[discia] GCSE Maths › Geometry & Measures › Similar Shapes (Lengths) › Practise
  1. Either paper 1 mark

    Here are five rectangles.

    21 cm12 cmXDiagram NOT accurately drawn 23 cm14 cmADiagram NOT accurately drawn 14 cm9 cmBDiagram NOT accurately drawn 21 cm10 cmCDiagram NOT accurately drawn 28 cm16 cmDDiagram NOT accurately drawn

    Which shape is similar to shape \(X\)?

    Circle the correct letter.

    Show answer
    D

    Shape \(X\) is \(21\) cm by \(12\) cm. A shape is similar to \(X\) only if both pairs of matching sides give the same scale factor.

    Shape \(A\): \(23 \div 21 = 1\frac{2}{21}\) and \(14 \div 12 = 1\frac{1}{6}\), different scale factors, so not similar.

    Shape \(B\): \(14 \div 21 = \frac{2}{3}\) and \(9 \div 12 = \frac{3}{4}\), different scale factors, so not similar.

    Shape \(C\): \(21 \div 21 = 1\) and \(10 \div 12 = \frac{5}{6}\), different scale factors, so not similar.

    Shape \(D\): \(28 \div 21 = 1\frac{1}{3}\) and \(16 \div 12 = 1\frac{1}{3}\), the same scale factor, so similar.

    The shape similar to \(X\) is D.

  2. Either paper 2 marks

    Here are two rectangles, \(A\) and \(B\). Each is drawn to fill its own frame, so the sizes on the page are not to the same scale.

    15 cm9 cmA 3 cm5 cmB

    Show that these two rectangles are similar.

    Show answer
    Long sides \(5 : 15 = 1 : 3\); short sides \(3 : 9 = 1 : 3\). The ratios are equal, so the rectangles are similar.

    Match the sides by size, not by where they are drawn: the long side of \(A\) goes with the long side of \(B\), and the short side with the short side.

    Long sides, \(B : A\): \(5 : 15 = 1 : 3\)

    Short sides, \(B : A\): \(3 : 9 = 1 : 3\)

    Both ratios are \(1 : 3\), so every length of \(B\) is in the same ratio to the matching length of \(A\): \(B\) is a reduction of \(A\) with a single scale factor.

    The rectangles are similar.

    Alternatively, long : short is \(15 : 9 = 5 : 3\) in \(A\) and \(5 : 3 = 5 : 3\) in \(B\), the same in both.

    • M1both ratios of matching sides, \(5 : 15\) and \(3 : 9\), each \(1 : 3\) (or long : short in each rectangle, both \(5 : 3\))
    • A1the two ratios are equal, so the rectangles are similar
  3. Either paper 4 marks

    Rectangle \(A\) and rectangle \(B\) are similar.

    11 cm4 cmADiagram NOT accurately drawn 44 cmx cmBDiagram NOT accurately drawn
    (a)

    Work out the scale factor of the enlargement from rectangle \(A\) to rectangle \(B\).

    (b)

    Work out the length marked \(x\).

    Show answer
    (a) \(4\)
    (b) \(16\ \text{cm}\)

    Take a pair of sides that correspond — the long side of each rectangle — and divide the new length by the old one.

    \(44 \div 11 = 4\)

    The scale factor is \(4\).

    In similar shapes every length is multiplied by the same scale factor, so multiply the short side of rectangle \(A\) by \(4\).

    \(4 \times 4 = 16\)

    \(x = 16\) cm

    • (a) M1new length divided by old length \(44 \div 11\)
    • (a) A1scale factor \(4\)
    • (b) M1short side multiplied by the scale factor \(4 \times 4\)
    • (b) A1length \(x = 16\) cm
  4. Either paper 3 marks

    Triangle \(ABC\) is similar to triangle \(PQR\), so that \(A\) corresponds to \(P\), \(B\) to \(Q\) and \(C\) to \(R\).

    7 cm8 cmABCDiagram NOT accurately drawn 14 cmx cmPQRDiagram NOT accurately drawn

    Find the length of \(QR\).

    Show answer
    \(16\ \text{cm}\)

    \(AB\) and \(PQ\) are corresponding sides, because \(A\) goes with \(P\) and \(B\) goes with \(Q\). Use them to find the scale factor.

    \(14 \div 7 = 2\)

    \(QR\) corresponds to \(BC\), so multiply \(BC\) by the same scale factor.

