Triangles: questions with answers

Join $A$ to $R$, meeting the middle parallel at $X$. In $\triangle ACR$, $BX \parallel CR$ gives $\frac{AB}{BC} = \frac{AX}{XR}$. In $\triangle RPA$, $XQ \parallel PA$ gives $\frac{RX}{XA} = \frac{RQ}{QP}$. The ratio on $AR$ is common to both, so it cancels, giving $\frac{AB}{BC} = \frac{PQ}{QR}$ and $QR = 32$ cm. Without this join there are only two trapeziums, and no triangle gives $QR = 8$ cm directly.

In $\triangle ADC$ the line $EG$ is parallel to $DC$, so $\frac{AE}{ED} = \frac{AG}{GC}$. In $\triangle CAB$ the line $GF$ is parallel to $AB$, so $\frac{CG}{GA} = \frac{CF}{FB}$. The ratio on the diagonal appears in both and cancels, leaving $\frac{AE}{ED} = \frac{BF}{FC}$ and so $FC = 20$ cm. Neither $6$ nor $12$ enters the answer at all: they fix the shape of the trapezium, not how $EF$ divides its slanted sides. Using them anyway -- $FC = BF \times \frac{DC}{AB}$ -- gives $16$ cm, and stopping after the first triangle, where nothing about $F$ has yet been said, gives $FC = BF = 8$ cm.

$\frac{BC}{ FD }$. Every pair of corresponding sides in a similarity has the same ratio. $B \to F$ and $C \to D$, so $BC$ goes with $FD$. The rotation is only a fact about the drawing. It changes nothing about the correspondence, which the letters fixed before any drawing was made. The drawn pairing looks tempting because it is what the eye sees first. Here that gives $\frac{BC}{ EF }$, a ratio between two sides that do not correspond.

$\angle B = \angle Y$. The same question as its sibling and a different answer, because the ratios name different sides: $AB$ and $BC$ meet at $B$, not at $A$. Answering $\angle A = \angle X$ here -- the sibling's answer -- is what happens when the vertex is remembered rather than located, and it is why the pair is worth doing together. The rule never changes: find the vertex the two named sides share, in each triangle, and that is the angle SAS requires.

The stem gives the WHOLE side $AB$, not the lower piece, so a subtraction comes first: $DB = 9 - 2 = 7$ cm. Then $\frac{AD}{DB} = \frac{AE}{EC}$ gives $EC = 14$ cm. Using $9$ where $7$ belongs gives $18$ -- the answer to $\frac{AD}{AB} = \frac{AE}{EC}$, which pairs a part of one side with the whole of the other and is a valid form of nothing. Note that $AB$ is not twice $AD$ here, so forgetting the subtraction and assuming the line bisects give different answers, and the two mistakes stay distinguishable.

The theorem itself says nothing about $DE$ -- it is about the two sides the line CUTS. $DE$ comes from the similarity of $\triangle ADE$ and $\triangle ABC$, which share the angle at $A$ and have equal corresponding angles at $D$ and $B$: $\frac{DE}{BC} = \frac{AD}{AB}$. The denominator is the WHOLE side $AB = 14$, not the lower piece, so $DE = 12$ cm. Using $DB$ there gives $21$, and using $\frac{DB}{AB}$ gives $16$ -- the fraction of $BC$ cut off by the wrong part of the side.

Nothing in the stem says $EF \parallel BC$ -- that has to be EARNED. The two ratios agree, so the converse of the proportionality theorem gives $EF \parallel BC$, and only then does $\triangle AEF \sim \triangle ABC$ follow by AA. Now the scale factor: it is $\frac{AE}{AB}$, a part over the WHOLE, and $\frac{AE}{EB} = \frac{ 3 }{ 2 }$ makes that $\frac{ 3 }{ 5 }$. So $BC = \frac{ 5 }{ 3 } \times 6 = 10$ cm. Using $\frac{ 3 }{ 2 }$ itself as the scale factor gives $4$ cm. That is the answer most often written down, because the given ratio sits right there in the stem.

Set $\frac{AD}{DB}=\frac{AE}{EC}$. Substituting gives $\frac{ 6 }{ 10 }=\frac{ 6 }{EC}$, so $EC=\frac{ 6\times 10 }{ 6 }=10$ cm. The ratios are equal, so the converse of the Basic Proportionality Theorem gives $DE\parallel BC$. Comparing $AD/DB$ with $AE/AC$ mixes a part-to-part ratio with a part-to-whole ratio.

Sides opposite equal angles of a triangle are equal. Here, $AC$ is opposite $\angle B$, and $AB$ is opposite $\angle C$. Since $\angle B = \angle C$, $AB = AC$. Appearance does not prove equality. The fact about angles opposite equal sides starts with equal sides, which are what we need to prove here.

The reason is the converse. The ratio is given and the parallel is what must be proved, so the statement needed is the one whose hypothesis is the ratio. Naming the Basic Proportionality Theorem would assume $DE \parallel BC$ to prove $DE \parallel BC$. The ratio $\frac{ 3 }{ 7 }$ appears in both statements, so check which fact the stem gives you first.

The ratios $\frac{AB}{PQ}$ and $\frac{AC}{PR}$ are equal. The included angles, $\angle A$ and $\angle P$, are also equal. Therefore, $\triangle ABC \sim \triangle PQR$ by SAS similarity. SAS requires proportional corresponding sides, so the ratio $\frac{ 3 }{ 5 }$ need not equal $1$. AA cannot be applied directly because only one pair of equal angles is given.

The proportional sides meet at $B$ and $E$, so their included angles are $\angle B$ and $\angle E$. Their equality is not given. The given equal angles, $\angle A$ and $\angle D$, are not between these sides. SAS similarity requires proportional corresponding sides and equal included angles; it does not require equal side lengths.

Every angle of an equilateral triangle is $60^\circ$, so the corresponding angles are equal. All three sides of the first are $3$ cm and all three of the second are $9$ cm, so every pair of corresponding sides is in the ratio $3:9$. Both conditions are met, so the triangles are similar. They are not congruent, since $3$ cm and $9$ cm differ.

Each object, its shadow and the ray of sunlight form a right triangle. The two triangles are similar because both have a right angle at the ground and the sun's rays meet both at the same angle. So each height goes with its own shadow: $\frac{ 6 }{ 9 } = \frac{h}{ 18 }$, giving $h = 12$ m. Turning the proportion over gives $3$ m, which is shorter than the pole. That is wrong since the tower's shadow is the longer of the two. Pairing the pole's height against the TOWER's shadow gives $27$ m.

The perimeter of $\triangle ABC$ is $15 + 20 + 25 = 60$ cm. Since $AB$ corresponds to $PQ$, the scale factor is $\frac{ 60 }{ 15 } = 4$. The perimeters of similar triangles are in the same ratio as their corresponding sides. Therefore, the perimeter of $\triangle PQR$ is $4 \times 60 = 240$ cm. Squaring the scale factor incorrectly gives $960$ cm. Dividing by the scale factor instead of multiplying incorrectly gives $15$ cm.