Hard GCSE Similarity Questions

Challenging, exam-style GCSE Similarity questions with worked solutions. Stretch yourself on the hardest similar triangles, parallel lines, missing length, alternate angles problems.

similar trianglesparallel linesmissing lengthalternate anglesratioscale factor
GCSE Foundation34 questionsStep-by-step solutions
Question 1
5 markschallenging
In triangles ABCABC and PQRPQR, angle BAC=55∘BAC = 55^\circ, angle ABC=60∘ABC = 60^\circ, angle QPR=55∘QPR = 55^\circ and angle PQR=70∘PQR = 70^\circ. No side lengths are given. Which of these statements about triangles ABCABC and PQRPQR is correct?
Show worked solution

Worked solution

  1. Recall the angle test for similar triangles

    AA\text{AA}

    If two pairs of corresponding angles are equal the triangles are similar. No side lengths are needed at all.

  2. Recall the angle sum of a triangle

    a+b+c=180∘a + b + c = 180^\circ

    This lets you fill in the third angle of each triangle before comparing them.

  3. Note what the question gives you

    four angles, no lengths\text{four angles, no lengths}

    Only angles are given, so any reason that talks about the ratio of sides cannot be used here.

  4. Plan the comparison

    complete both triangles→compare\text{complete both triangles} \rightarrow \text{compare}

    Work out the missing angle in each triangle, then compare the two sets of three.

  5. Note that the correspondence matters

    A↔P,B↔Q,C↔RA \leftrightarrow P, \quad B \leftrightarrow Q, \quad C \leftrightarrow R

    The letters pair the vertices, so compare angle BACBAC with angle QPRQPR, not with angle PQRPQR.

  6. Work out the third angle of triangle ABC

    ∠ACB=180−55−60=65\angle ACB = 180 - 55 - 60 = 65

    The angles of a triangle add to 180∘180^\circ, so angle ACB=65∘ACB = 65^\circ.

  7. Work out the third angle of triangle PQR

    ∠PRQ=180−55−70=55\angle PRQ = 180 - 55 - 70 = 55

    In the same way angle PRQ=55∘PRQ = 55^\circ.

  8. Compare the two sets of three angles

    {55,60,65}≠{55,70,55}\{55, 60, 65\} \ne \{55, 70, 55\}

    The two triangles do not have the same three angles, so they are not the same shape.

  9. State the correct conclusion

    ∠ACB=65≠55=∠PRQ\angle ACB = 65 \ne 55 = \angle PRQ

    The third angles differ, so the triangles cannot be similar.

  10. Rule out the SSS reason

    no side lengths are given\text{no side lengths are given}

    SSS needs the three pairs of sides to be in the same ratio, but the question gives no side lengths at all, so this reason cannot be used.

  11. Rule out the SAS reason

    SAS needs side lengths too\text{SAS needs side lengths too}

    SAS needs two pairs of sides in the same ratio, and again there are no side lengths in the question, so this reason is not available either.

  12. Rule out the congruence claim

    equal angles≠equal size\text{equal angles} \ne \text{equal size}

    Doubling every side of a triangle keeps all three angles the same but changes the triangle, so equal angles alone can never prove congruence.

  13. Rule out the remaining option

    ∠BAC=55,∠QPR=55\angle BAC = 55, \quad \angle QPR = 55

    The claim that two pairs of angles are equal is false here: angle ABC=60∘ABC = 60^\circ but angle PQR=70∘PQR = 70^\circ.

  14. Note that similar shapes can be any size

    same angles⇒same shape\text{same angles} \Rightarrow \text{same shape}

    Similarity fixes the shape, not the size — that is the whole point of a scale factor.

  15. State the option you have chosen

    △ABC≁△PQR\triangle ABC \not\sim \triangle PQR

    So the triangles are not similar, because their third angles are 65∘65^\circ and 55∘55^\circ, which are different.

