AIME 2021 I · 第 9 题

Solution 1 (Similar Triangles and Pythagorean Theorem)

Let $\overline{AE}, \overline{AF},$ and $\overline{AG}$ be the perpendiculars from $A$ to $\overleftrightarrow{BC}, \overleftrightarrow{CD},$ and $\overleftrightarrow{BD},$ respectively. Next, let $H$ be the intersection of $\overline{AF}$ and $\overline{BD}.$

We set $AB=x$ and $AH=y,$ as shown below.

From here, we obtain $HF=18-y$ by segment subtraction, and $BG=\sqrt{x^2-10^2}$ and $HG=\sqrt{y^2-10^2}$ by the Pythagorean Theorem.

Since $\angle ABG$ and $\angle HAG$ are both complementary to $\angle AHB,$ we have $\angle ABG = \angle HAG,$ from which $\triangle ABG \sim \triangle HAG$ by AA. It follows that $\frac{BG}{AG}=\frac{AG}{HG},$ so $BG\cdot HG=AG^2,$ or

$$\sqrt{x^2-10^2}\cdot\sqrt{y^2-10^2}=10^2. \hspace{10mm} (1)$$ Since $\angle AHB = \angle FHD$ by vertical angles, we have $\triangle AHB \sim \triangle FHD$ by AA, with the ratio of similitude $\frac{AH}{FH}=\frac{BA}{DF}.$ It follows that $DF=BA\cdot\frac{FH}{AH}=x\cdot\frac{18-y}{y}.$

Since $\angle EBA = \angle ECD = \angle FDA$ by angle chasing, we have $\triangle EBA \sim \triangle FDA$ by AA, with the ratio of similitude $\frac{EA}{FA}=\frac{BA}{DA}.$ It follows that $DA=BA\cdot\frac{FA}{EA}=x\cdot\frac{18}{15}=\frac{6}{5}x.$

By the Pythagorean Theorem on right $\triangle ADF,$ we have $DF^2+AF^2=AD^2,$ or

$$\left(x\cdot\frac{18-y}{y}\right)^2+18^2=\left(\frac{6}{5}x\right)^2. \hspace{7mm} (2)$$ Solving this system of equations ($(1)$ and $(2)$), we get $x=\frac{45\sqrt2}{4}$ and $y=\frac{90}{7},$ so $AB=x=\frac{45\sqrt2}{4}$ and $CD=AB+2DF=x+2\left(x\cdot\frac{18-y}{y}\right)=\frac{81\sqrt2}{4}.$ Finally, the area of $ABCD$ is

$$K=\frac{AB+CD}{2}\cdot AF=\frac{567\sqrt2}{2},$$ from which $\sqrt2 \cdot K=\boxed{567}.$

~MRENTHUSIASM

Remark

Instead of solving the system of equations $(1)$ and $(2),$ which can be time consuming, by noting that $\triangle ACF \sim \triangle ABG$ by AA, we could find out $\frac{AB}{AG} = \frac{AC}{AF}$, which gives $AC = \frac{9}{5}x$. We also know that $EB = \sqrt{x^2 - 15^2}$ by Pythagorean Theorem on $\triangle ABE$. From $BC = AD = \frac{6}{5}x,$ we apply the Pythagorean Theorem to $\triangle ACE$ and obtain

$$AC^2 = (EB+BC)^2 + AE^2.$$ Substituting, we get

$$\frac{81}{25}x^2 = \left(\sqrt{x^2 -225}+\frac{6}{5}x\right)^2+225 \iff x = 3\sqrt{x^2 - 15^2},$$ from which $x = \frac{45\sqrt{2}}{4}.$

~Chupdogs

Solution 2 (Similar Triangles and Pythagorean Theorem)

