Euclidean Geometry

Toolkit

Area

•  $A_{\text{Rectangle}}=l\times w$
  
•  $A_{\text{Parallelogram}}=b\times h$
  
•  $A_{\text{Triangle}}=\frac{1}{2}(b\times h)$
  
•  $A_{\text{Trapezoid}}=\frac{1}{2}(a+b)h$
  
•  $A_{\text{Circle}}=\pi r^2$
  
  Note: The perimeter of a circle is $2\pi r$.

• The area of an equilateral triangle of side length $s$ is ${\displaystyle {\displaystyle \frac{\sqrt{3}{s}^2}{4}}}$.

• Heron’s formula - The area of a triangle with side lengths $a,b,$ and $c$ is $\sqrt{s(s-a)(s-b)(s-c)}$ where $s={\displaystyle \frac{a+b+c}{2}}$.

Volume

•  $V_{\text{Prism}}=A_{\text{Base}}\times h$
  
•  $V_{\text{Pyramid}}=\frac{1}{3}{A}_{\text{Base}}\times h$
  
•  $V_{\text{Cylinder}}=\pi r^{2}h$
  
•  $V_{\text{Cone}}=\frac{1}{3}\pi r^2\times h$
  
•  $V_{\text{Sphere}}=\frac{4}{3}\pi r^3$
  

Pythagorean Theorem

• For a right triangle with side lengths, $a$, $b$ and $c$, where $c$ is the length of the hypotenuse, we have $a^2+b^2=c^2$.

• Ordered triples of integers $(a,b,c)$ which satisfy this relationship are called Pythagorean Triples. The triples $(3,4,5)$, $(7,24,25)$ and $(5,12,13)$ are common examples.

Similarity and Congruence

• When two polygons have equal corresponding angles, they are similar. When the corresponding side lengths are also equal, they are congruent.

• Two triangles are similar when they satisfy any of the following rules:

  • Angle-angle similarity (two corresponding angles are equal)

  • Side-angle-side similarity (two pairs of corresponding sides are in the same proportion and the contained angles are equal)

  • Side-side-side similarity (all three pairs of corresponding sides are in the same proportion)

Angles

• For a polygon with $n$ sides, the sum of the interior angles is given by $((n-2)\times 180)\mathrm{°}$ or $(180n-360)\mathrm{°}$.

• For a regular polygon, all interior angles are equal in measure.

• A trapezoid has two pairs of supplementary angles.

• A parallelogram has opposite angles that are equal and adjacent angles that are supplementary.

• Angles that form a straight line and share a vertex add to $180\mathrm{°}$.

• Two angles that sum to $90\mathrm{°}$ are called complementary.

• Opposite angles are equal.

  

• Corresponding angles are equal.

  

• Alternate angles are equal.

  

• Co-interior angles sum to $180\mathrm{°}$.

  

Circle Properties

• The area of the sector formed by a central angle of $\theta \mathrm{°}$ is given by $A_{\text{sector}}={\displaystyle \frac{\theta }{360}}\pi r^2$, where $r$ is the radius of the circle. The arc formed has length $s={\displaystyle \frac{\theta }{360}}2\pi r$.

• The angle inscribed in a semicircle is a right angle.

  

• When an arc subtends an inscribed angle and a central angle, the measure of the central angle is twice the measure of the inscribed angle.

  

• Angles subtended by the same arc are equal.

  

• A tangent to a circle and the radius drawn to the point of tangency meet at $90\mathrm{°}$.

  

• If two chords $AB$ and $CD$ of a circle intersect at the point $P$ then $(PA)(PB)=(PC)(PD)$.

  

• Tangent segments from an external point to a circle are equal.

  

Sample Problems

1. In the diagram, a sector of a circle with centre $O$, a radius of $5$ and $\mathrm{\angle }AOB=72\mathrm{°}$ is shown.

   

   What is the perimeter of the sector?

