We think you are located in United States. Is this correct?

Solving more geometric problems

12.6 Solving more geometric problems

When solving geometry problems involving lines, you need to use some facts about quadrilaterals and triangles.

  • The sum of the interior angles of a triangle is \(180^\circ\).
  • The exterior angle of a triangle is equal to the sum of the interior opposite angles.
exterior angle of a triangle
the angle formed outside a triangle when one of its sides is produced

  • The interior angles of a quadrilateral add up to \(360^{\circ}\).
    \(\hat{P}+\hat{Q}+\hat{R}+\hat{S}= 360^{\circ}\)
interior angle
an angle that lies inside a shape

Worked example 12.9: Identifying angles on parallel lines

Calculate the size of angles \(a\), \(b\), \(c\) and \(d\).

Examine the parallel lines. Identify how many transversals there are.

Notice that there are two traversals: \(SQ\) and \(UR\). Now look for the angles that form F or Z pattern along each transversal.

  • Transversal \(UR\): We can see that \(b\) is a corresponding angle to \(48^{\circ}\), making an F pattern. There are no alternate angles formed by this transversal.
  • Transversal \(SQ\): We can see that \(a\) is alternate to \(c\), making a Z pattern. There are no corresponding angles formed by this transversal.

Look for any adjacent supplementary angles that can help you find one of the angles.

Angle \(c\) is an adjacent supplementary angle to 154°.

Write out your solution.

\[\begin{align} b & = 48^{\circ}&&(\text{corresp }\angle\text{s; } TS \parallel PR) \\ c & = 180^{\circ} - 154^{\circ}&&(\angle\text{s on a straight line}) \\ c & = 26^{\circ} \\ a & = 26^{\circ}&&(\text{alt }\angle\text{s; } TS \parallel PR) \\ d & = 180^{\circ}-48^{\circ} - 26^{\circ} &&(\angle\text{s on a straight line}) \\ d & = 106^{\circ} \end{align}\]

Worked example 12.10: Using angles on parallel lines

Calculate the size of \(x\).

The two given angles lie between parallel lines, and are corresponding angles, therefore they are equal. We will use this fact to form an equation to help us solve for \(x\).

\[\begin{align} x+50^{\circ} &= 3x-10^{\circ} &&(\text{corresp }\angle\text{s}) \\ x-3x &= -10^{\circ} -50^{\circ} \\ -2x &= -60^{\circ} \\ x &= 30^{\circ} \end{align}\]

Worked example 12.11: Using angles on parallel lines

Calculate the size of \(x\).

Notice that there is only one transversal in this diagram. The two given angles make a C pattern.

The two angles form co-interior angles, which add up to \(180^{\circ}\), because the lines are parallel. So we can write out an equation using this information.

\[\begin{align} 2x+3x &= 180^{\circ} &&(\angle\text{s on a straight line}) \\ 5x &= 180^{\circ} \\ x &= 30^{\circ} \\ \end{align}\]

Worked example 12.12: Solving geometric problems involving properties of triangles

Calculate the sizes of \(\hat{1}\) to \(\hat{6}\).

Examine the diagram carefully. Look for the parallel lines and all transversals.

Notice that there are two transversals; \(AD\) and \(OB\):

  • Looking at \(AD\), notice that \(\hat{2}\) and \(\hat{6}\) are alternate angles and therefore equal. We also see that \(\hat{1}\) and \(\hat{6}\) are corresponding angles and therefore equal.
  • Looking at \(OB\), we notice that \(\hat{4}\) and \(\hat{5}\) are alternate angles.

Look for all the adjacent angles on a straight line.

You may need to use the fact they are supplementary later in your calculation. How many supplementary angles can you find?

The supplementary angles are:

\[\begin{align} 132^\circ+\hat{2}&=180^\circ \\ 132° + \hat{1}&=180^\circ \\ \hat{2}+\hat{3} + \hat{4}&=180^\circ \\ \hat{1} + \hat{3}+ \hat{4}&=180^\circ \end{align}\]

Look for vertically opposite angles.

How many can you find? (\(\hat{1}\) is vertically opposite \(\hat{2}\)).

Write out your solution.

\[\begin{align} \hat{1}&= 180^{\circ} -132^{\circ} &&(\angle\text{s on a straight line}) \\ \hat{1}&= 48^{\circ} \\ \hat{2}&= 48^{\circ} &&(\text{vert opp }\angle\text{s}) \\ \hat{6}&= 48^{\circ} &&(\text{alt }\angle\text{s; KT} \parallel {AB}) \end{align}\]

Hint: Remember that in an isoceles triangle, two sides are the same length.