First, draw the diagram. Then, notice that since $ABCD$ is isosceles, $\Delta ABD \cong \Delta BAC$, and the length of the altitude from $B$ to $AC$ is also $10$. Let the foot of this altitude be $F$, and let the foot of the altitude from $A$ to $BC$ be denoted as $E$. Then, $\Delta BCF \sim \Delta ACE$. So, $\frac{BC}{AC} = \frac{BF}{AE} = \frac{2}{3}$. Now, notice that $[ABC] = \frac{10 \cdot AC} {2} = \frac{AB \cdot 18}{2} \implies AC = \frac{9 \cdot AB}{5}$, where $[ABC]$ denotes the area of triangle $ABC$. Letting $AB = x$, this equality becomes $AC = \frac{9x}{5}$. Also, from $\frac{BC}{AC} = \frac{2}{3}$, we have $BC = \frac{6x}{5}$. Now, by the Pythagorean theorem on triangles $ABF$ and $CBF$, we have $AF = \sqrt{x^{2}-100}$ and $CF = \sqrt{ \left( \frac{6x}{5} \right) ^{2}-100}$. Notice that $AC = AF + CF$, so $\frac{9x}{5} = \sqrt{x^{2}-100} + \sqrt{ \left( \frac{6x}{5} \right) ^{2}-100}$. Squaring both sides of the equation once, moving $x^{2}-100$ and $\left( \frac{6x}{5} \right) ^{2}-100$ to the right, dividing both sides by $2$, and squaring the equation once more, we are left with $\frac{32x^{4}}{25} = 324x^{2}$. Dividing both sides by $x^{2}$ (since we know $x$ is positive), we are left with $\frac{32x^{2}}{25} = 324$. Solving for $x$ gives us $x = \frac{45}{2\sqrt{2}}$.

Now, let the foot of the perpendicular from $A$ to $CD$ be $G$. Then let $DG = y$. Let the foot of the perpendicular from $B$ to $CD$ be $H$. Then, $CH$ is also equal to $y$. Notice that $ABHG$ is a rectangle, so $GH = x$. Now, we have $CG = GH + CH = x + y$. By the Pythagorean theorem applied to $\Delta AGC$, we have $(x+y)^{2}+18^{2}= \left( \frac{9x}{5} \right) ^{2}$. We know that $\frac{9x}{5} = \frac{9}{5} \cdot \frac{45}{2\sqrt{2}} = \frac{81}{2\sqrt{2}}$, so we can plug this into this equation. Solving for $x+y$, we get $x+y=\frac{63}{2\sqrt{2}}$.

Finally, to find $[ABCD]$, we use the formula for the area of a trapezoid: $K = [ABCD] = \frac{b_{1}+b_{2}}{2} \cdot h = \frac{AB+CD}{2} \cdot 18 = \frac{x+(CG+DG)}{2} \cdot 18 = \frac{2x+2y}{2} \cdot 18 = (x+y) \cdot 18 = \frac{63}{2\sqrt{2}} \cdot 18 = \frac{567}{\sqrt{2}}$. The problem asks us for $K \cdot \sqrt{2}$, which comes out to be $\boxed{567}$.

~advanture

Solution 3 (Similar Triangles and Pythagorean Theorem)

Make $AE$ perpendicular to $BC$; $AG$ perpendicular to $BD$; $AF$ perpendicular $DC$.

It's obvious that $\triangle{AEB} \sim \triangle{AFD}$. Let $EB=5x; AB=5y; DF=6x; AD=6y$. Then make $BQ$ perpendicular to $DC$, it's easy to get $BQ=18$.

Since $AB$ parallel to $DC$, $\angle{ABG}=\angle{BDQ}$, so $\triangle{ABG} \sim \triangle{BDQ}$. After drawing the altitude, it's obvious that $FQ=AB=5y$, so $DQ=5y+6x$. According to the property of similar triangles, $AG/BQ=BG/DQ$. So, $\frac{5}{9}=\frac{GB}{(6x+5y)}$, or $GB=\frac{(30x+25y)}{9}$.

Now, we see the $\triangle AEB$, pretty easy to find that $15^2+(5x)^2=(5y)^2$, then we get $x^2+9=y^2$, then express $y$ into $x$ form that $y=\sqrt{x^2+9}$ we put the length of $BG$ back to $\triangle AGB$: $BG^2+100=AB^2$. So,

$$\frac{[30x+25\sqrt{(x^2+9)}]^2}{81}+100=(5\sqrt{x^2+9})^2.$$ After calculating, we can have a final equation of $x^2+9=\sqrt{x^2+9}\cdot3x$. It's easy to find $x=\frac{3\sqrt{2}}{4}$ then $y=\frac{9\sqrt{2}}{4}$. So,

$$\sqrt{2}\cdot K = \sqrt{2}\cdot(5y+5y+6x+6x)\cdot9=\boxed{567}.$$ ~bluesoul

Solution 4 (Cyclic Quadrilaterals, Similar Triangles, Pythagorean Theorem)

Let the foot of the altitude from $A$ to $BC$ be $P$, to $CD$ be $Q$, and to $BD$ be $R$.