Answer
△ABC≁△PQR\triangle ABC \not\sim \triangle PQR
Question 2
5 markschallenging
In triangles ABCABC and PQRPQR, angle BAC=72∘BAC = 72^\circ, angle ABC=44∘ABC = 44^\circ, angle QPR=72∘QPR = 72^\circ and angle PQR=44∘PQR = 44^\circ. No side lengths are given. Which of these statements about triangles ABCABC and PQRPQR is correct?
Show worked solution

Worked solution

  1. Recall the angle test for similar triangles

    AA\text{AA}

    If two pairs of corresponding angles are equal the triangles are similar. No side lengths are needed at all.

  2. Recall the angle sum of a triangle

    a+b+c=180∘a + b + c = 180^\circ

    This lets you fill in the third angle of each triangle before comparing them.

  3. Note what the question gives you

    four angles, no lengths\text{four angles, no lengths}

    Only angles are given, so any reason that talks about the ratio of sides cannot be used here.

  4. Plan the comparison

    complete both triangles→compare\text{complete both triangles} \rightarrow \text{compare}

    Work out the missing angle in each triangle, then compare the two sets of three.

  5. Note that the correspondence matters

    A↔P,B↔Q,C↔RA \leftrightarrow P, \quad B \leftrightarrow Q, \quad C \leftrightarrow R

    The letters pair the vertices, so compare angle BACBAC with angle QPRQPR, not with angle PQRPQR.

  6. Work out the third angle of triangle ABC

    ∠ACB=180−72−44=64\angle ACB = 180 - 72 - 44 = 64

    The angles of a triangle add to 180∘180^\circ, so angle ACB=64∘ACB = 64^\circ.

  7. Work out the third angle of triangle PQR

    ∠PRQ=180−72−44=64\angle PRQ = 180 - 72 - 44 = 64

    In the same way angle PRQ=64∘PRQ = 64^\circ.

  8. Compare the two sets of three angles

    {72,44,64}={72,44,64}\{72, 44, 64\} = \{72, 44, 64\}

    The two triangles have the same three angles, so they are the same shape.

  9. State the correct reason

    ∠BAC=∠QPR,∠ABC=∠PQR⇒△ABC∼△PQR\angle BAC = \angle QPR, \quad \angle ABC = \angle PQR \Rightarrow \triangle ABC \sim \triangle PQR

    Two pairs of equal corresponding angles is exactly the AA test, so the triangles are similar.

  10. Rule out the SSS reason

    no side lengths are given\text{no side lengths are given}

    SSS needs the three pairs of sides to be in the same ratio, but the question gives no side lengths at all, so this reason cannot be used.

  11. Rule out the SAS reason

    SAS needs side lengths too\text{SAS needs side lengths too}

    SAS needs two pairs of sides in the same ratio, and again there are no side lengths in the question, so this reason is not available either.

  12. Rule out the congruence claim

    equal angles≠equal size\text{equal angles} \ne \text{equal size}

    Doubling every side of a triangle keeps all three angles the same but changes the triangle, so equal angles alone can never prove congruence.

  13. Rule out the remaining option

    ∠ACB=64=∠PRQ\angle ACB = 64 = \angle PRQ

    The third angles are both 64∘64^\circ, so the claim that they are different is false.

  14. Note that similar shapes can be any size

    same angles⇒same shape\text{same angles} \Rightarrow \text{same shape}

    Similarity fixes the shape, not the size — that is the whole point of a scale factor.

  15. State the option you have chosen

    △ABC∼△PQR(AA)\triangle ABC \sim \triangle PQR \quad (\text{AA})

    So the triangles are similar, and the reason is AA: two pairs of corresponding angles are equal.

Answer
△ABC∼△PQR(AA)\triangle ABC \sim \triangle PQR \quad (\text{AA})
Question 3
5 markschallenging
Triangle ABCABC has sides AB=9AB = 9 cm, BC=12BC = 12 cm and AC=15AC = 15 cm. Triangle PQRPQR has sides PQ=12PQ = 12 cm, QR=16QR = 16 cm and PR=21PR = 21 cm. AA corresponds to PP, BB to QQ and CC to RR. Which statement is true?
Show worked solution

Worked solution

  1. Recall the side test for similar triangles

    PQAB=QRBC=PRAC\frac{PQ}{AB} = \frac{QR}{BC} = \frac{PR}{AC}

    Two triangles are similar if all three pairs of corresponding sides are in the same ratio (SSS). If even one ratio differs they are not similar.

