# Thread: Parallelogram's area, not sufficient information

1. ## Parallelogram's area, not sufficient information

Hi Forum.
I have found a test with some strange questions, mainly lacking information.
So this is one of them

The diagonals of a parallelogram are 16 and 20; and one side of the parallelogram is 12. Find the area of the parallelogram.

Now, the thing is, we can calculate all of the measures.
using $2x^2+2(12)^2=16^2+20^2$

But, how are we going to use it?

The proposed solution to this is to separate the parallelogram into 4 triangles.
2 types of triangles here, right?
We could use Heron's Formula in each triangle and then multiply it by 2.
The thing is, that this is an exam question, where time is limited, things can require a lot of work this way.

If we would have 4 equal triangles we would have a rhombus, wouldn't we?
But that is exactly what the solution does.
So, this probably doesn't have enough information,

Either that, or I am not used to the "side" word here.
Parallelograms and rectangles have different sides, right?
That wouldn't mean that they are equal.

Can someone check if the way I am thinking is right?
Thanks a lot.

2. ## Re: Parallelogram's area, not sufficient information

Hello, Zellator!

If we are allowed to use Trig, I have a solution.

The diagonals of a parallelogram are 16 and 20.
One side of the parallelogram is 12.
Find the area of the parallelogram.
Code:
             D                         C
o  *  *  *  *  *  *  *  o
*  *                 *  *
*     * 8         *     *
*        *  E  * 10     *
*           ♥           *
*    10  *     *        *
*     *         8 *     *
*  *                 *  *
o  *  *  *  *  *  *  *  o
A           12            B
We have parallelogram $ABCD$
. . with diagonals $BD = 16,\:AC = 20$
. . which are bisected at $E.$

Consider triangle $EAB.$

Code:
                      E
o
10    *   : *  8
*       :h  *
* @         :     *
A o * * * * * * * * * * * o B
12
Let $h$ be the altitude from $E.$
Let $\theta = \angle EAB.$

$\text{Law of Cosines: }\:\cos\theta \:=\:\frac{10^2 + 12^2 - 8^2}{2\!\cdot\!10\!\cdot\!12} \:=\:\frac{3}{4}$

$\sin^2\!\theta + \cos^2\!\theta \:=\:1 \quad\Rightarrow\quad \sin^2\!\theta + \left(\tfrac{3}{4}\right)^2 \:=\:1$

. . $\sin^2\!\theta \:=\:\tfrac{7}{16} \quad\Rightarrow\quad \sin\theta \:=\:\tfrac{\sqrt{7}}{4}$

$\sin\theta \:=\:\frac{h}{10} \quad\Rightarrow\quad h \:=\:10\sin\theta \:=\:10\left(\tfrac{\sqrt{7}}{4}\right) \:=\:\tfrac{5\sqrt{7}}{2}$

$\text{Hence, the height of the parallelogram is: }\:H \:=\:2h \:=\:2\left(\tfrac{5\sqrt{7}}{2}\right) \:=\:5\sqrt{7}$

$\text{The base of the parallelogram is: }\:B \:=\:12$

$\text{Therefore, its area is: }\:BH \:=\:(12)(5\sqrt{7}) \:=\:60\sqrt{7}$

3. ## Re: Parallelogram's area, not sufficient information

Wow! Impressive Soroban!
That was great.

The way you used 12 as a base was great, since there are a lot of congruent triangles there it wouldn't matter if it would be the base or the side.
I'm still kind of bugged by the solution that gave 4 congruent triangles and just multiplied the area of one by 4, though.

Using E with a heart was a nice touch too! hahaha
That was really appreciated! Thanks for your help!

4. ## Re: Parallelogram's area, not sufficient information

If one side is 12, then we have 2 opposite triangles of sides 12, 8 and 10,
whether or not the side of length 12 is the base or left or right side of the parallelogram.

The parallelogram area is 12(height).
The height is twice the perpendicular height of the 12, 10, 8 triangle.
Split this triangle into 2 back-to-back right-angled triangles,
such that the base lengths of these right-angled triangles are x and 12-x.
Let "h" be the height of these.

$x^2+h^2=10^2$

$(12-x)^2+h^2=8^2$

$h^2=100-x^2=64-144+24x-x^2$

$100-x^2=-80+24x-x^2\Rightarrow\ 180=24x$

$x=7.5$

$h^2=10^2-x^2=100-56.25=43.75$

The parallelogram area is

$2h(12)=2\sqrt{43.75}(12)$

5. ## Re: Parallelogram's area, not sufficient information

Hi Archie Meade! How are you?

Thanks for the reply, you guys sure have a great amount of willingless to be helping in the morning! hahaha
That would be the way I would proceed as well, now that I am used to paralellograms. Soroban's solution is great too.

The bad thing is that we would both get a fractioned result, since this is a test question it would be better to used $\frac{15}{2}$ don't you think?
That's a great way of solving it too.

Hey, thanks!

6. ## Re: Parallelogram's area, not sufficient information

Originally Posted by Zellator
Hi Archie Meade! How are you?

Thanks for the reply, you guys sure have a great amount of willingless to be helping in the morning! hahaha
That would be the way I would proceed as well, now that I am used to paralellograms. Soroban's solution is great too.

The bad thing is that we would both get a fractioned result, since this is a test question it would be better to used $\frac{13}{2}$ don't you think?
That's a great way of solving it too.

Hey, thanks!
Yes, it's preferable to work with exact values.

$x=\frac{15}{2}$

$h^2=\frac{400-225}{4}=\frac{175}{4}$

$A=2(12)\sqrt{\frac{175}{4}}$

7. ## Re: Parallelogram's area, not sufficient information

$43.75=43\frac{3}{4}$

8. ## Re: Parallelogram's area, not sufficient information

Haahaha that was one ugly mistake, really sorry about that.
I will look up and not do that again.

Thanks for your replies, this gets easier and easier.
I'm not so familiar with using Also sprach Zarathustra method, I've seen it a couple of times, but I prefer to stick to the old way.
Thanks anyway.

All the best!

9. ## Re: Parallelogram's area, not sufficient information

NO trig required; using Soroban's diagram:
Code:
            D                         C
o  *  *  *  *  *  *  *  o
*  *                 *  *
*     * 8         *     *
*        *  E  * 10     *
*           ♥           *
*    10  *     *        *
*     *         8 *     *
*  *                 *  *
o  *  *  *  *  *  *  *  o
A           12            B
Draw height lines CX and BY ; so CX = BY and BX = CY.
Let h = CX and BY, x = BX and CY

Right triangle ACX:
h^2 + (x + 12)^2 = 20^2 : h^2 = 256 - x^2 - 24x
Right triangle BDY:
h^2 + (12 - x)^2 = 16^2 : h^2 = 112 - x^2 + 24x

So 256 - x^2 - 24x = 112 - x^2 + 24x
48x = 144 : x = 3
Which makes h = SQRT(175) = 5SQRT(7)