, 2023

Solution 1

We make two copies of the given triangle, labelling them $\triangle ABC$ and $\triangle DEF$, as shown: [[IMAGE0]] The combined area of these two triangles is $2 $\cdot 770\text{}$ cm}^2 = $1540\text{}$
cm}^2, and the shaded area in each triangle is the same. Next, we rotate

$\triangle$ DEF$by$$180^\circ$: [[IMAGE1]] and join the two triangles together: [[IMAGE2]] We note that

BC$and$AE$(which was$FE) are equal in length (since they were copies of each other) and parallel (since they are

$180^\circ$ rotations of each other). The same is true for

AB$and$EC$. Therefore,$ABCE$is a parallelogram. Further,$ABCE is divided into 11 identical parallelograms (6 shaded and 5 unshaded) by the horizontal lines. (Since the sections of the two triangles are equal in height, the horizontal lines on both sides of

AC$align.) The total area of parallelogram$ABCE$is$$1540\text{}$ cm}^2$. Thus, the shaded area of$ABCE$is$$\frac{6}{11} \cdot 1540\text{}$ cm}^2 =
$840\text{}$ cm}^2. Since this shaded area is equally divided between the two halves of the parallelogram, then the combined area of the shaded regions of

$\triangle$ ABC$is$$\frac{1}{2} \cdot 840\text{}$ cm}^2 = $420\text{}$ cm}^2$.

Solution 2

We label the points where the horizontal lines touch $AB$ and $AC$ as shown: [[IMAGE3]] We use the notation $\$|\triangle$
ABC|$to represent the area of$$\triangle$ ABC and use similar notation for the area of other triangles and quadrilaterals. Let

$\mathcal{A}$ be equal to the total area of the shaded regions. Thus,

$\mathcal{A} = |\triangle AB_1C_1| + |B_2B_3C_3C_2| + |B_4B_5C_5C_4| + |B_6B_7C_7C_6| + |B_8B_9C_9C_8| + |B_{10}BCC_{10}|$ The area of each of these quadrilaterals is equal to the difference of the area of two triangles. For example,

$|B_2B_3C_3C_2| = |\triangle AB_3C_3| - |\triangle AB_2C_2| = - |\triangle AB_2C_2| + |\triangle AB_3C_3|$$Therefore,$$\text{A = | AB 1C 1| - | AB 2C 2| + | AB 3C 3| - | AB 4C 4| + | AB 5C 5| - | AB 6C 6| + | AB 7C 7|- | AB 8C 8| + | AB 9C 9| - | AB 10 C 10 | + | ABC|}$$Each of$$\triangle$ AB_1C_1$,$$\triangle$ AB_2C_2$,$$\ldots$$,$$\triangle AB_{10}C_{10}$$is similar to$$\triangle$ ABC because their two base angles are equal due. Suppose that the height of

$\triangle$
ABC$from$A$to$BC$is$h. Since the height of each of the 11 regions is equal in height, then the height of

$\triangle$ AB_1C_1$is$$\frac{1}{11}h$$, the height of$$\triangle$ AB_2C_2$is$$\frac{2}{11}h$, and so on. When two triangles are similar, their heights are in the same ratio as their side lengths: To see this, suppose that

$\triangle$
PQR$is similar to$$\triangle$ STU and that altitudes are drawn from

P$and$S$to$V$and$W$. [[IMAGE4]] Since$$\angle$ PQR = $\angle$ STU$, then$$\triangle$ PQV$is similar to$$\triangle$ STW (equal angle; right angle), which means that

$\dfrac{PQ}{ST} = \dfrac{PV}{SW}$. In other words, the ratio of sides is equal to the ratio of heights. Since the height of

$\triangle$
AB_1C_1$is$$\frac{1}{11}h$$, then$B_1C_1 = $\frac{1}{11}BC$$. Therefore,$$|\triangle$ AB_1C_1| =
$\frac{1}{2} \cdot$ B_1C_1 $\cdot \frac{1}{11}h = \frac{1}{2} \cdot \frac{1}{11}BC \cdot \frac{1}{11}h = \frac{1^2}{11^2} \cdot \frac{1}{2} \cdot BC \cdot$ h = $\frac{1^2}{11^2} |\triangle$ ABC|$. Similarly, since the height of$$\triangle$
AB_2C_2$is$$\frac{2}{11}h$$, then$B_2C_2 = $\frac{2}{11}BC$$. Therefore,$$|\triangle$ AB_2C_2| =
$\frac{1}{2} \cdot$ B_2C_2 $\cdot \frac{2}{11}h = \frac{1}{2} \cdot \frac{2}{11}BC \cdot \frac{2}{11}h = \frac{2^2}{11^2} \cdot \frac{1}{2} \cdot BC \cdot$ h = $\frac{2^2}{11^2} |\triangle$ ABC|. This result continues for each of the triangles. Therefore,

