Similarity and congruence – Year 11 GCSE maths worksheet answers to download as PDF

← Back to the worksheet

1. Answer to Question 1

   1. SSS: all three sides are equal.
   2. SAS: two sides and the included angle.
   3. ASA: two angles and the side between them.
   4. RHS: right angle, hypotenuse and one side. They are congruent.
   5. Not congruent: equal angles (AAA) give similar triangles of any size, so we cannot conclude they are congruent.

2. Answer to Question 2

   1. \(\dfrac{10}{4}=\) 2.5. Divide the corresponding widths: large \(\div\) small.
   2. \(6\times2.5=\) 15 cm. Multiply the corresponding length by the scale factor.
   3. Small: \(2(4+6)=\) 20 cm. Large: \(2(10+15)=\) 50 cm. The perimeter is twice the sum of the length and the width.
   4. \(50:20=2.5:1\), the same as the scale factor: perimeters scale like lengths.

3. Answer to Question 3

   1. \(\dfrac96=\) 1.5. Divide the corresponding shortest sides: \(9\div6\).
   2. \(8\times1.5=\) 12 cm and \(10\times1.5=\) 15 cm. Multiply each side of \(PQR\) by the scale factor.
   3. \(9^2+12^2=81+144=225=15^2\), so by the converse of Pythagoras it is right-angled. Similar shapes have the same angles, and \(PQR\) is a 3-4-5 type triangle, so \(XYZ\) has a right angle too.

4. Answer to Question 4

   1. Angle \(A\) is common, and angles \(ADE=ABC\) and \(AED=ACB\) (corresponding angles between parallel lines). So the triangles have equal angles and are similar (AA).
   2. \(AB=4+2=6\), so the scale factor is \(\dfrac64=\) 1.5.
   3. \(BC=5\times1.5=\) 7.5 cm.

5. Answer to Question 5

   1. \(2^2:3^2=\) \(4:9\). For similar solids the area scale factor is the square of the length scale factor.
   2. \(2^3:3^3=\) \(8:27\). The volume scale factor is the cube of the length scale factor.
   3. \(36\times\dfrac94=\) 81 cm². Going from small to large we multiply by the area factor \(\tfrac94=2.25\).
   4. \(40\times\dfrac{27}8=\) 135 cm³. Going from small to large we multiply by the volume factor \(\tfrac{27}{8}=3.375\).

6. Answer to Question 6

   1. \(AB=AC\) (given), \(BM=CM\) (\(M\) is the midpoint), \(AM\) is common. So the triangles are congruent by SSS.
   2. Corresponding angles in congruent triangles are equal, so angle \(ABM=\) angle \(ACM\) (base angles of an isosceles triangle).
   3. Angle \(AMB=\) angle \(AMC\) (congruent), and they add to \(180^\circ\) on the straight line \(BC\), so each is \(90^\circ\).

7. Answer to Question 7

   1. \(432\div24=\) 18 cm. Change 4.32 m into 432 cm, then divide by 24.
   2. Area scale factor \(=24^2=576\). \(25\times576=14\,400\) cm² \(=\) 1.44 m².
   3. Volume scale factor \(=24^3=13\,824\). \(400\) litres \(=400\,000\) cm³, and \(400\,000\div13\,824=\) 28.9 cm³. Change litres into cm³ first: 1 litre is 1000 cm³. Divide because we go from real to model.

8. Answer to Question 8

   1. \(\dfrac{24}{15}=\) 1.6. Divide the new width by the old width.
   2. \(10\times1.6=\) 16 cm. Multiply the old height by the scale factor. Check: \(24:16=15:10\).
   3. \(1.6^2=\) 2.56. The area scale factor is the square of the length scale factor. Check: \(\tfrac{24\times16}{15\times10}=\tfrac{384}{150}=2.56\).
   4. \(3\times2.56=\) 7.68 g. The ink covers the area, so it scales with the area factor.

9. Answer to Question 9

   1. The sun's rays are parallel, so the angles of elevation are equal; both objects are vertical. The triangles have equal angles.
   2. Scale factor \(=\dfrac{9.6}{2.4}=4\). Height \(=1.7\times4=\) 6.8 m.
   3. \(\dfrac{12.5}{1.7}\times2.4=\) 17.6 m.

10. Answer to Question 10

    1. \(\dfrac{30}{300}=\) \(\dfrac1{10}\). Change 3 m into 300 cm so that both heights are in the same units.
    2. Volume factor \(=\left(\dfrac1{10}\right)^3=\dfrac1{1000}\). Mass \(=500\div1000=\) 0.5 kg. Mass depends on volume, so it scales with the cube of the length factor.
    3. Area factor \(=\dfrac1{100}\). \(12\div100=0.12\) m² \(=\) 1 200 cm². Area scales with the square of the length factor. 1 m² is 10 000 cm².
    4. Statue: \(12\times4=\) £48. Model: \(0.12\times4=\) £0.48. Multiply each area by the price per square metre.

11. Answer to Question 11

    1. \(\dfrac{12}{8}=\) 1.5. Divide the corresponding heights.
    2. \(250\times1.5^3=250\times3.375=\) 843.75 ml. Volume scales with the cube of the length scale factor.
    3. \(90\times1.5^2=\) 202.5 cm². Area scales with the square of the length scale factor.
    4. Small: \(\dfrac{80}{250}=0.32\)p per ml. Large: \(\dfrac{140}{843.75}=0.166\)p per ml. The larger can is better value.