  2. Write down the pairs of corresponding sides

    AB↔PQ,BC↔QR,AC↔PRAB \leftrightarrow PQ, \quad BC \leftrightarrow QR, \quad AC \leftrightarrow PR

    The letters give the correspondence, so pair the sides that join matching vertices.

  3. Plan the check

    three ratios→compare\text{three ratios} \rightarrow \text{compare}

    Work out each ratio as an exact fraction and see whether all three agree.

  4. Keep the ratios exact

    fractions, not decimals\text{fractions, not decimals}

    A ratio like 54\frac{5}{4} and a ratio like 43\frac{4}{3} are close as decimals but they are not equal, so work in fractions.

  5. Note that the order matters

    A↔P,B↔Q,C↔RA \leftrightarrow P, \quad B \leftrightarrow Q, \quad C \leftrightarrow R

    Dividing the wrong pair of sides is the usual way this check goes wrong.

  6. Work out the ratio of the first pair of corresponding sides

    PQAB=129=43\frac{PQ}{AB} = \frac{12}{9} = \frac{4}{3}

    ABAB and PQPQ are corresponding sides, so their ratio is 43\frac{4}{3}.

  7. Work out the ratio of the second pair

    QRBC=1612=43\frac{QR}{BC} = \frac{16}{12} = \frac{4}{3}

    BCBC and QRQR are corresponding sides, so their ratio is 43\frac{4}{3}.

  8. Work out the ratio of the third pair

    PRAC=2115=75\frac{PR}{AC} = \frac{21}{15} = \frac{7}{5}

    ACAC and PRPR are corresponding sides, so their ratio is 75\frac{7}{5}.

  9. Compare the three ratios and decide

    43,43,75\frac{4}{3}, \quad \frac{4}{3}, \quad \frac{7}{5}

    The ratios are not all the same — 43\frac{4}{3} against 75\frac{7}{5} — so the triangles are not similar.

  10. Rule out wrong option number 1

    PRAC=75≠43\frac{PR}{AC} = \frac{7}{5} \ne \frac{4}{3}

    A scale factor of 43\frac{4}{3} from ABCABC to PQRPQR would need PRAC=43\frac{PR}{AC} = \frac{4}{3}, but that ratio is really 75\frac{7}{5}, so the option is false.

  11. Rule out wrong option number 2

    PQAB=43≠75\frac{PQ}{AB} = \frac{4}{3} \ne \frac{7}{5}

    A scale factor of 75\frac{7}{5} from ABCABC to PQRPQR would need PQAB=75\frac{PQ}{AB} = \frac{7}{5}, but that ratio is really 43\frac{4}{3}, so the option is false.

  12. Rule out wrong option number 3

    ACPR=57≠34\frac{AC}{PR} = \frac{5}{7} \ne \frac{3}{4}

    A scale factor of 34\frac{3}{4} from PQRPQR to ABCABC would need ACPR=34\frac{AC}{PR} = \frac{3}{4}, but that ratio is really 57\frac{5}{7}, so the option is false.

  13. Rule out wrong option number 4

    AB=9≠12=PQAB = 9 \ne 12 = PQ

    Congruent means identical, so every pair of corresponding sides would have to be equal. Here AB=9AB = 9 cm but PQ=12PQ = 12 cm, so it is false.

  14. Note the common mistake

    PQAB≠ABPQ\frac{PQ}{AB} \ne \frac{AB}{PQ}

    The scale factor must be new over old. Turning the fraction the wrong way up is the most common error.

  15. State the option you have chosen

    not similar\text{not similar}

    The three ratios are not all equal, so the triangles are not similar.