   Solution:

   The perimeter of the sector is made up of two line segments (of total length $5+5=10$) and one arc of a circle. Since ${\displaystyle \frac{72\mathrm{°}}{360\mathrm{°}}}={\displaystyle \frac{1}{5}}$, the length of the arc is ${\displaystyle \frac{1}{5}}$ of the total circumference of a circle of radius $5$. Thus, the length of the arc is ${\displaystyle \frac{1}{5}}(2\pi (5))=2\pi$. Therefore, the perimeter of the sector is $10+2\pi$.

2. In the diagram, $AB$ and $BC$ are chords of the circle with $AB<BC$. If $D$ is the point on the circle such that $AD$ is perpendicular to $BC$ and $E$ is the point on the circle such that $DE$ is parallel to $BC$, prove that $\mathrm{\angle }EAC+\mathrm{\angle }ABC=90\mathrm{°}$.

   

   Solution 1:

   Join $A$ to $E$ and $C$, and $B$ to $E$.

   

   Since $DE$ is parallel to $BC$ and $AD$ is perpendicular to $BC$, $AD$ is perpendicular to $DE$, ie. $\mathrm{\angle }ADE=90\mathrm{°}$. Therefore, $AE$ is a diameter. Now $\mathrm{\angle }EAC=\mathrm{\angle }EBC$, since both are subtended by $EC$. Therefore, $\mathrm{\angle }EAC+\mathrm{\angle }ABC=\mathrm{\angle }EBC+\mathrm{\angle }ABC=\mathrm{\angle }EBA$ which is indeed equal to $90\mathrm{°}$, as required, since $AE$ is a diameter.

   Solution 2:

   Join $A$ to $E$ and $C$.

   

   Since $DE$ is parallel to $BC$ and $AD$ is perpendicular to $BC$, $AD$ is perpendicular to $DE$, ie. $\mathrm{\angle }ADE=90\mathrm{°}$. Therefore, $AE$ is a diameter. Thus, $\mathrm{\angle }ECA=90\mathrm{°}$. Now $\mathrm{\angle }ABC=\mathrm{\angle }AEC$ since both are subtended by $AC$. Now $\mathrm{\angle }EAC+\mathrm{\angle }ABC=\mathrm{\angle }EAC+\mathrm{\angle }AEC=180\mathrm{°}-\mathrm{\angle }ECA$ using the sum of the angles in $\mathrm{△}AEC$. But $\mathrm{\angle }ECA=90\mathrm{°}$, so $\mathrm{\angle }EAC+\mathrm{\angle }AEC=90\mathrm{°}$.

   Solution 3:

   Join $A$ to $E$ and $C$, and $C$ to $D$.

   

   Since $DE$ is parallel to $BC$ and $AD$ is perpendicular to $BC$, $AD$ is perpendicular to $DE$, ie. $\mathrm{\angle }ADE=90\mathrm{°}$. Therefore, $AE$ is a diameter. Now $\mathrm{\angle }ABC=\mathrm{\angle }ADC$ since both are subtended by $AC$. Also $\mathrm{\angle }EAC=\mathrm{\angle }EDC$ since both are subtended by $EC$.

   So $\mathrm{\angle }EAC+\mathrm{\angle }ABC=\mathrm{\angle }EDC+\mathrm{\angle }ADC=\mathrm{\angle }ADE=90\mathrm{°}$.

3. In the isosceles trapezoid $ABCD$, $AB=CD=x$. A circle with centre $O$ is drawn inside the trapezoid such that it is tangent to all four sides of the trapezoid.

   

   The area of the trapezoid is $80$. Given that the radius of the circle is $4$, determine the value of $x$.

   Solution:

   We label the points of tangency $P,Q,R$ and $S$ and connect them to the centre of the circle.

   

   Each of the these line segments is a radius of the circle and so each is perpendicular to the sides of the trapezoid, since the sides of the trapezoid are tangent to the circle. The sides $BC$ and $AD$ are parallel and therefore, $QO$ and $OS$ are parallel and thus, $QS$ is a diameter of the circle. Therefore, $QS=2(4)=8$ and so the height of the trapezoid is $8$.

