might appear a little too easy for those of you who are pros but still .. It's 24th ex. and everything that was before and after this turned out quite easy for me. But for some reason I got stuck with this one :/

Since I doubt I can translate the contest of the exercise decently, I scanned what I wrote. I hope you can understand the symbols at least? I mean, isn't maths the international language and all I did start and I did try but kjaskjgkagf

Therefore, I hope you can help me^^~~~

P.S. for Area (A), in my country we use symbol (S). And in case it's not obvious (because damn I'm not good at drafting) enough, it's a parallelogram

2. ## Re: Help me solve this exercise, please? Parallelogram Area

Originally Posted by buujiji
Hello ^^

P.S. for Area (A), in my country we use symbol (S). And in case it's not obvious (because damn I'm not good at drafting) enough, it's a parallelogram
Hello,

I've modified your sketch a little bit:

As you can see the exterior angle at the right lower vertex has the same size as $\alpha$

Therefore.

$h_2 = b \cdot \sin(\alpha) ~\implies~ 4 = b \cdot \frac{13}{17}$

Determine b and consequently the value of the area.

You should come out with $A = 68$

3. ## Re: Help me solve this exercise, please? Parallelogram Area

oh my god! thank you so much! I couldn't see that with the angle. Thank you so much for actually taking time to answer! Thanks a lot! This was a great help!
Hope you'll help me again someday if I need it! Thank you again ^^

P.S. I wont double-post anymore, I promise ^^

4. ## Re: Help me solve this exercise, please? Parallelogram Area

Hello, buujiji!

A slightly different approach . . .

$\text{Parallelgram }PQRS\!:\;PS = a,\:PQ = b$
$\text{Altitude from }Q\text{ to }RS\!:\:h_1 = QA = 13$
$\text{Altitude from }Q\text{ to }PS\!:\:h_2 = QB = 4$
$\alpha = \angle AQB,\;\sin\alpha = \tfrac{13}{17}$

Code:
              : - - - - a - - - - :
Q *-------------------* R
/. α . θ'         θ /
/ .       .         /
/θ'.       13  .    /
b /   .4              .
/    .              / A
/     .             /
/ θ    .            /
P *-------.-----------* S
B
: - - - - a - - - - :
$\text{Find the area of parallelogram }PQRS.$

Let $\theta \,=\,\angle P \,=\,\angle R.$
Then $\angle BQP \,=\, \angle AQR \,=\, 90^o-\theta \,=\,\theta'$

$\angle PQR$ is the supplement of $\angle R.$

. . $\begin{array}{c}\theta' + \alpha + \theta' \;=\;180^o - \theta \\ \\ (90^o-\theta) + \alpha + (90^o - \theta) \;=\;180^o - \theta \\ \\ \alpha \;=\;\theta \end{array}$

In right triangle $QAR\!:\:\sin\theta \,=\,\frac{13}{a} \quad\Rightarrow\quad a \,=\,\frac{13}{\sin\theta}$
. . $a \:=\:\frac{13}{\sin\alpha} \:=\:\frac{13}{\frac{13}{17}} \:=\:17$

Therefore: . $\text{Area} \:=\:a\cdot h_2 \:=\:17\cdot 4 \:=\:68$