Answer
not similar\text{not similar}
Question 4
5 markschallenging
Triangle ABCABC has sides AB=8AB = 8 cm, BC=10BC = 10 cm and AC=14AC = 14 cm. Triangle PQRPQR has sides PQ=12PQ = 12 cm, QR=15QR = 15 cm and PR=21PR = 21 cm. AA corresponds to PP, BB to QQ and CC to RR. Which statement is true?
Show worked solution

Worked solution

  1. Recall the side test for similar triangles

    PQAB=QRBC=PRAC\frac{PQ}{AB} = \frac{QR}{BC} = \frac{PR}{AC}

    Two triangles are similar if all three pairs of corresponding sides are in the same ratio (SSS). If even one ratio differs they are not similar.

  2. Write down the pairs of corresponding sides

    AB↔PQ,BC↔QR,AC↔PRAB \leftrightarrow PQ, \quad BC \leftrightarrow QR, \quad AC \leftrightarrow PR

    The letters give the correspondence, so pair the sides that join matching vertices.

  3. Plan the check

    three ratios→compare\text{three ratios} \rightarrow \text{compare}

    Work out each ratio as an exact fraction and see whether all three agree.

  4. Keep the ratios exact

    fractions, not decimals\text{fractions, not decimals}

    A ratio like 54\frac{5}{4} and a ratio like 43\frac{4}{3} are close as decimals but they are not equal, so work in fractions.

  5. Note that the order matters

    A↔P,B↔Q,C↔RA \leftrightarrow P, \quad B \leftrightarrow Q, \quad C \leftrightarrow R

    Dividing the wrong pair of sides is the usual way this check goes wrong.

  6. Work out the ratio of the first pair of corresponding sides

    PQAB=128=32\frac{PQ}{AB} = \frac{12}{8} = \frac{3}{2}

    ABAB and PQPQ are corresponding sides, so their ratio is 32\frac{3}{2}.

  7. Work out the ratio of the second pair

    QRBC=1510=32\frac{QR}{BC} = \frac{15}{10} = \frac{3}{2}

    BCBC and QRQR are corresponding sides, so their ratio is 32\frac{3}{2}.

  8. Work out the ratio of the third pair

    PRAC=2114=32\frac{PR}{AC} = \frac{21}{14} = \frac{3}{2}

    ACAC and PRPR are corresponding sides, so their ratio is 32\frac{3}{2}.

  9. Compare the three ratios and decide

    32=32=32=32\frac{3}{2} = \frac{3}{2} = \frac{3}{2} = \frac{3}{2}

    All three ratios are equal, so the triangles are similar (SSS) with scale factor 32\frac{3}{2} from triangle ABCABC to triangle PQRPQR.

  10. Rule out wrong option number 1

    PQAB=32≠23\frac{PQ}{AB} = \frac{3}{2} \ne \frac{2}{3}

    A scale factor of 23\frac{2}{3} from ABCABC to PQRPQR would need PQAB=23\frac{PQ}{AB} = \frac{2}{3}, but that ratio is really 32\frac{3}{2}, so the option is false.

  11. Rule out wrong option number 2

    ABPQ=23≠32\frac{AB}{PQ} = \frac{2}{3} \ne \frac{3}{2}

    A scale factor of 32\frac{3}{2} from PQRPQR to ABCABC would need ABPQ=32\frac{AB}{PQ} = \frac{3}{2}, but that ratio is really 23\frac{2}{3}, so the option is false.

  12. Rule out wrong option number 3

    32=32=32\frac{3}{2} = \frac{3}{2} = \frac{3}{2}

    This option says the ratios differ, but all three ratios came out as 32\frac{3}{2}, so it is false.

  13. Rule out wrong option number 4

    AB=8≠12=PQAB = 8 \ne 12 = PQ

    Congruent means identical, so every pair of corresponding sides would have to be equal. Here AB=8AB = 8 cm but PQ=12PQ = 12 cm, so it is false.

  14. Note the common mistake

    PQAB≠ABPQ\frac{PQ}{AB} \ne \frac{AB}{PQ}

    The scale factor must be new over old. Turning the fraction the wrong way up is the most common error.

  15. State the option you have chosen

    similar,k=32\text{similar}, \quad k = \frac{3}{2}

    All three ratios equal 32\frac{3}{2}, so the triangles are similar with scale factor 32\frac{3}{2} from triangle ABCABC to triangle PQRPQR.