Note that all isosceles trapezoids are cyclic quadrilaterals; thus, $A$ is on the circumcircle of $\triangle BCD$ and we have that $PRQ$ is the Simson Line from $A$. As $\angle QAB = 90^\circ$, we have that $\angle QAR = 90^\circ - \angle RAB =\angle ABR = \angle APR = \angle APQ$, with the last equality coming from cyclic quadrilateral $APBR$. Thus, $\triangle QAR \sim \triangle QPA$ and we have that $\frac{AQ}{AR} = \frac{PQ}{PA}$ or that $\frac{18}{10} = \frac{QP}{15}$, which we can see gives us that $QP = 27$. Further ratios using the same similar triangles gives that $QR = 12$ and $RP = 15$.

We also see that quadrilaterals $APBR$ and $ARQD$ are both cyclic,it is clear that $\angle AQR =\angle ADR$ and $\angle APR =\angle ABR$,which shows $\triangle ABD \sim \triangle APQ$. we wish to find the ratio of similitude between the two triangles.

To do this, we use the one number we have for $\triangle ABD$: we know that the altitude from $A$ to $BD$ has length $10$. As the two triangles are similar, if we can find the height from $A$ to $PQ$, we can take the ratio of the two heights as the ratio of similitude. To do this, we once again note that $QP = 27$. Using this, we can drop the altitude from $A$ to $QP$ and let it intersect $QP$ at $H$. Then, let $QH = x$ and thus $HP=27-x$. We then have by the Pythagorean Theorem on $\triangle AQH$ and $\triangle APH$:

$$\begin{aligned} 18^2 - x^2 &= 15^2 - (27-x)^2 \\ 324 - x^2 &= 225 - (x^2-54x+729) \\ 54x &= 828 \\ x &= \frac{46}{3}. \end{aligned}$$ Then, $RH = QH - QR = \frac{46}{3} - 12 = \frac{10}{3}$. This gives us then from right triangle $\triangle ARH$ that $AH = \frac{20\sqrt{2}}{3}$ and thus the ratio of $\triangle APQ$ to $\triangle ABD$ is $\frac{3\sqrt{2}}{4}$. From this, we see then that

$$AB = AP \cdot \frac{3\sqrt{2}}{4} = 15 \cdot \frac{3\sqrt{2}}{4} = \frac{45\sqrt{2}}{4}$$ and

$$AD = AQ \cdot \frac{3\sqrt{2}}{4} = 18 \cdot \frac{3\sqrt{2}}{4} = \frac{27\sqrt{2}}{2}.$$ The Pythagorean Theorem on $\triangle AQD$ then gives that

$$QD = \sqrt{AD^2 - AQ^2} = \sqrt{\left(\frac{27\sqrt{2}}{2}\right)^2 - 18^2} = \sqrt{\frac{81}{2}} = \frac{9\sqrt{2}}{2}.$$ Then, we have the height of trapezoid $ABCD$ is $AQ = 18$, the top base is $AB = \frac{45\sqrt{2}}{4}$, and the bottom base is $CD = \frac{45\sqrt{2}}{4} + 2\cdot\frac{9\sqrt{2}}{2}$. From the equation of a trapezoid, $K = \frac{b_1+b_2}{2} \cdot h = \frac{63\sqrt{2}}{4} \cdot 18 = \frac{567\sqrt{2}}{2}$, so the answer is $K\sqrt{2} = \boxed{567}$.

~lvmath

~Corrected by dongjiu0728(Former Ans made the wrong QH and wrong QR、RP but somehow came to the right answer,which trapped me for a weekend)

Solution 5 (Cyclic Quadrilaterals, Similar Triangles, Pythagorean Theorem)

Let $E,F,$ and $G$ be the feet of the altitudes from $A$ to $BC,CD,$ and $DB$, respectively.

Claim: We have $2$ pairs of similar right triangles: $\triangle AEB \sim \triangle AFD$ and $\triangle AGD \sim \triangle AEC$.

Proof: Note that $ABCD$ is cyclic. We need one more angle, and we get this from this cyclic quadrilateral:

$$\begin{aligned} \angle ABE &= 180^\circ - \angle ABC =\angle ADC = \angle ADG, \\ \angle ADG &= \angle ADB =\angle ACB = \angle ACE. \hspace{20mm} \square \end{aligned}$$ Let $AD=a$. We obtain from the similarities $AB = \frac{5a}{6}$ and $AC=BD=\frac{3a}{2}$.

By Ptolemy, $\left(\frac{3a}{2}\right)^2 = a^2 + \frac{5a}{6} \cdot CD$, so $\frac{5a^2}{4} = \frac{5a}{6} \cdot CD$.

