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A quadrilateral $ABCD$ with $AB = 10$ , $BC = 6$ , $CD = 8$ and $DA = 2$ has diagonals $AC$ and $BD$ meeting at $O$, such that $\angle COB = 45^\circ$. Find the area of $ABCD$.

What I Tried: Here is a picture :-

The only thing missing to find the area is one of the diagonals, and I can find the area easily by Heron's Formula. Also I thought that since some angles are there, I might have to use trigonometry, and I tried but without success.

We have that $\angle DOC = \angle AOB = 135^\circ$ . I know that the area of a triangle can be found from the formula :- $$\rightarrow \frac{1}{2}ab\sin \gamma$$

Where $\gamma$ is the angle between $a$ and $b$, but I cannot use it sadly as I don't know any of the lengths of $AO$ , $BO$ , $CO$ and $DO$. Next I thought of the formula :- $$\rightarrow \frac{a}{\sin a} = \frac{b}{\sin b} = \frac{c}{\sin c}$$

But this dosen't work here either due to the same reason.

I think that Law of Cosines might work, but I am actually new to Trigonometry and not sure if it will really work or not, and then I have to consider $4$ equations for $4$ sides, which will be complicated.

Can anyone help me? Thank You.

Note:- Solutions without Trigonometry and mostly welcome. I hate using Trigonometry.

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Here is a solution without trigonometry. Drop altitudes DE and BF to the base AC, with E and F as the feet. Then, the area is

\begin{align} Area&= \frac12 AC \cdot ( DE + BF)\\ & =\frac12(AO \cdot ED +CO\cdot ED+AO\cdot BF + CO\cdot BF)\tag1 \end{align}

Per ED = EO and the Pythagorean’s theorem \begin{align} 2AO\cdot ED &= -(AO -EO)^2 +AO^2 +ED^2\\ &=-AE^2 +AO^2 +ED^2\\ &= -AD^2 +AO^2+ 2ED^2 \end{align} Similarly $$ 2 CO\cdot ED= CD^2 -CO^2-2ED^2$$ $$2AO\cdot BF = AB^2 -AO^2- 2BF^2 $$ $$ 2 CO\cdot BF= -BC^2 +CO^2+2BF^2$$ Then, substitute above into (1) to obtain the area $$Area= \frac{-AD^2+AB^2-BC^2+CD^2}4$$

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  • $\begingroup$ Nice! Same route but in OP's preference. +1 $\endgroup$
    – cosmo5
    Dec 7 '20 at 17:42
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Hint :

$$[ABCD] = \dfrac{1}{2} AC \cdot BD \cdot \sin 45$$

$$[BAD] = \dfrac{1}{2} \cdot OA \sin 45 \cdot BD$$ $$[BCD] = \dfrac{1}{2} \cdot OC \sin 45 \cdot BD$$

Hint 2 : Use cosine-rule in all four small triangles and $$\cos (180 - \theta) = -\cos \theta$$.

$$OD^2+OA^2-2\cdot OD\cdot OA\cos 45 = AD^2$$ $$OA^2+OB^2+2\cdot OA\cdot OB\cos 45 = AB^2$$ $$OB^2+OC^2-2\cdot OB\cdot OC\cos 45 = BC^2$$ $$OC^2+OD^2+2\cdot OC\cdot OD\cos 45 = CD^2$$

Obtain

$$4[ABCD] = -AD^2+AB^2-BC^2+CD^2$$

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  • 2
    $\begingroup$ Taken trigonometry? I am in grade $8$ now, just a teenager :P . $\endgroup$
    – Anonymous
    Dec 7 '20 at 16:20
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    $\begingroup$ Nice solution! and @Anonymous, you are in class 8? wow! Also which book are you doing these geo problems from ? $\endgroup$ Dec 7 '20 at 16:24
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    $\begingroup$ @Anonymous, your teacher has wonderful collection of problems! And I commend you for being passionate in solving them! Maybe you can ask him to give your brief intro to trigo? :) $\endgroup$
    – cosmo5
    Dec 7 '20 at 16:25
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    $\begingroup$ @cosmo5 thanks for the compliment !! and yes, I already have a brief intro to trig, I just need to practice right now to strengthen it, I feel a little short in cases where to use trig and where not to. $\endgroup$
    – Anonymous
    Dec 7 '20 at 16:27
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    $\begingroup$ Yes. Fixed, thanks! $\endgroup$
    – cosmo5
    Dec 7 '20 at 16:34

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