   We let $BQ=b$ and $CQ=a$. Since tangent segments from an external point to a circle are equal, we obtain the following equalities. $$\begin{array}{rl}BQ& =BP=b\\ AP& =AS=x-b\\ CQ& =CR=a\\ DR& =DS=x-a\end{array}$$ Therefore, $$\begin{array}{rl}BC+AD& =BQ+QC+AS+SD\\ & =b+a+(x-b)+(x-a)\\ & =2x\end{array}$$ The area of trapezoid $ABCD$ is equal to $${\displaystyle \frac{1}{2}}(BC+AD)\times QS={\displaystyle \frac{1}{2}}(2x)\times 8=8x$$ Since the area of the $ABCD$ is $80$, we have $8x=80$, which gives us that $x=10$.

4. Three squares have dimensions as indicated in the diagram.

   

   What is the area of the shaded quadrilateral?

   Solution

   We make use of similar triangles. We start by labelling the diagram as shown.

   

   We want to calculate the lengths $EB$ and $FC$, which will allow us to calculate the area of $\mathrm{△}AEB$ and $\mathrm{△}AFC$ . Since $FC$ is parallel to $GD$, we have $\mathrm{\angle }AFC=\mathrm{\angle }AGD$ and $\mathrm{\angle }ACF=\mathrm{\angle }ADG$. In addition, we note that the angle at $A$ is common to $\mathrm{△}AFC$ and $\mathrm{△}AGD$. Therefore, by angle-angle (AA), $\mathrm{△}AFC$ and $\mathrm{△}AGD$ are similar. So $${\displaystyle \frac{AC}{AD}}={\displaystyle \frac{FC}{GD}}={\displaystyle \frac{5}{10}}={\displaystyle \frac{1}{2}}$$ Therefore, $FC={\displaystyle \frac{1}{2}}GD={\displaystyle \frac{1}{2}}(5)={\displaystyle \frac{5}{2}}$.

   Using similar reasoning, $\mathrm{△}AEB$ and $\mathrm{△}AFC$ are also similar. Therefore, $${\displaystyle \frac{AB}{AC}}={\displaystyle \frac{EB}{FC}}={\displaystyle \frac{2}{5}}$$ Therefore, $EB={\displaystyle \frac{2}{5}}FC={\displaystyle \frac{2}{5}}\left({\displaystyle \frac{5}{2}}\right)=1$.

   Thus, the shaded area is equal to the area of $\mathrm{△}AFC$ minus the area of $\mathrm{△}AEB$. Therefore, the area of the shaded quadrilateral is ${\displaystyle \frac{1}{2}}(5)\left({\displaystyle \frac{5}{2}}\right)-{\displaystyle \frac{1}{2}}(2)(1)={\displaystyle \frac{21}{4}}$. (Alternately, quadrilateral $EBCF$ is a trapezoid, where we know the base length is 3 and the two heights are $1$ and ${\displaystyle \frac{5}{2}}$, and so the area is ${\displaystyle \frac{1}{2}}(1+{\displaystyle \frac{5}{2}})(3)={\displaystyle \frac{21}{4}}$.)

5. In the diagram, $PABCD$ is a pyramid with square base $ABCD$ and with $PA=PB=PC=PD$. Suppose that $M$ is the midpoint of $PC$ and that $\mathrm{\angle }BMD=90\mathrm{°}$. Triangular-based pyramid $MBCD$ is removed by cutting along the triangle defined by the points $M$, $B$ and $D$. The volume of the remaining solid $PABMD$ is $288$. What is the length of $AB$?

   

   Solution: (This is a challenging problem.)

   Let the side length of the square base $ABCD$ be $2a$ and the height of the pyramid (that is, the distance of $P$ above the base) be $2h$.

   Let $F$ be the point of intersection of the diagonals $AC$ and $BD$ of the base. By symmetry, $P$ is directly above $F$; that is, $PF$ is perpendicular to the plane of square $ABCD$.

   Note that $AB=BC=CD=DA=2a$ and $PF=2h$. We want to determine the value of $2a$.

   Let $G$ be the midpoint of $FC$.

   Join $P$ to $F$ and $M$ to $G$.