We obtain $CD=\frac{3a}{2}$, so $DF=\frac{CD-AB}{2}=\frac{a}{3}$.

Applying the Pythagorean theorem on $\triangle ADF$, we get $324=a^2 - \frac{a^2}{9}=\frac{8a^2}{9}$.

Thus, $a=\frac{27}{\sqrt{2}}$, and $[ABCD]=\frac{AB+CD}{2} \cdot 18 = \frac{\frac{5a}{6} +\frac{9a}{6}}{2} \cdot 18 = 18 \cdot \frac{7}{6} \cdot \frac{27}{\sqrt{2}} = \frac{567}{\sqrt{2}}$, yielding $\sqrt2\cdot[ABCD]=\boxed{567}$.

Solution 6 (Similar Triangles and Trigonometry)

Let $AD=BC=a$. Draw diagonal $AC$ and let $G$ be the foot of the perpendicular from $B$ to $AC$, $F$ be the foot of the perpendicular from $A$ to line $BC$, and $H$ be the foot of the perpendicular from $A$ to $DC$.

Note that $\triangle CBG\sim\triangle CAF$, and we get that $\frac{10}{15}=\frac{a}{AC}$. Therefore, $AC=\frac32 a$. It then follows that $\triangle ABF\sim\triangle ADH$. Using similar triangles, we can then find that $AB=\frac{5}{6}a$. Using the Law of Cosines on $\triangle ABC$, We can find that the $\cos\angle ABC=-\frac{1}{3}$. Since $\angle ABF=\angle ADH$, and each is supplementary to $\angle ABC$, we know that the $\cos\angle ADH=\frac{1}{3}$. It then follows that $a=\frac{27\sqrt{2}}{2}$. Then it can be found that the area $K$ is $\frac{567\sqrt{2}}{2}$. Multiplying this by $\sqrt{2}$, the answer is $\boxed{567}$.

~happykeeper

Solution 7 (Similar Triangles and Trigonometry)

Draw the distances in terms of $B$, as shown in the diagram. By similar triangles, $\triangle{AEC}\sim\triangle{BIC}$. As a result, let $AB=u$, then $BC=AD=\frac{6}{5}u$ and $2AC=3BC$. The triangle $ABC$ is $6-5-9$ which $\cos(\angle{ABC})=-\frac{1}{3}$. By angle subtraction, $\cos(180-\theta)=-\cos\theta$. Therefore, $AB=\frac{45}{2\sqrt{2}}=\frac{45\sqrt{2}}{4}$ and $AD=BC=\frac{27}{\sqrt{2}}$. By trapezoid area formula, the area of $ABCD$ is equal to $(AB+DF)\cdot 18=567\cdot \frac{\sqrt{2}}{2}$ which $\sqrt{2}\cdot k=\boxed{567}$.

~math2718281828459

Solution 8 (Heron's Formula)

Let the points formed by dropping altitudes from $A$ to the lines $BC$, $CD$, and $BD$ be $E$, $F$, and $G$, respectively.

We have

$$\triangle ABE \sim \triangle ADF \implies \frac{AD}{18} = \frac{AB}{15} \implies AD = \frac{6}{5}AB$$ and

$$BD\cdot10 = 2[ABD] = AB\cdot18 \implies BD = \frac{9}{5}AB.$$ For convenience, let $AB = 5x$. By Heron's formula on $\triangle ABD$, we have sides $5x,6x,9x$ and semiperimeter $10x$, so

$$\sqrt{10x\cdot5x\cdot4x\cdot1x} = [ABD] = \frac{AB\cdot18}{2} = 45x \implies 10\sqrt{2}x^2 = 45x \implies x= \frac{45}{10\sqrt{2}},$$ so $AB = 5x = \frac{45}{2\sqrt{2}}$.

Then,

$$BE = \sqrt{AB^2 - CA^2} = \sqrt{\left(\frac{45}{2\sqrt{2}}\right)^2 - 15^2} = \sqrt{\frac{225}{8}} = \frac{15}{2\sqrt{2}}$$ and

$$\triangle ABE \sim \triangle ADF \implies DF = \frac{6}{5}BE = \frac{6}{5}\cdot\frac{15}{2\sqrt{2}} = \frac{18}{2\sqrt{2}}.$$ Finally, recalling that $ABCD$ is isosceles,

$$K = [ABCD] = \frac{18}{2}(AB + (AB + 2DF)) = 18(AB + DF) = 18\left(\frac{45}{2\sqrt{2}} + \frac{18}{2\sqrt{2}}\right) = \frac{567}{\sqrt{2}},$$ so $\sqrt{2}\cdot K = \boxed{567}$.