    \(8 \times 2 = 16\)

    \(QR = 16\) cm

    • M1scale factor \(14 \div 7 = 2\)
    • M1\(8 \times 2\)
    • A1\(QR = 16\) cm
  5. Either paper 3 marks

    Priya has a photograph that is \(6\) cm wide and \(9\) cm high.

    She enlarges it. The enlarged photograph is \(14\) cm wide, and it is similar to the original.

    Calculate the height of the enlarged photograph.

    Show answer
    \(21\ \text{cm}\)

    The widths are corresponding lengths, so they give the scale factor.

    \(14 \div 6 = \frac{ 14 }{ 6 } = \frac{7}{3}\)

    Every length on the enlargement is \(\frac{7}{3}\) times the matching length on the original, so multiply the height by the same amount.

    \(9 \times \frac{7}{3} = 21\)

    The enlarged photograph is \(21\) cm high.

    • M1scale factor \(14 \div 6 = \frac{7}{3}\)
    • M1\(9 \times \frac{7}{3}\)
    • A1height \(21\) cm
  6. Either paper 3 marks

    One afternoon Priya measures the shadows in a park.

    A bollard \(2\) m tall casts a shadow \(5\) m long. At the same moment a tree casts a shadow \(25\) m long.

    The sun's rays are parallel, so the two triangles below are similar.

    5 m2 mDiagram NOT accurately drawn 25 mx mDiagram NOT accurately drawn

    Find the height of the tree.

    Show answer
    \(10\ \text{m}\)

    The two shadows are corresponding lengths, so they give the scale factor.

    \(25 \div 5 = 5\)

    The tree and its shadow are in the same triangle as the bollard and its shadow, enlarged by \(5\), so the height is multiplied by \(5\) as well.

    \(2 \times 5 = 10\)

    The tree is \(10\) m tall.

    • M1scale factor from the shadows \(25 \div 5 = 5\)
    • M1height multiplied by the scale factor \(2 \times 5\)
    • A1height \(10\) m
  7. Either paper 4 marks

    Triangle \(ABC\) is similar to triangle \(PQR\), so that \(A\) corresponds to \(P\), \(B\) to \(Q\) and \(C\) to \(R\).

    \(AB = 9\) cm and \(PQ = 36\) cm.

    Dev says:

    "The scale factor of the enlargement from \(ABC\) to \(PQR\) is \(27\), because \(36 - 9 = 27\)."

    (a)

    Explain why Dev is wrong.

    (b)

    State the correct scale factor.

    (c)

    \(BC = 10\) cm. Work out the length of \(QR\).

    Show answer
    (a) A scale factor multiplies rather than adds, so it comes from dividing corresponding lengths: \(36 \div 9 = 4\), not \(36 - 9 = 27\).
    (b) \(4\)
    (c) \(40\ \text{cm}\)

    An enlargement multiplies every length by the same number. Dev has subtracted instead, which would add the same amount to every length rather than multiplying it.

    You can see the difference straight away by testing his rule on the pair he started from: adding \(27\) to \(9\) does give \(36\), but if the shape had a side of \(1\) cm it would become \(27 + 1\) cm, and that is not \(27\) times bigger. Only multiplying keeps every side in the same ratio, which is what "similar" means.

    Divide a length on the new shape by the corresponding length on the old one.

    \(36 \div 9 = 4\)

    The scale factor is \(4\).

    \(QR\) corresponds to \(BC\), so multiply by the scale factor.

    \(10 \times 4 = 40\)

    \(QR = 40\) cm

    Dev's method would have given \(10 + 27 = 37\) cm, which is wrong.

    • (c) M1multiplies \(10\) by the scale factor \(4\)
    • (c) A1QR \(40\) cm
  8. Either paper 6 marks

    Triangle \(ABC\) is similar to triangle \(PQR\), so that \(A\) corresponds to \(P\), \(B\) to \(Q\) and \(C\) to \(R\).

    27 cmx cm21 cmABCDiagram NOT accurately drawn 9 cm8 cmy cmPQRDiagram NOT accurately drawn
    (a)

    Write down, as a fraction in its simplest form, the scale factor of the enlargement from triangle \(ABC\) to triangle \(PQR\).

    (b)

    Work out the length marked \(y\).

    (c)

    Work out the length marked \(x\).

    Show answer
    (a) \(\frac{1}{3}\)
    (b) \(7\ \text{cm}\)
    (c) \(24\ \text{cm}\)

    \(AB\) and \(PQ\) correspond, so divide the new length by the old one.

