In trapezoid $ABCD$, $AB \parallel CD$ , $AB = 4$ cm and $CD = 10$ cm. Suppose the lines $AD$ and $BC$ intersect at right angles and the lines $AC$ and $BD$ when extended at point $Q$ form an angle of $45^\circ$. Compute the area of $ABCD$.

What I Tried :- Here is the picture :-

Now to find the area of $ABCD$, I just need to find its height, but I cannot find it.

I can see that $\Delta AOB \sim \Delta COD$. So :- $$\frac{AB}{CD} = \frac{AO}{OD} = \frac{BO}{OC} = \frac{2}{5}$$ So I assumed $AO = 2x$ , $BO = 2y$ , $CO = 5y$ , $DO = 5x$.
Now in $\Delta AOB$, by Pythagorean Theorem :-
$AO^2 + OB^2 = AB^2$
$\rightarrow 4x^2 + 4y^2 = 16$
$\rightarrow x^2 + y^2 = 4$

Also $\Delta QAB \sim \Delta QDC$. So:- $$\frac{QA}{AC} = \frac{QB}{BD}$$
I get $AC$ and $BD$ by Pythagorean Theorem again, which gives me :- $$\frac{QA}{\sqrt{4x^2 + 25y^2}} = \frac{QB}{\sqrt{25x^2 + 4y^2}}$$

I don't know how to proceed next, as this result only gives me that $\left(\frac{QA}{OB}\right)^2 = \frac{21y^2 + 16}{21x^2 + 16}$ . Also I couldn't think of any way to use the $45^\circ$ angle, except that I can figure out that the triangle is cyclic.

Can anyone help?


Let $OC = a$, $OD = b$. So $OA=\frac{2}{5}OC$, $OB = \frac{2}{5} OD$.

(Note you have swapped labels $C$ and $D$ in figure)

Also let $AD=3x$, $BC=3y$, so that $QA=2x$, $QB=2y$.

We have $a^2+b^2=100$

By Pythagoras, $$ (OA^2+OD^2)+(OB^2+OC^2)=9(x^2+y^2) $$ $$ \Rightarrow x^2+y^2=116/9 $$

By cosine-rule in $\triangle QAB$, $$ 4x^2+4y^2-4\sqrt{2}xy=4^2 $$ $$\Rightarrow xy=\dfrac{40\sqrt{2}}{9}$$

So $$ \begin{align} [ABCD] &= (1-\dfrac{4}{25})[QDC] \\ &=\dfrac{21}{25}(\frac{1}{2}\cdot QD\cdot QC\cdot\sin 45^{\circ})\\ &=\dfrac{21}{25}(\frac{1}{2}\cdot 5x\cdot5y\cdot\frac{1}{\sqrt{2}}) \\ &=\boxed{\dfrac{140}{3}} \end{align} $$

  • $\begingroup$ Oh so I had to use cosine with the $45^\circ$ angle, I get it. Also sorry for the swapped $C$ and $D$. $\endgroup$
    – Anonymous
    Oct 14 '20 at 4:37
  • $\begingroup$ No problem. The trapezium configuration is interesting though. Other nice problems also possible, just within the trapezium. $\endgroup$
    – cosmo5
    Oct 14 '20 at 4:40
  • 1
    $\begingroup$ @Anonymous You can always change the question so that the diagram matches, which is what I have done with my edit. $\endgroup$
    – Toby Mak
    Oct 14 '20 at 9:54

By your work $$(2x)^2+(2y)^2=4^2,$$ which gives $$x^2+y^2=4.$$ Also, $$AD=\sqrt{DO^2+AO^2}=\sqrt{25y^2+4x^2}$$ and $$BC=\sqrt{25x^2+4y^2}.$$ Now, let $PABC$ be parallelogram.

Thus, $P\in DC$, $AP=BC$, $$DP=DC-PC=10-4=6$$ and $$\measuredangle DAP=\measuredangle Q=45^{\circ}.$$ Thus, by the law of cosines for $\Delta DAP$ we obtain: $$\frac{25y^2+4x^2+25x^2+4y^2-36}{2\sqrt{(25x^2+4y^2)(25y^2+4x^2)}}=\frac{1}{\sqrt2}$$ or $$\frac{29\cdot4-36}{\sqrt2}=\sqrt{(25x^2+4y^2)(25y^2+4x^2)}$$ or $$3200=641x^2y^2+100(x^4+y^4)$$ or $$3200=441x^2y^2+100(x^2+y^2)^2,$$ which gives $$xy=\frac{40}{21}.$$ Id est, $$S_{ABCD}=\frac{1}{2}AC\cdot DB=\frac{1}{2}\cdot7x\cdot7y=\frac{140}{3}.$$


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