   

   Consider $\mathrm{△}PCF$ and $\mathrm{△}MCG$. Since $M$ is the midpoint of $PC$, we have $MC={\displaystyle \frac{1}{2}}PC$. Since $G$ is the midpoint of $FC$, we have $GC={\displaystyle \frac{1}{2}}FC$.

   Since $\mathrm{△}PCF$ and $\mathrm{△}MCG$ share an angle at $C$ and the two pairs of corresponding sides adjacent to this angle are in the same ratio, $\mathrm{△}PCF$ is similar to $\mathrm{△}MCG$.

   Since $PF$ is perpendicular to $FC$, $MG$ is perpendicular to $GC$.

   Also, $MG={\displaystyle \frac{1}{2}}PF=h$ since the side lengths of $\mathrm{△}MCG$ are half those of $\mathrm{△}PCF$.

   The volume of the square-based pyramid $PABCD$ equals ${\displaystyle \frac{1}{3}}(A{B}^2)(PF)={\displaystyle \frac{1}{3}}(2a{)}^2(2h)={\displaystyle \frac{8}{3}}{a}^{2}h.$

   Triangular-based pyramid $MBCD$ can be viewed as having right-angled $\mathrm{△}BCD$ as its base and $MG$ as its height.

   Thus, its volume equals ${\displaystyle \frac{1}{3}}({\displaystyle \frac{1}{2}}\cdot BC\cdot CD)(MG)={\displaystyle \frac{1}{6}}(2a{)}^{2}h={\displaystyle \frac{2}{3}}{a}^{2}h$.

   Therefore, the volume of solid $PABMD$, in terms of $a$ and $h$, equals ${\displaystyle \frac{8}{3}}{a}^{2}h-{\displaystyle \frac{2}{3}}{a}^{2}h=2a^{2}h.$

   Since the volume of $PABMD$ is $288$, we have $2a^{2}h=288$ or $a^{2}h=144$.

   We have not yet used the information that $\mathrm{\angle }BMD=90\mathrm{°}$.

   Since $\mathrm{\angle }BMD=90\mathrm{°}$, $\mathrm{△}BMD$ is right-angled at $M$ and so $B{D}^2=B{M}^2+M{D}^2$.

   By symmetry, $BM=MD$ and so $B{D}^2=2B{M}^2$.

   Since $\mathrm{△}BCD$ is right-angled at $C$, we have $B{D}^2=B{C}^2+C{D}^2=2(2a{)}^2=8a^2.$

   Since $\mathrm{△}BGM$ is right-angled at $G$, we have $B{M}^2=B{G}^2+M{G}^2=B{G}^2+h^2.$

   Since $\mathrm{△}BFG$ is right-angled at $F$ (the diagonals of square $ABCD$ are equal and perpendicular), it follows $$\begin{array}{rl}B{G}^2& =B{F}^2+F{G}^2\\ & ={\left(\frac{1}{2}BD\right)}^2+{\left(\frac{1}{4}AC\right)}^2\\ & =\frac{1}{4}B{D}^2+\frac{1}{16}A{C}^2\\ & =\frac{1}{4}B{D}^2+\frac{1}{16}B{D}^2\\ & =\frac{5}{16}B{D}^2\\ & =\frac{5}{2}{a}^2\end{array}$$ Since $2B{M}^2=B{D}^2$, it follows that $2(B{G}^2+h^2)=8a^2$ which gives ${\displaystyle \frac{5}{2}}{a}^2+h^2=4a^2$ or $h^2={\displaystyle \frac{3}{2}}{a}^2$ or $a^2={\displaystyle \frac{2}{3}}{h}^2$.

   Since $a^{2}h=144$, we have ${\displaystyle \frac{2}{3}}{h}^2\cdot h=144$ or $h^3=216$ which gives $h=6$. From $a^{2}h=144$, we obtain $6a^2=144$ or $a^2=24$.

   Since $a>0$, it follows that $a=2\sqrt{6}$ and so $AB=2a=4\sqrt{6}$.

Problem Set

1. In the diagram, $\mathrm{△}ABC$ is right-angled at $B$ and $\mathrm{△}ACD$ is right-angled at $A$. Also, $AB=3$,$BC=4$, and $CD=13$.

