# In trapezoid $ABCD$, $AB \parallel CD$ , $AB = 4$ cm and $CD = 10$ cm.

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}

• Oh so I had to use cosine with the $45^\circ$ angle, I get it. Also sorry for the swapped $C$ and $D$. Oct 14 '20 at 4:37
• No problem. The trapezium configuration is interesting though. Other nice problems also possible, just within the trapezium. Oct 14 '20 at 4:40
• @Anonymous You can always change the question so that the diagram matches, which is what I have done with my edit. 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, $$S_{ABCD}=\frac{1}{2}AC\cdot DB=\frac{1}{2}\cdot7x\cdot7y=\frac{140}{3}.$$