$\text{A = 1 2 11 2 | ABC| - 2 2 11 2 | ABC| + 3 2 11 2 | ABC| - 4 2 11 2 | ABC| + 5 2 11 2 | ABC| - 6 2 11 2 | ABC| + 7 2 11 2 | ABC| - 8 2 11 2 | ABC| + 9 2 11 2 | ABC| - 10 2 11 2 | ABC| + 11 2 11 2 | ABC| = 1 11 2 | ABC| (11 2 - 10 2 + 9 2 - 8 2 + 7 2 - 6 2 + 5 2 - 4 2 + 3 2 - 2 2 + 1) = 1 11 2 (770 cm 2) ((11+10)(11-10) + (9+8)(9-8) + + (3+2)(3-2) + 1) = 1 11 2 (770 cm 2) (11+10+9+8+7+6+5+4+3+2+1) = 1 11 (70 cm 2) 66 = 420 cm 2}$ Therefore, the combined area of the shaded regions of

$\triangle$
ABC$is$$420\text{}$ cm}^2$.
Solution 1

We label five additional points in the diagram:

[[IMAGE5]]

Since $PQ=QR=RS=1$, then $PS = 3$ and $PR = 2$.

Since $\angle PST = 90^\circ$, then $PT = $\sqrt{PS^2}$ + ST^2} = $\sqrt{3^2}$ + 1^2} =
$\sqrt{10}$ by the Pythagorean Theorem. We are told that

ABCD$is a square. Thus,$PT$is perpendicular to$QC$and to$RB$. Thus,$$\triangle$ PDQ$is right-angled at$D$and$$\triangle$ PAR$is right-angled at$A$. Since$$\triangle$ PDQ$,$$\triangle$ PAR$and$$\triangle$ PST are all right-angled and all share an angle at

P, then these three triangles are similar. This tells us that

$\dfrac{PA}{PS} = \dfrac{PR}{PT}$$and so$PA = $\dfrac{3 \cdot 2}{\sqrt{10}}$$. Also,$$\dfrac{PD}{PS} = \dfrac{PQ}{PT}$$and so$PD = $\dfrac{1 \cdot 3}{\sqrt{10}}$$. Therefore,$$DA = PA - PD = \dfrac{6}{\sqrt{10}} - \dfrac{3}{\sqrt{10}} = \dfrac{3}{\sqrt{10}}$ This means that the area of square

ABCD$is equal to$DA^2 = $\left(\dfrac{3}{\sqrt{10}}\right)^2 = \dfrac{9}{10}\$.$

Solution 2

We add coordinates to the diagram as shown:

[[IMAGE6]]

We determine the side length of square $ABCD$ by determining the coordinates of $D$ and $A$ and then calculating the distance between these points. The slope of the line through $(0,3)$ and $(3,2)$ is $\dfrac{3-2}{0-3} = -\dfrac{1}{3}$. This equation of this line can be written as $y = -\dfrac{1}{3}x + 3$. The slope of the line through $(0,0)$ and $(1,3)$ is 3. The equation of this line can be written as $y = 3x$. The slope of the line through $(1,0)$ and $(2,3)$ is also 3. The equation of this line can be written as $y = 3(x-1) = 3x - 3$. Point $D$ is the intersection point of the lines with equations $y =
$-\dfrac{1}{3}x$ + 3$and$y = 3x$.

Equating expressions for $y$, we obtain $-\dfrac{1}{3}x + 3 = 3x$ and so $\dfrac{10}{3}x = 3$ which gives $x = \dfrac{9}{10}$. Since $y = 3x$, we get $y = \dfrac{27}{10}$ and so the coordinates of $D$ are $\left(\dfrac{9}{10}, \dfrac{27}{10}\right)$.

Point $A$ is the intersection point of the lines with equations $y =
$-\dfrac{1}{3}x$ + 3$and$y = 3x - 3$.

Equating expressions for $y$, we obtain $-\dfrac{1}{3}x + 3 = 3x - 3$ and so $\dfrac{10}{3}x = 6$ which gives $x = \dfrac{18}{10}$. Since $y = 3x - 3$, we get $y = \dfrac{24}{10}$ and so the coordinates of $A$ are $\left(\dfrac{18}{10}, \dfrac{24}{10}\right)$. (It is easier to not reduce these
fractions.)

Therefore, $$DA = \sqrt{\left(\dfrac{9}{10} - \dfrac{18}{10}\right)^2 + \left(\dfrac{27}{10} - \dfrac{24}{10}\right)^2} = \sqrt{\left(-\dfrac{9}{10}\right)^2 + \left(\dfrac{3}{10}\right)^2} = \sqrt{\dfrac{90}{100}} = \sqrt{\dfrac{9}{10}}$$ This means that the area of square $ABCD$ is equal to $DA^2 = $\left(\sqrt{\dfrac{9}{10}}\right)^2 = \dfrac{9}{10}\$.$