   

   What is the area of quadrilateral $ABCD$?

2. One of the faces of a rectangular prism has area $27{\text{ cm}}^2$. Another face has area $32{\text{ cm}}^2$. If the volume of the prism is $144{\text{ cm}}^3$, determine the surface area of the prism in ${\text{cm}}^2$.

3. The sum of the radii of two circles is $10$ cm. The circumference of the larger circle is $3$ cm greater than the circumference of the smaller circle. Determine the difference between the area of the larger circle and the area of the smaller circle.

4. In the diagram, $ABC$ is a quarter of a circular pizza with centre $A$ and radius $20$ cm. The piece of pizza is placed on a circular pan with $A$, $B$ and $C$ touching the circumference of the pan, as shown.

   

   What fraction of the pan is covered by the piece of pizza?

5. In rectangle $ABCD$, point $E$ is on side $DC$. Line segments $AE$ and $BD$ are perpendicular and intersect at $F$.

   

   If $AF=4$ and $DF=2$, determine the area of quadrilateral $BCEF$.

6. In the diagram, $ACDF$ is a rectangle with $AC=200$ and $CD=50$.

   

   Also, $\mathrm{△}FBD$ and $\mathrm{△}AEC$ are congruent triangles which are right-angled at $B$ and $E$, respectively. What is the area of the shaded region?

7. In the diagram, $\mathrm{△}XYZ$ is isosceles with $XY=XZ=a$ and $YZ=b$ where $b<2a$. A larger circle of radius $R$ is inscribed in the triangle (that is, the circle is drawn so that it touches all three sides of the triangle). A smaller circle of radius $r$ is drawn so that it touches $XY$, $XZ$ and the larger circle.

   

   Determine an expression for ${\displaystyle \frac{R}{r}}$ in terms of $a$ and $b$.

8. In the diagram, line segments $AC$ and $DF$ are tangent to the circle at $B$ and $E$, respectively. Also, $AF$ intersects the circle at $P$ and $R$, and intersects $BE$ at $Q$, as shown.

   

   If $\mathrm{\angle }CAF=35\mathrm{°}$, $\mathrm{\angle }DFA=30\mathrm{°}$, and $\mathrm{\angle }FPE=25\mathrm{°}$, determine the measure of $\mathrm{\angle }PEQ$.

9. In the diagram, $ABCD$ and $PNCD$ are squares of side length $2$, and $PNCD$ is perpendicular to $ABCD$. Point $M$ is chosen on the same side of $PNCD$ as $AB$ so that $\mathrm{△}PMN$ is parallel to $ABCD$, so that $\mathrm{\angle }PMN=90\mathrm{°}$, and so that $PM=MN$.

   

   Determine the volume of the convex solid $ABCDPMN$.

10. In the diagram, $ABCD$ is a parallelogram. Point $E$ is on $DC$ with $AE$ perpendicular to $DC$, and point $F$ is on $CB$ with $AF$ perpendicular to $CB$. If $AE=20$, $AF=32$, and $\cos (\mathrm{\angle }EAF)=13$, determine the exact value of the area of quadrilateral $AECF$.

    

11. In the diagram, $C$ lies on $BD$. Also, $\mathrm{△}ABC$ and $\mathrm{△}ECD$ are equilateral triangles. If $M$ is the midpoint of $BE$ and $N$ is the midpoint of $AD$, prove that $\mathrm{△}MNC$ is equilateral.

    

12. $ABCD$ is a trapezoid with parallel sides $AB$ and $DC$. Also, $BC$ is perpendicular to $AB$ and to $DC$. The line $PQ$ is parallel to $AB$ and divides the trapezoid into two regions of equal area. If $AB=x$, $DC=y$, and $PQ=r$, prove that $x^2+y^2=2r^2$.

    

13. Square $WXYZ$ has side length $6$ and is drawn, as shown, completely inside a larger square $EFGH$ with side length $10$, so that the squares do not touch and so that $WX$ is parallel to $EF$.

    

    Prove that the sum of the areas of trapezoid $EFXW$ and trapezoid $GHZY$ does not depend on the position of $WXYZ$ inside $EFGH$.