~emerald_block

Solution 9 (Three Heights)

Let $\overline{AE}, \overline{AF},$ and $\overline{AG}$ be the perpendiculars from $A$ to ${BC}, {CD},$ and ${BD},$ respectively. $AE = 15, AF = 18, AG =10$. Denote by $G'$ the base of the perpendicular from $B$ to $AC, H$ be the base of the perpendicular from $C$ to $AB$. Denote $\theta = \angle{CBH}.$ It is clear that

$$BG' = AG, CH = AF, \triangle CBH \ =\triangle ADF,$$ the area of $ABCD$ is equal to the area of the rectangle $AFCH.$

The problem is reduced to finding $AH$.

In triangle $ABC$ all altitudes are known:

$$AB : BC : AC = \frac{1}{CH}\ : \frac{1}{AE}\ : \frac{1}{BG'}\ =$$ $$= \frac{1}{AF}\ : \frac{1}{AE}\ : \frac{1}{AG}\ = 5 : 6 : 9.$$ We apply the Law of Cosines to $\triangle ABC$ and get$:$

$$\begin{aligned} 2\cdot AB\cdot BC \cdot \cos\theta = AC^2 – AB^2 – BC^2, \end{aligned}$$ $$\begin{aligned} 2\cdot 5\cdot 6\cdot \cos\theta = 60 \cos\theta = 9^2 – 5^2 – 6^2 = 20, \cos\theta =\frac{1}{3}. \end{aligned}$$ $$\begin{aligned} BH = BC \cos\theta = \frac{BC}{3}.\end{aligned}$$ We apply the Pythagorean Law to $\triangle HBC$ and get$:$

$$\begin{aligned} HC^2 = 18^2 = BC^2 – BH^2 = 9\cdot BH^2 – BH^2 = 8 BH^2.\end{aligned}$$ $$\begin{aligned} BH = \frac{9}{\sqrt2}, AH = (\frac{5}{2} + 1)\cdot BH = \frac{63}{2\cdot \sqrt2}. \end{aligned}$$ Required area is

$$\begin{aligned} K = \frac{63}{2\cdot \sqrt{2}} \cdot 18 = \frac{567}{\sqrt{2}} \implies \sqrt{2} K=\boxed{567}. \end{aligned}$$ vladimir.shelomovskii@gmail.com, vvsss

Solution 10 (Area)

Let $F$ be on $DC$ such that $AF \| DC$. Let $G$ be on $BD$ such that $AG \| BD$.

Let $m$ be the length of $AB$. Let $n$ be the length of $AD$.

The area of $\triangle ABD$ can be expressed in three ways: $\frac{1}{2}(15)(BC) = \frac{1}{2}(15)(n)$, $\frac{1}{2}(18)(m)$, and $\frac{1}{2}(10)(BD)$.

$$\frac{1}{2}(15)(n) = \frac{1}{2}(18)(m)$$ $$15n = 18m$$ $$5n = 6m$$ $$n = \frac{6}{5}m$$ Now, $BD = BG + GD = \sqrt{m^2-100} + \sqrt{n^2-100}$. We can substitute in $n = \frac{6}{5}m$ to get

$BD = \sqrt{m^2-100} + \sqrt{(\frac{6}{5}m)^2-100}$.

We have

$$\frac{1}{2}(10)\left(\sqrt{m^2-100} + \sqrt{(\frac{6}{5}m)^2-100}\right) = \frac{1}{2}(18)(m)$$ After a fairly straightforward algebraic bash, we get $m = \frac{45\sqrt{2}}{4}$, and $n = (\frac{6}{5})(\frac{45\sqrt{2}}{4}) = \frac{27\sqrt{2}}{2}$. By the Pythagorean Theorem on $\triangle ADF$, $DF^2 = n^2 - 18^2 = \frac{729}{2} - 324 = \frac{81}{2}$, and $DF = \frac{9\sqrt{2}}{2}$.

Thus, $DC = 2DF + AB = 9\sqrt{2}+\frac{45\sqrt{2}}{4} = \frac{81\sqrt{2}}{4}$. Therefore, $K = \frac{1}{2}(\frac{45\sqrt{2}}{4}+\frac{81\sqrt{2}}{4}) \cdot 18 = \frac{63\sqrt{2}}{2} \cdot 18 = \frac{567\sqrt{2}}{2}$. The requested answer is $K \cdot \sqrt{2} = \boxed{567}$.

