Level 4 — ApplicationBasic Geometry

Basic Geometry

⏱ 60 minutes60 marksprintable — key stays hidden on paper

Level: 4 (Application — novel problems, no hints) Time limit: 60 minutes Total marks: 60

Use $\pi = 3.14159$ (or the $\pi$ button) unless told otherwise. Show all working. Diagrams are not to scale.

Question 1. (12 marks)

A straight road runs horizontally. A transversal power line crosses it, and a second parallel road also crosses the same power line.

(a) At the first road, the acute angle between the road and the power line is $(3x + 10)^\circ$ . At the parallel second road, the co-interior (allied) angle on the same side of the power line measures $(5x - 30)^\circ$ . Form an equation and solve for $x$ . (4)

(b) Hence state the size of the acute angle the power line makes with the roads. (2)

(c) A third line is drawn perpendicular to the power line. This third line makes an angle $\theta$ with the first road. Using your value of $x$ , find $\theta$ , giving a reason. (3)

(d) State whether the angle in part (b) and its supplement can both be complementary to some angle. Justify with a calculation. (3)

Question 2. (13 marks)

In triangle $PQR$ , the side $QR$ is extended to a point $S$ . The exterior angle $\angle PRS = 128^\circ$ . Angle $\angle PQR = (2y)^\circ$ and angle $\angle QPR = (3y - 12)^\circ$ .

(a) Use the exterior angle theorem to form an equation and find $y$ . (4)

(b) Find all three interior angles of triangle $PQR$ . (3)

(d) A second triangle is drawn that is right-angled and has one angle equal to the smallest angle of $PQR$ . Find its third angle. (3)

Question 3. (12 marks)

A garden is shaped as a trapezium $ABCD$ with parallel sides $AB = 14\text{ m}$ and $DC = 8\text{ m}$ , and perpendicular height $6\text{ m}$ . A semicircular flower bed of diameter $8\text{ m}$ is cut out and removed from inside the garden (the flat edge of the semicircle lies along $DC$ ).

(a) Find the area of the trapezium garden. (3)

(b) Find the area of the semicircular flower bed. (3)

(d) A path of uniform width runs around the outside perimeter of the whole trapezium. The trapezium's slant sides are each $6.7\text{ m}$ . Find the perimeter of the trapezium (its total boundary length). (4)

Question 4. (13 marks)

A solid is made by placing a cone exactly on top of a cylinder (they share a circular face). The cylinder has radius $5\text{ cm}$ and height $12\text{ cm}$ . The cone has the same radius $5\text{ cm}$ and slant height $13\text{ cm}$ .

(a) Find the vertical height of the cone. (2)

(b) Find the total volume of the solid. (4)

(c) Find the total external surface area of the solid (the two circular faces where cone meets cylinder are internal and are not counted; the bottom circular face of the cylinder is counted). (5)

(d) State how many faces, edges and vertices a triangular prism has, and describe what the net of a cylinder consists of. (2)

Question 5. (10 marks)

A shape $F$ has vertices $A(1,1)$ , $B(4,1)$ , $C(4,3)$ .

(a) $F$ is reflected in the line $y = x$ to give $F'$ . Write the coordinates of the images $A'$ , $B'$ , $C'$ . (3)

(b) $F$ is then translated by the vector $\begin{pmatrix} -2 \\ 5 \end{pmatrix}$ . Write the coordinates of the image of $A$ under this translation. (2)

(c) A regular hexagon and a rectangle (non-square) are drawn. State the order of rotational symmetry and the number of lines of symmetry for each. (3)

(d) Triangle $F$ is enlarged by scale factor $3$ about the origin. Find the coordinates of the image of $C(4,3)$ , and state how the area of the triangle changes. (2)

End of paper.

Answer keyMark scheme & solutions

Question 1 (12)

(a) Co-interior angles between parallel lines are supplementary (sum to $180^\circ$ ). The acute angle at the first road is $(3x+10)$ ; the co-interior angle at the second road is $(5x-30)$ . $(3x+10) + (5x-30) = 180$ (1 for equation) $8x - 20 = 180 \Rightarrow 8x = 200 \Rightarrow x = 25$ (2 for algebra, 1 final) $x = 25$

(b) Acute angle $= 3x+10 = 3(25)+10 = 85^\circ$ . (1) Check: co-interior $=5(25)-30=95^\circ$ ; $85+95=180$ ✓. (1) $85^\circ$

(c) The third line is perpendicular to the power line, so it makes $90^\circ$ with it. The road makes $85^\circ$ with the power line. The angle between road and the perpendicular line $= 90^\circ - 85^\circ = 5^\circ$ . (2 method, 1 answer) $\theta = 5^\circ$ (angles about the power line).

