Congruence, Similarity and Geometrical Proof: past paper questions & worked solutions

The scale factor from the largest quadrilateral to the middle quadrilateral is \[ \frac{7}{28}=\frac14 \]

So the side corresponding to \(60\text{ cm}\) in the middle shape is \[ 60\times \frac14=15 \]

The scale factor from the middle quadrilateral to the smallest quadrilateral is \[ \frac{6}{18}=\frac13 \]

**Step 1: Identify corresponding sides from the similar shapes.** Looking at the given angles/orientations:

Left side *A**B* corresponds to *E**F* and *I**J*.

Top side *B**C* corresponds to *F**G* and *J**K*.

Right side *C**D* corresponds to *G**H* and *K**L*.

We are given *C**D* = 60 (wait, diagram check: *B**C* is the top side labeled 60, *A**B* = 28. On middle shape *E**F* = 7, *G**H* = 18. On right shape *K**L* = 6, *J**K* is unknown). Let’s use the explicit lengths: $$\\begin{aligned} AB = 28 \\quad &\\text{corresponds to} \\quad EF = 7 \\\\ BC = 60 \\quad &\\text{corresponds to} \\quad FG \\\\ CD \\quad &\\text{corresponds to} \\quad GH = 18 \\quad \\text{and} \\quad KL = 6 \\end{aligned}$$

**Step 2: Find scale factors.** Scale factor from *A**B**C**D* to *E**F**G**H* using the left sides: $$\\text{SF}\_1 = \\frac{EF}{AB} = \\frac{7}{28} = \\frac{1}{4}$$ So, $FG = BC \\times \\frac{1}{4} = 60 \\times \\frac{1}{4} = 15 \\text{ cm}$.

Scale factor from *E**F**G**H* to *I**J**K**L* using the right sides (*G**H* and *K**L*): $$\\text{SF}\_2 = \\frac{KL}{GH} = \\frac{6}{18} = \\frac{1}{3}$$

**Step 3: Calculate *J**K*.** Side *J**K* corresponds to *F**G*. $$JK = FG \\times \\text{SF}\_2 = 15 \\times \\frac{1}{3} = 5 \\text{ cm}$$

Since \(AB\) is parallel to \(DE\), triangles \(ACB\) and \(ECD\) are similar.

So corresponding sides are in the same ratio: \[ \frac{AC}{CE}=\frac{BC}{CD} \]

Substitute the lengths: \[ \frac{1.2}{6}=\frac{2x}{2x+9} \]

The scale factor from triangle \(ABC\) to triangle \(PQR\) is \[ \frac{PQ}{AB}=\frac{12}{4}=3 \]

So \[ BC=\frac{RQ}{3}=\frac{16.5}{3}=5.5 \]

Also \(PR\) corresponds to \(AC\), so \[ y=3x \]

Each congruent isosceles triangle has two angles of \(66^\circ\).

So the angle at the point between the two polygons is \[ 180^\circ-66^\circ-66^\circ=48^\circ \]

The two regular polygons have equal interior angles. Around the point: \[ 48^\circ+I+I=360^\circ \]

The scale factor from \(ABCD\) to \(EFGH\) is \[ \frac{FG}{BC}=\frac{10}{4}=2.5 \]

Since \(AB\) corresponds to \(EF\), \[ x=EF=5\times 2.5=12.5 \]

Since \(CD\) corresponds to \(GH\), \[ 2.5y=24 \]

The scale factor from \(ABCD\) to \(EFGH\) is

So

and

Since \(BE\parallel CD\), triangles \(ABE\) and \(ACD\) are similar.

Also \[ CD=1.5BE \] so \[ \frac{BE}{CD}=\frac{1}{1.5}=\frac23 \]

Therefore \[ \frac{AE}{AD}=\frac23 \]

Given \(AE=10\), \[ \frac{10}{AD}=\frac23 \]

So \[ ED=AD-AE=15-10=5 \]

For part (b), \[ AB=2x+5, \qquad BC=3x-5 \]

Using similarity, \[ \frac{AB}{AC}=\frac23 \]

The scale factor from \(EFGH\) to \(ABCD\) is \[ \frac{9.6}{6}=1.6 \]

Since \(GH\) corresponds to \(CD\), \[ GH=\frac{4}{1.6}=2.5 \]

Since \(FG\) corresponds to \(BC\), \[ BC=5.7\times 1.6=9.12 \]

The scale factor from triangle \(DEF\) to triangle \(ABC\) is \[ \frac{AC}{DF}=\frac{45}{20}=2.25 \]

Given \(BC=54\), \[ EF=\frac{54}{2.25}=24 \]

In the larger triangle, \[ DE^2=51^2-24^2 \]

The scale factor from the large triangle to the small triangle is \[ \frac{7.5}{45}=\frac16 \]

So \[ x=24\times \frac16=4 \]