Answer
similar,k=32\text{similar}, \quad k = \frac{3}{2}
Question 5
6 markschallenging
Triangle ABCABC is similar to triangle PQRPQR, with AA corresponding to PP, BB to QQ and CC to RR. AB=12AB = 12 cm, BC=15BC = 15 cm, AC=21AC = 21 cm and PQ=8PQ = 8 cm. Work out the perimeter of triangle PQRPQR.
Show worked solution

Worked solution

  1. Write down the pairs of corresponding sides

    AB↔PQ,BC↔QR,AC↔PRAB \leftrightarrow PQ, \quad BC \leftrightarrow QR, \quad AC \leftrightarrow PR

    The letters give the correspondence, so ABAB and PQPQ are a corresponding pair.

  2. Recall how a perimeter behaves under an enlargement

    PPQR=k×PABCP_{PQR} = k \times P_{ABC}

    A perimeter is a sum of lengths, and every length is multiplied by kk, so the perimeter is multiplied by kk as well.

  3. Write that out to see why

    PQ+QR+PR=k×AB+k×BC+k×ACPQ + QR + PR = k \times AB + k \times BC + k \times AC

    Factorising the right-hand side gives k(AB+BC+AC)k(AB + BC + AC), which is kk times the perimeter of triangle ABCABC.

  4. Decide which triangle you are enlarging from

    ABC→PQRABC \rightarrow PQR

    The perimeter of triangle ABCABC is known and the perimeter of triangle PQRPQR is wanted, so the scale factor must run that way.

  5. Say why one pair of sides is enough

    k=PQABk = \frac{PQ}{AB}

    One pair of corresponding sides fixes the scale factor for every length in the figure.

  6. Work out the scale factor from triangle ABC to triangle PQR

    k=PQAB=812=23k = \frac{PQ}{AB} = \frac{8}{12} = \frac{2}{3}

    Dividing the new length by the old one gives k=23k = \frac{2}{3}.

  7. Add the three sides to get the perimeter of triangle ABC

    PABC=12+15+21=48P_{ABC} = 12 + 15 + 21 = 48

    The perimeter of triangle ABCABC is 4848 cm.

  8. Multiply the perimeter by the scale factor

    PPQR=48×23P_{PQR} = 48 \times \frac{2}{3}

    Every length is 23\frac{2}{3} times as long, so the perimeter is too.

  9. Work out the perimeter of triangle PQR

    PPQR=32 cmP_{PQR} = 32 \text{ cm}

    So the perimeter of triangle PQRPQR is 3232 cm.

  10. Check the ratio of the two perimeters

    PPQRPABC=3248=23\frac{P_{PQR}}{P_{ABC}} = \frac{32}{48} = \frac{2}{3}

    The perimeters are in the same ratio as the sides — the same scale factor appears again.

  11. Note that the perimeter uses the length scale factor

    P→kPP \rightarrow kP

    Perimeter is a length, so it scales by kk. Only areas would scale by kk squared.

  12. Write the scale factor the other way round

    ABPQ=128=32\frac{AB}{PQ} = \frac{12}{8} = \frac{3}{2}

    Going back the other way the scale factor is 32\frac{3}{2}, and 32×32=4832 \times \frac{3}{2} = 48.

  13. Note the common mistake

    PPQR≠48+(8−12)P_{PQR} \ne 48 + (8 - 12)

    Adding the difference of one pair of sides onto the perimeter is wrong: every side grows, not just one.

  14. Sense check the size of the answer

    23<1⇒PPQR<PABC\frac{2}{3} < 1 \Rightarrow P_{PQR} < P_{ABC}

    The scale factor is smaller than 11, so the new perimeter must be smaller — and it is.

  15. State the final answer

    PPQR=32 cm,k=23P_{PQR} = 32 \text{ cm}, \quad k = \frac{2}{3}

    So the perimeter of triangle PQRPQR is 3232 cm.

Answer
PPQR=32 cmP_{PQR} = 32 \text{ cm}

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