Level 3 — ProductionBasic Geometry

Basic Geometry

⏱ 45 minutes60 marksprintable — key stays hidden on paper

Level: 3 (Production — from-scratch derivations, reasoning out loud) Time limit: 45 minutes Total marks: 60

Instructions: Show all reasoning. Where a question says "derive" or "explain", full justification is required, not just a final value. Use $\pi = \dfrac{22}{7}$ unless otherwise stated.

Question 1 — Derive the triangle angle sum (12 marks)

(a) Using a line drawn through one vertex parallel to the opposite side, derive from scratch that the interior angles of any triangle sum to $180°$ . State clearly which parallel-line angle property you use at each step. (6)

(b) Using your result in (a), derive the Exterior Angle Theorem (an exterior angle equals the sum of the two remote interior angles). (3)

(c) A triangle has angles $x$ , $2x$ , and an exterior angle at the third vertex equal to $150°$ . Find $x$ and classify the triangle by its angles. (3)

Question 2 — Circle measures from memory (10 marks)

(a) State from memory the formulas for the circumference and area of a circle of radius $r$ , and explain in one sentence why circumference is linear in $r$ but area is quadratic. (3)

(b) A circular sector has radius $14\text{ cm}$ and central angle $90°$ . Derive expressions for, then compute: (i) the arc length, (ii) the sector area, (iii) the perimeter of the sector. (7)

Question 3 — Parallel lines and a transversal (10 marks)

Two parallel lines are cut by a transversal. One of the co-interior (same-side interior) angles is $(3x + 20)°$ and the other is $(5x - 40)°$ .

(a) Explain out loud the property that relates co-interior angles, and write the equation it gives. (3) (b) Solve for $x$ . (3) (c) Find the size of each co-interior angle, and state the size of the alternate angle to the first one. Justify. (4)

Question 4 — Surface area & volume, derived (12 marks)

A solid cylinder has radius $7\text{ cm}$ and height $10\text{ cm}$ .

(a) Starting from its net, derive the formula for the total surface area of a cylinder, explaining where each term comes from. (4) (b) Compute the total surface area. (3) (c) Compute the volume. (2) (d) The cylinder is melted and recast into a sphere of the same volume. Derive an expression for the sphere's radius in terms of the cylinder's dimensions, then compute it (give to 2 d.p., using $\pi = 3.1416$ ). (3)

Question 5 — Composite area & perimeter (10 marks)

A trapezium has parallel sides $12\text{ cm}$ and $20\text{ cm}$ , a perpendicular height of $6\text{ cm}$ , and the two non-parallel (slant) sides are each $10\text{ cm}$ .

(a) Derive the area formula for a trapezium by splitting it into a rectangle and two triangles (or a parallelogram), then compute the area. (6) (b) Compute the perimeter. (2) (c) A semicircle of diameter $12\text{ cm}$ is attached along the shorter parallel side. Find the total area of the composite shape. (2)

Question 6 — Symmetry & transformations, explain-out-loud (6 marks)

(a) State the order of rotational symmetry and the number of lines of symmetry for: a square, a rhombus (non-square), and an equilateral triangle. (3)

(b) The point $A(3, 1)$ is reflected in the line $y = x$ to give $A'$ , then $A'$ is translated by the vector $\begin{pmatrix} -2 \\ 4 \end{pmatrix}$ to give $A''$ . Find the coordinates of $A''$ , explaining each step. (3)

End of paper.

Answer keyMark scheme & solutions

Question 1 (12)

(a) Draw a triangle $ABC$ . Through vertex $A$ draw line $\ell$ parallel to $BC$ . (1 for construction)

Angle between $\ell$ and $AB$ = angle $ABC$ (alternate angles, $\ell \parallel BC$ ). (2)
Angle between $\ell$ and $AC$ = angle $ACB$ (alternate angles). (1)
The three angles on line $\ell$ at $A$ form a straight angle: $\angle B + \angle A + \angle C = 180°$ (angles on a straight line). (2) ∴ interior angles sum to $180°$ .

(b) Let exterior angle at $C$ be $\angle ACD$ (on straight line $BCD$ ). Then $\angle ACB + \angle ACD = 180°$ (straight line). But $\angle A + \angle B + \angle ACB = 180°$ (from (a)). Equating: $\angle ACD = \angle A + \angle B$ = sum of the two remote interior angles. (3)

(c) Exterior angle at third vertex = sum of remote interior angles $x + 2x = 150°$ $3x = 150 \Rightarrow x = 50°$ . (2) Angles: $50°, 100°$ , and third interior $= 180° - 150° = 30°$ . Largest is $100°$ → obtuse triangle. (1)

Question 2 (10)

(a) $C = 2\pi r$ , $A = \pi r^2$ . (2) Circumference scales with the single length $r$ (linear); area covers a 2-D region so it scales with $r^2$ (quadratic). (1)