(d) Complementary means summing to $90^\circ$ . The angle is $85^\circ$ ; a complement would be $90-85 = 5^\circ$ (valid, positive). Its supplement is $95^\circ$ ; a complement would be $90-95 = -5^\circ$ (impossible). (2) So no — an obtuse angle ( $95^\circ$ ) cannot have a complement, so they cannot both be complementary to an angle. (1)

Question 2 (13)

(a) Exterior angle theorem: exterior angle = sum of two opposite interior angles. $128 = 2y + (3y - 12)$ (1 equation) $128 = 5y - 12 \Rightarrow 5y = 140 \Rightarrow y = 28$ (3) $y = 28$

(b) $\angle PQR = 2y = 56^\circ$ ; $\angle QPR = 3y-12 = 72^\circ$ ; $\angle PRQ = 180 - 128 = 52^\circ$ (angles on a straight line). (3 — 1 each) Check: $56+72+52 = 180$ ✓.

(c) Angles are $52^\circ, 56^\circ, 72^\circ$ — all less than $90^\circ$ , so the triangle is acute-angled. (2) All three angles are different, so no two sides are equal → cannot be isosceles (isosceles requires two equal angles). (1)

(d) Smallest angle of $PQR$ is $52^\circ$ . Right-angled triangle has $90^\circ$ ; third angle $= 180 - 90 - 52 = 38^\circ$ . (2 method, 1 answer) $38^\circ$

Question 3 (12)

(a) Trapezium area $= \tfrac12(a+b)h = \tfrac12(14+8)(6) = \tfrac12(22)(6) = 66\text{ m}^2$ . (3)

(b) Diameter $8$ → radius $4$ . Semicircle area $= \tfrac12 \pi r^2 = \tfrac12 \pi (4)^2 = 8\pi \approx 25.13\text{ m}^2$ . (3)

(c) Grass $= 66 - 8\pi \approx 66 - 25.13 = 40.87\text{ m}^2$ (≈ $40.9\text{ m}^2$ ). (2)

(d) Perimeter $= AB + DC +$ two slant sides $= 14 + 8 + 6.7 + 6.7 = 35.4\text{ m}$ . (2 method, 2 answer) $35.4\text{ m}$

Question 4 (13)

(a) Cone height: $h = \sqrt{l^2 - r^2} = \sqrt{13^2 - 5^2} = \sqrt{169-25} = \sqrt{144} = 12\text{ cm}$ . (2)

(b) Cylinder volume $= \pi r^2 h = \pi(25)(12) = 300\pi$ . Cone volume $= \tfrac13 \pi r^2 h = \tfrac13 \pi (25)(12) = 100\pi$ . (2) Total $= 400\pi \approx 1256.6\text{ cm}^3$ . (2)

(c) Curved surface of cylinder $= 2\pi r h = 2\pi(5)(12) = 120\pi$ . Bottom circle of cylinder $= \pi r^2 = 25\pi$ . Curved surface of cone $= \pi r l = \pi(5)(13) = 65\pi$ . (3) Total external $= 120\pi + 25\pi + 65\pi = 210\pi \approx 659.7\text{ cm}^2$ . (2)

(d) Triangular prism: 5 faces, 9 edges, 6 vertices. (1) Net of a cylinder: two circles (the ends) and one rectangle (the curved surface, of length = circumference). (1)

Question 5 (10)

(a) Reflection in $y=x$ swaps coordinates: $(x,y)\to(y,x)$ . $A(1,1)\to A'(1,1)$ ; $B(4,1)\to B'(1,4)$ ; $C(4,3)\to C'(3,4)$ . (3 — 1 each)

(b) Translation $\binom{-2}{5}$ : $A(1,1)\to(1-2,\,1+5) = (-1,6)$ . (2)

(c) Regular hexagon: rotational symmetry order 6, 6 lines of symmetry. Rectangle (non-square): rotational order 2, 2 lines of symmetry. (3)

(d) Enlargement s.f. 3 about origin: $C(4,3)\to(12,9)$ . (1) Area is multiplied by $3^2 = 9$ (becomes 9 times larger). (1)

[
  {"claim":"Q1(a) x=25 and acute angle 85 with co-interior 95 supplementary","code":"x=25; a=3*x+10; b=5*x-30; result = (x==25) and (a==85) and (b==95) and (a+b==180)"},
  {"claim":"Q2 angles solve to 56,72,52 summing 180 from y=28","code":"y=28; q=2*y; p=3*y-12; r=180-128; result = (y==28) and (q==56) and (p==72) and (r==52) and (q+p+r==180)"},
  {"claim":"Q3 grass area = 66 - 8*pi and perimeter 35.4","code":"area=Rational(1,2)*(14+8)*6; semi=Rational(1,2)*pi*16; grass=area-semi; per=14+8+6.7+6.7; result = (area==66) and (simplify(semi-8*pi)==0) and (abs(float(grass)-40.867)<0.01) and (abs(per-35.4)<1e-9)"},
  {"claim":"Q4 cone height 12, total volume 400pi, external SA 210pi","code":"h=sqrt(13**2-5**2); vcyl=pi*25*12; vcone=Rational(1,3)*pi*25*12; V=vcyl+vcone; SA=2*pi*5*12+pi*25+pi*5*13; result = (h==12) and (simplify(V-400*pi)==0) and (simplify(SA-210*pi)==0)"}
]