Edexcel IGCSE Math B 6 Geometry Questions
Practise angle facts, congruence, similarity, loci, symmetry, Pythagoras, and circle geometry using diagrams and written reasons.
- Syllabus
- First assessment 2018
- Course
- Math B 4MB1
Practise angle facts, congruence, similarity, loci, symmetry, Pythagoras, and circle geometry using diagrams and written reasons.
Each interior angle of a regular polygon is 172∘
Calculate the number of sides of the polygon.
The length of each side of the regular polygon is 5.2 cm to 2 significant figures.
180−172360=45
Final answer:
45
Marking notes:
Equivalent equation (n-2)180=172n is allowed.
Diagram NOT accurately drawn.
Figure 2 shows a quadrilateral ABCD with
AB=(4.58.5)andAC=(515).
The diagonals AC and BD meet at the point E.
Given that ABCD is a kite, find the coordinates of the point D.
OB=(1−3)+(4.58.5)=(5.55.5)BE=(−4.5−8.5)+(−13)+(46)=(−1.50.5)OD=OE+BE=(46)+(−1.50.5)=(2.56.5)
Final answer:
(6.5,2.5)
Write down
the number of lines of symmetry of the figure,
0
the order of rotational symmetry of the figure.
1
Figure 3 shows a solid right prism ABCDEFGH.
Diagram NOT accurately drawn.
The base of the prism, ABCD, is a rectangle and is horizontal. Trapezium BCHG is a cross section of the prism.
AB=4 m,BC=3 m,CH=2.4 m,GB=1.5 m∠GBC=∠BCH=∠ABG=∠DCH=90∘
Calculate the length, in metres to 3 significant figures, of AH.
AH=42+32+2.42=30.76
Final answer:
5.55 m
Figure 1 shows triangle OAB in which OA=4a and OB=8b
P is the point on OA such that O P: O A=1: 4
The area of APQB is 150 cm².
Calculate the area of triangle OPQ.
Triangles OAB and OPQ are similar.
∣△OAB∣=42∣△OPQ∣
The area of APQB is:
150=∣△OAB∣−∣△OPQ∣150=42∣△OPQ∣−∣△OPQ∣150=15∣△OPQ∣∣△OPQ∣=10
Final answer:
10 cm2
M1: similar-triangle area scale factor.
M1 (dep): using the difference of the two triangle areas for APQB.
A1.
Figure 2
Diagram NOT accurately drawn
Figure 2 shows a kite ABCD in which
AEC and BED are straight lines.
Prove that triangle ABE is similar to triangle BCE.
Method 1 - Show 2 of
3
A C B= A B E oe or
B A C= C B D oe or
A E B= B E C=90^ oe
M2 Show either by argument or calculation that 2 pairs of angles are equal or 1 pair of angles are equal and stating A E B= B E C=90
If using lengths AE or EC the calculation must be seen for one of them. Calculation for one angle must be seen. These may be seen in part(c)
M1 for showing 1 pair of angles are equal or stating
A E B= B E C=90
Method 2
B E/A E=4/3 and B C/A B=4/3 and C E/B E=4/3 oe
M2 Show either by argument or calculation that at all 3 pairs of corresponding sides are in the same ratio.
Need 4/3 or 3/4 (must be exact) If using lengths AE or EC the calculation must be seen for one of them. M1 for showing 2 pairs of corresponding sides are in the same ratio
Method 3
Show two of these
B E/A E=4/3 and B C/A B=4/3 and C E/B E=4/3 oe
and the angle between them is the same
M2 Show either by argument or calculation that 2 pairs of corresponding sides are equal and the angle between is the same If using lengths AE or EC the calculation must be seen for one of them.
M1 for showing 2 pairs of corresponding sides are in the same ratio
NB If angles shown equal by calculation the calculation must be given as an exact value eg.
A C B=6/8
Both triangles are similar as 2 angles are equal or sides pair in same ratio or SAS
A1 dep on M2 being awarded. A written conclusion with their reason.
9
The diagram shows a square ABCD.
M is the midpoint of AD.
N is the midpoint of AB.
Prove that △DAN is congruent to △CDM.
Valid congruence statements include:
AD=DC
because ABCD is a square.
AN=DM
because N and M are midpoints of AB and AD.
∠DAN=∠CDM=90∘
because the angles are right angles in the square.
Therefore △DAN≅△CDM by SAS.
M2 for two correct statements with reasons.
M1 for one correct statement with reason, or all three statements with no reasons.
A1 for SAS/RHS/SSS stated with all three correct reasons.
Diagram NOT accurately drawn
Figure 3 shows a circle with centre O and radius 10 cm.
The points A and B lie on the circle such that ∠AOB=100∘ The point C is such that AC and BC are tangents to the circle.
Giving reasons, calculate the size, in degrees, of ∠ACB.
∠OAC=∠OBC=90∘∠ACB=80∘
Reasons may use tangent-radius angles and angles in a quadrilateral, or split into two triangles using tangent, triangle angle sum, and the isosceles/bisector facts.
Diagram NOT accurately drawn
In the diagram, A, B, C and D are points on a circle, centre O. ∠CBO=62∘ and ∠BCD=126∘
Find the size, in degrees, of ∠BAD.
Give a reason for your answer.
CFDE is cyclic, so opposite angles sum to 180∘:
∠CDE=180∘−126∘=54∘.
Answer: 54∘.
Marks: B1.
The scale drawing opposite shows the positions, A, B and C, of three lighthouses. The points A, B and C lie at sea level in the same horizontal plane with B due north of A.
The scale of the drawing is 1 cm represents 50 km.
A ship S is equidistant from A and C and 225 km from B.
Showing all your construction lines, find and mark each of the two possible positions of S on the scale drawing with a cross (×).
Constructs perpendicular bisector of line AC
Draws circle (or two arcs crossing their perpendicular bisector) with radius 4.5 cm from centre B2
B1
ABCD is a quadrilateral.
Leaving in all of your construction lines, construct the locus of all points inside the quadrilateral that are
equidistant from B and C,
Correct perpendicular bisector of BC.
Marks: M1 for arcs centred on B and C intersecting twice, A1 dependent for correct perpendicular bisector.
equidistant from the lines DA and DC.
Correct angle bisector of ∠ADC.
Marks: M1 for construction arcs from D and from points on DA, DC, A1 dependent for correct angle bisector.