Because \(\triangle ABO\) is an isoceles triangle:

\[\begin{align} \hat{5}&= 48^{\circ} &&(\angle\text{s equal opp sides}) \\ \hat{3} &= 180^{\circ}- \hat{5}- \hat{6} &&(\angle\text{s in a }\triangle) \\ \hat{3} &= 180^{\circ}-48^{\circ}-48^{\circ} \\ \hat{3} &= 84^{\circ} \end{align}\]

Worked example 12.13: Solving geometric problems involving properties of triangles and quadrilaterals

\(ABCD)\ is a parallelogram. Calculate the sizes of \(A\hat{B}D\), \(A\hat{D}B\), \(\hat{C}\) and \(C\hat{B}D\).

Use the properties of the parallelogram \(ABCD\) to expand the given information.

Both pairs of opposite angles are equal in a parallelogram, therefore we can say that \(\hat{A}=\hat{C}\).

Look for any alternate or corresponding angles for both sets of parallel lines. We also want to make of a note of any co-interior angles formed by the parallel lines. These supplementary angles may also be useful in helping us to find the missing angles.

We are given two sets of parallel lines: \(AD \parallel BC\) and \(DC \parallel AB\).

  • \(DC \parallel AB\): Looking at transversal \(BD\), we notice that \(20^{\circ}\) and \(\hat{DBA}\) are alternate angles formed by these parallel lines, and therefore are equal.
  • \(AD \parallel BC\): Looking at transversal \(BD\), we notice that \(\hat{ADB}\) and \(\hat{DBC}\) are alternate angles formed by these parallel lines, and therefore are equal.

Use the properties of triangles to find angles.

We are also given two triangles: \(\triangle{ABD}\) and \(\triangle{DBC}\). We can use the fact that the sum of \(\angle\)s of a \(\triangle\) = \(180^\circ\) to help find remaining angles. 

Write out your solution.

\[\begin{align} \hat{ABD}&= 20^{\circ} &&(\text{alt }\angle\text{s; AB} \parallel {DC}) \\ \hat{ADB}&= 180^{\circ} -20^{\circ}-82^{\circ} &&(\angle\text{s in a }\triangle) \\ \hat{ADB}&= 78^{\circ} \\ \hat{C} &= 82^{\circ} &&(\text{opp sides of parallelogram}) \\ \hat{CBD}&= 78^{\circ} &&(\text{alt }\angle\text{s; AB} \parallel {DC}) \end{align}\]
Exercise 12.7

\(RSTU\) is a trapezium. Calculate the sizes of \(\hat{T}\) and \(\hat{R}\).

\[\begin{align} \hat{R} &=180°-143° &&(\text{co-int}\angle\text{s}; UT \parallel RS) \\ \hat{R} &= 37° \\ \hat{T} &=180°-112° &&(\text{co-int}\angle\text{s}; UT \parallel RS) \\ \hat{T} &= 68° \end{align}\]

\(JKLM\) is a rhombus. Calculate the sizes of of \(J\hat{M}L\), \(\hat{M}_2\) and \(\hat{K}_1\).

Hint: Remember, \(JKLM\) is a rhombus, therefore all four sides of \(JKLM\) are equal.

\[\begin{align} \hat{J} &=102° &&(\text{opp }\angle\text{s of rhombus)} \\ \hat{M_2} &= \hat{K_2} &&(\text{opp }\angle\text{s; } \text { opp = sides; ML = KL}) \\ \hat{M_2} + \hat{K_2} +102° &= 180° \\ \hat{M_2} + \hat{K_2} +102° &= 180° - 102° \\ \hat{M_2} + \hat{K_2} +102° &= 78° \\ \hat{M_2} = \hat{K_2} &= 39° \\ \text{Similarly, } \hat{M_1} = \hat{K_1} &= 39° \\ J\hat{M}L &= \hat{M_1} + \hat{M_2} \\ &= 78° \end{align}\]

In the diagram, \(AB \parallel CD\). Calculate the sizes of \(F\hat{H}G, \hat{F},\hat{C}\) and \(\hat{D}\). Give reasons for your answers.

\[\begin{align} \hat{FHG} &=53° && (\text{vert opp }\angle\text{s}) \\ \hat{F}+ \hat{FHG}+ 90° &=180°&&(\text{sum of }\angle\text{s in } \triangle )\\ \hat{F}+53°+ 90° &=180° \\ \hat{F}&=180° -90°-53° \\ \hat{F}&=37° \\ \\ \hat{CIE} &= 53° &&(\text{corresp }\angle\text{s}; AB \parallel CD) \\ \hat{C}+ \hat{E}+ \hat{CIE} &=180°&&(\text{sum of }\angle\text{s in } \triangle)\\ \hat{C}+85°+53° &=180° \\ \hat{C}&=180° -85°-53° \\ \hat{C}&=42° \\ \\ \hat{AGD} &= 180°-110 &&(\angle\text{s on a str line})\\ \hat{AGD} &= 70° \\ \hat{D} &= 180°-70° &&(\text{co-int }\angle\text{s}; AB \parallel CD) \\ \hat{D} &= 110° \end{align}\]