(b) Fraction of circle $= \dfrac{90}{360} = \dfrac14$ . (i) Arc $= \frac14 \cdot 2\pi r = \frac14 \cdot 2\cdot\frac{22}{7}\cdot 14 = 22\text{ cm}$ . (2) (ii) Sector area $= \frac14 \pi r^2 = \frac14 \cdot \frac{22}{7}\cdot 196 = 154\text{ cm}^2$ . (3) (iii) Perimeter $=$ arc $+ 2r = 22 + 28 = 50\text{ cm}$ . (2)

Question 3 (10)

(a) Co-interior (same-side interior) angles between parallel lines are supplementary (sum to $180°$ ). (2) Equation: $(3x+20)+(5x-40)=180$ . (1)

(b) $8x - 20 = 180 \Rightarrow 8x = 200 \Rightarrow x = 25$ . (3)

(c) First angle $= 3(25)+20 = 95°$ ; second $= 5(25)-40 = 85°$ (check $95+85=180$ ✓). (2) The alternate angle to the first co-interior angle equals the other interior angle's supplement... more precisely: the alternate angle to a co-interior angle is supplementary to it, so alternate to $95°$ is $85°$ . (2)

Question 4 (12)

(a) Net = two circles (top & bottom) + one rectangle (curved surface unrolled). Rectangle width = circumference $2\pi r$ , height $= h$ , so curved area $=2\pi r h$ ; two circles $=2\pi r^2$ . Total: $A = 2\pi r^2 + 2\pi r h = 2\pi r(r+h)$ . (4)

(b) $A = 2\cdot\frac{22}{7}\cdot 7 \cdot (7+10) = 2\cdot 22\cdot 17 = 748\text{ cm}^2$ . (3)

(c) $V = \pi r^2 h = \frac{22}{7}\cdot 49 \cdot 10 = 1540\text{ cm}^3$ . (2)

(d) $\frac43\pi R^3 = \pi r^2 h \Rightarrow R = \left(\dfrac{3 r^2 h}{4}\right)^{1/3} = \left(\dfrac{3\cdot 49\cdot 10}{4}\right)^{1/3} = (367.5)^{1/3} \approx 7.16\text{ cm}$ . (3)

Question 5 (10)

(a) Split trapezium into a central rectangle (width = shorter side $12$ ) and two right triangles. The base overhang total $= 20-12 = 8$ , so each triangle base $=4$ . Area $=$ rectangle $+ 2$ triangles $=12\cdot6 + 2\cdot(\frac12\cdot4\cdot6) = 72 + 24 = 96\text{ cm}^2$ . Equivalent to $\frac12(a+b)h = \frac12(12+20)(6)=96$ . (6)

(b) Perimeter $= 12 + 20 + 10 + 10 = 52\text{ cm}$ . (2)

(c) Semicircle radius $=6$ ; area $=\frac12\pi r^2 = \frac12\cdot\frac{22}{7}\cdot36 = \frac{396}{7}\approx 56.57\text{ cm}^2$ . Total $=96 + 56.57 = 152.57\text{ cm}^2$ . (2)

Question 6 (6)

(a) Square: order $4$ , $4$ lines. Rhombus (non-square): order $2$ , $2$ lines. Equilateral triangle: order $3$ , $3$ lines. (3, 1 each)

(b) Reflect $A(3,1)$ in $y=x$ : swap coords → $A'(1,3)$ . (1) Translate by $(-2,4)$ : $A''=(1-2,\ 3+4)=(-1,7)$ . (2)

[
  {"claim":"Q1c: 3x=150 gives x=50 and third angle 30, obtuse largest 100","code":"x=Rational(150,3); third=180-150; result=(x==50 and third==30 and max(50,100,30)==100)"},
  {"claim":"Q2b: arc=22, sector area=154, perimeter=50 with pi=22/7,r=14","code":"pi=Rational(22,7); r=14; arc=Rational(1,4)*2*pi*r; area=Rational(1,4)*pi*r**2; per=arc+2*r; result=(arc==22 and area==154 and per==50)"},
  {"claim":"Q4b/c: cylinder SA=748, V=1540 with r=7,h=10,pi=22/7","code":"pi=Rational(22,7); r=7; h=10; SA=2*pi*r*(r+h); V=pi*r**2*h; result=(SA==748 and V==1540)"},
  {"claim":"Q4d: sphere radius from V=1540 approx 7.16","code":"R=(Rational(3*49*10,4))**Rational(1,3); result=(abs(float(R)-7.157)<0.02)"},
  {"claim":"Q5a: trapezium area=96, perimeter=52","code":"A=Rational(1,2)*(12+20)*6; P=12+20+10+10; result=(A==96 and P==52)"}
]