12. Answer to Question 12

    1. \(50\,000\) cm \(=0.5\) km. Every centimetre on the map is 50 000 cm in real life, and 100 000 cm is 1 km.
    2. 1 cm² on the map is \(0.5\times0.5=0.25\) km². So \(3\times0.25=\) 0.75 km². For areas, the scale is squared.
    3. \(2\div0.25=\) 8 cm². Divide the real area by the area that 1 cm² represents.

13. Answer to Question 13

    1. \(\dfrac{1.8}{1.2}=\) 1.5. Divide the corresponding depths.
    2. \(30\times1.5^2=\) 67.5 m². Surface area scales with the square of the length factor, \(1.5^2=2.25\).
    3. \(36\times1.5^3=\) 121.5 m³. Volume scales with the cube of the length factor, \(1.5^3=3.375\).
    4. Extra area \(=67.5-30=37.5\) m². Cost \(=37.5\times28=\) £1 050. Multiply the extra floor area by the cost per square metre.

14. Answer to Question 14

    1. Radii 4 in and 6 in: \(\pi\times16=\) 50.27 in² and \(\pi\times36=\) 113.10 in².
    2. Ratio of diameters \(8:12=2:3\), so the area ratio is \(2^2:3^2=4:9\). (\(16:36=4:9\).)
    3. Price per square inch: small \(\dfrac6{50.27}=£0.119\); large \(\dfrac{11}{113.10}=£0.097\). The 12-inch pizza is better value.

15. Answer to Question 15

    1. \(AB=CD\) and \(BC=DA\) (opposite sides of a parallelogram) and \(AC\) is common. So the triangles are congruent by SSS.
    2. Corresponding angles are equal: angle \(ABC=\) angle \(CDA\).
    3. \(AB=CD\); angle \(MAB=\) angle \(MCD\) and angle \(MBA=\) angle \(MDC\) (alternate angles, \(AB\parallel DC\)). So triangles \(AMB\) and \(CMD\) are congruent (ASA), giving \(AM=CM\) and \(BM=DM\).

16. Answer to Question 16

    1. \(\dfrac{AD}{DB}=\dfrac{AE}{EC}\), so \(\dfrac64=\dfrac9{EC}\) and \(EC=\) 6 cm.
    2. \(\dfrac{DE}{BC}=\dfrac{AD}{AB}=\dfrac6{10}\), so \(DE=0.6\times15=\) 9 cm.
    3. Area scale factor from \(ADE\) to \(ABC\) is \(\left(\dfrac{10}6\right)^2=\dfrac{100}{36}=\dfrac{25}9\). If \(ADE\) has area 9 units then \(ABC\) has 25 and \(DBCE\) has \(25-9=16\). The ratio is \(9:16\).

17. Answer to Question 17

    1. Area ratio \(\dfrac{45}{20}=\dfrac94\). The length scale factor is \(\sqrt{\dfrac94}=\) 1.5.
    2. \(24\times1.5=\) 36 cm. Perimeter is a length, so it scales with the length factor.
    3. Volume ratio \(\dfrac{128}{54}=\dfrac{64}{27}\). The length scale factor is \(\sqrt[3]{\dfrac{64}{27}}=\) \(\dfrac43\).
    4. Area factor \(=\left(\dfrac43\right)^2=\dfrac{16}9\). \(36\times\dfrac{16}9=\) 64 cm². Surface area scales with the square of the length factor.

18. Answer to Question 18

    1. \(\sqrt[3]{\dfrac{216}{125}}=\dfrac65=\) 1.2. The volume ratio is \(\tfrac{216}{125}\), and the length factor is its cube root: \(\sqrt[3]{216}=6\) and \(\sqrt[3]{125}=5\).
    2. Area factor \(=1.2^2=1.44\). \(75\times1.44=\) 108 cm². Surface area scales with the square of the length factor.
    3. \(125:216\) (it cannot be simplified). 125 and 216 have no common factor except 1, so the ratio is already in its simplest form.

19. Answer to Question 19

    1. \(AB=AD\) (given), \(CB=CD\) (given) and \(AC\) is common. Congruent by SSS.
    2. Corresponding angles \(BAC=DAC\), so \(AC\) bisects angle \(BAD\).
    3. Similarly angle \(BCA=\) angle \(DCA\).
    4. Let \(AC\) meet \(BD\) at \(M\). Triangles \(ABM\) and \(ADM\) have \(AB=AD\), angle \(BAM=\) angle \(DAM\) and \(AM\) common, so they are congruent (SAS). Then angle \(AMB=\) angle \(AMD\), and they add to \(180^\circ\), so each is \(90^\circ\).

20. Answer to Question 20

    1. \(\sqrt{36+64}=\) 10 cm.
    2. Both triangles have a right angle and share the angle at \(B\), so they are similar (AA). \(\dfrac{BD}{BA}=\dfrac{BA}{BC}\), so \(BD=\dfrac{6^2}{10}=\) 3.6 cm.
    3. \(DC=10-3.6=\) 6.4 cm. \(AD=\sqrt{36-3.6^2}=\sqrt{23.04}=\) 4.8 cm.
    4. \(BD\times DC=3.6\times6.4=23.04=4.8^2=AD^2\). ✓