~ adam_zheng

Solution 11 (Coordinates)

By the given information, the height of the trapezoid is $18,$ but to avoid numerical calculations, call it $b.$ Label $A(-a, b)$, $B(a, b)$, $C(c, 0)$, $D(-c, 0),$ with $b>0, 0 We have the line equations$BD: y=\frac{b}{a+c}x + \frac{bc}{a+c}, BC: y=\frac{b}{a-c}x - \frac{bc}{a-c},$or rewriting them,$BD: bx-(a+c)y + bc=0, BC: bx - (a-c)y-bc = 0.$ Hence applying two projection theorems gives

$$10 = \frac{|-ab - b(a+c) + bc|}{\sqrt{b^2 + (a+c)^2}},$$ $$15 = \frac{|-ab - b(a-c) - bc|}{\sqrt{b^2 + (a-c)^2}}.$$ Simplifying and remembering $a,b>0,$

$$10 = \frac{36a}{\sqrt{324+(a+c)^2}},$$ $$15 = \frac{36a}{\sqrt{324+(a-c)^2}}.$$ Hence

$$324 + (a+c)^2 = \left(\frac{18a}{5}\right)^2,$$ $$324 + (a-c)^2 = \left(\frac{12a}{5}\right)^2.$$ Subtracting the two equations yield $a = \frac{5}{9}c.$ Plugging that back into the first equation gives $324 + \left(\frac{14}{9}c\right)^2 = 4c^2$ or $\frac{128}{81}c^2 = 324, c^2 = \frac{81^2}{32} \rightarrow c = \frac{81}{4\sqrt{2}}.$ Hence $a = \frac{45}{4\sqrt{2}},$ and the area of the trapezoid is $18 \left( \frac{2a + 2c}{2}\right) = 18(a+c) = \frac{18 \cdot 63}{2\sqrt{2}}=\frac{567}{\sqrt{2}}.$ Answer extraction gives $\boxed{567}.$

~anduran

Solution 12 (Trig Bash Using Heights)

We will use trig bash. After some trial, we can see that knowing the size of one angle of the trapezoid allows us to calculate every length of it easily. We will try to find $\angle C$ in this solution.

Let $\angle C=a$ and $\angle ABD=b$. $\angle ADB=a-b$ as $AB$ and $CD$ are parallel lines.

The height of the trapezoid is $18$. By trigs we have $AD=\frac{18}{\sin{a}}$.

Inside the triangle $ABD$, its height from point $A$ is $10$, its height from point $A$ is also $AD \sin{\angle ADB}=\frac{18}{\sin{a}}\sin(a-b)$. We obtain the equation $\frac{18}{\sin{a}}\sin(a-b)=10$.

By trigs, $AB$ is both $\frac{15}{\sin{a}}$ and $\frac{10}{\sin{b}}$. $\frac{15}{\sin{a}} = \frac{10}{\sin{b}}$.

Expanding the equation $\frac{18}{\sin{a}}\sin(a-b)=10$ using angle-sum formula, we have $18(\sin{a}\cos{b}-\cos{a}\sin{b})=10\sin{a}$. We can just plug in $\sin{b}=\frac{2}{3}\sin{a}$ and use the identity $\cos{a}=\sqrt{1-\sin^2{a}}$ now to solve this common, algebraic equation.The process of solving is long and boring. Eventually we will get $\sin{a}=\frac{2\sqrt{2}}{3}$.

In the end, by trigs we have $AB=\frac{15}{\sin{a}}$ and $CD=\frac{15}{\sin{a}}+2\frac{18}{\tan{a}}$

The area of $ABCD$ is $(AB+CD)\times \frac{18}{2} = 9(\frac{30}{\sin{a}}+\frac{36}{\tan{a}})$. Plugging in $\sin{a}=\frac{2\sqrt{2}}{3}$ and $\tan{a}=\frac{\sin{a}}{\sqrt{1-\sin^2{a}}}=2\sqrt{2}$, we find that the area of $ABCD$ is $\frac{567\sqrt{2}}{2}$.

The answer to the problem is $567$.

~AOPS Community — G2JFForever

Video Solution

https://youtu.be/uItEKVj-tF8

~Mathproblemsolvingskills.com

Video Solution

https://www.youtube.com/watch?v=6rLnl8z7lnM