    \(9 \div 27 = \frac{ 9 }{ 27 } = \frac{1}{3}\)

    The scale factor is \(\frac{1}{3}\). A scale factor smaller than 1 is still called an enlargement, even though the shape gets smaller.

    \(RP\) corresponds to \(CA\), so multiply \(CA\) by the scale factor.

    \(21 \times \frac{1}{3} = 7\)

    \(y = 7\) cm

    This one goes the other way: \(BC\) is on the larger triangle, and \(QR\) on the smaller one is known. Going from \(PQR\) back to \(ABC\) means dividing by the scale factor, which is the same as multiplying by it upside down.

    \(8 \div \frac{1}{3} = 8 \times 3 = 24\)

    \(x = 24\) cm

    • (a) M1\(9 \div 27\) from corresponding sides \(PQ\) and \(AB\)
    • (a) A1scale factor \(\frac{1}{3}\)
    • (b) M1\(21 \times \frac{1}{3}\)
    • (b) A1\(y = 7\) cm
    • (c) M1\(8 \div \frac{1}{3}\) or \(8 \times 3\)
    • (c) A1\(x = 24\) cm
  9. Either paper 3 marks

    The diagram shows two similar right-angled triangles, \(A\) and \(B\).

    12 cmh cmTriangle ADiagram NOT accurately drawn 20 cmTriangle BDiagram NOT accurately drawn

    The area of triangle \(B\) is 150 cm².

    Calculate the value of \(h\).

    Show answer
    \(9\ \text{cm}\)

    The area of a triangle is \(\frac{1}{2} \times \text{base} \times \text{height}\), so the height of triangle \(B\) is

    \(\text{height} = 2 \times 150 \div 20 = 15\) cm

    The bases are corresponding sides, so the scale factor from triangle \(B\) to triangle \(A\) is

    \(\frac{ 12 }{ 20 }\)

    \(h\) corresponds to the height of triangle \(B\), so multiply that height by the same scale factor.

    \(h = 15 \times \frac{ 12 }{ 20 } = 9\) cm

    • M1height of the given triangle from its area \(2 \times 150 \div 20 = 15\)
    • M1scale factor \(12 \div 20\) applied to the height
    • A1height \(h = 9\) cm
  10. Either paper 2 marks

    The diagram shows triangle \(FGH\) and triangle \(FJK\).

    \(J\) lies on \(FG\) and \(K\) lies on \(FH\). \(JK\) is parallel to \(GH\).

    FJGHK

    Show that triangle \(FGH\) and triangle \(FJK\) are similar.

    Give a reason for each step of your working.

    Show answer
    Angle \(GFH\) is the same as angle \(JFK\), because it is the angle at \(F\) shared by both triangles. Angle \(FGH\) = angle \(FJK\), because they are corresponding angles on the parallel lines \(JK\) and \(GH\). Angle \(FHG\) = angle \(FKJ\), because they are corresponding angles on the same parallel lines (or the third pair is equal by the angle sum). Two pairs of angles are equal, so the triangles are similar.

    Similar triangles have all their angles equal, so find the pairs of equal angles.

    Both triangles have a corner at \(F\), and it is the same corner: \(FJ\) lies along \(FG\) and \(FK\) lies along \(FH\). So angle \(GFH\) = angle \(JFK\), a common angle.

    \(JK\) is parallel to \(GH\), and \(FG\) crosses both lines. Angle \(FJK\) and angle \(FGH\) are in matching positions at the two crossings, so they are corresponding angles and are equal.

    Angle \(FHG\) and angle \(FKJ\) are also corresponding angles, at the crossings of \(FH\) with the same two parallel lines, so they are equal. (The angles in a triangle add up to \(180^\circ\), so the third pair is equal by the angle sum too.) Any two pairs of equal angles are enough.

    All three angles are equal, so triangle \(FGH\) and triangle \(FJK\) are similar.

    • B1a first pair of equal angles with its reason: the angle at \(F\) common to both triangles, or angle \(FGH\) = angle \(FJK\) as corresponding angles (\(JK\) parallel to \(GH\))
    • A1a second, different pair of equal angles with its own reason from the parallel lines or the shared corner (corresponding angles: angle \(FHG\) = angle \(FKJ\); or angle \(FGH\) = angle \(FJK\); or the common angle at \(F\)), so the triangles are similar. The angle sum alone does not earn this mark, since it needs two pairs already