Basic Geometry

Question 1

(a) Co-interior angles between parallel lines are supplementary: $(3x+10) + (5x-30) = 180$ $8x - 20 = 180 \Rightarrow 8x = 200 \Rightarrow x = 25.$ (3) Acute angle at $\ell_1$: $3(25)+10 = 85^\circ$. The road makes $85^\circ$ (acute) with each wall — same acute angle on both since corresponding angles are equal. (2)

(b) Let transversal cut parallels at $A$ (on $\ell_1$) and $B$ (on $\ell_2$). Consider co-interior angles $\alpha$ (at $A$) and $\beta$ (at $B$) on the same side.

• At $A$, angle $\alpha$ and the alternate interior angle $\alpha'$ at $B$ are equal (alternate angles, $\ell_1\parallel\ell_2$). (2)
• At $B$, $\alpha'$ and $\beta$ lie on a straight line, so $\alpha' + \beta = 180^\circ$ (angles on a straight line). (2)
• Substituting $\alpha' = \alpha$: $\alpha + \beta = 180^\circ$. Hence co-interior angles are supplementary. ∎ (1)

(c) Trapezium area $= \tfrac12(a+b)h = \tfrac12(12+18)(8) = \tfrac12(30)(8) = 120\text{ m}^2$. (2) Semicircle radius $= 3\text{ m}$; area $= \tfrac12\pi r^2 = \tfrac12(3.14159)(9) = 14.1372\text{ m}^2$. (2) Remaining area $= 120 - 14.1372 = 105.86\text{ m}^2$ (2 d.p.). (2)

(d) One-eighth of trapezium $= 120/8 = 15\text{ m}^2$. Pond area $=14.14\text{ m}^2 < 15\text{ m}^2$. (3) The claim is TRUE. (1)

Question 2

(a) Cone height by Pythagoras: $h_c = \sqrt{l^2 - r^2} = \sqrt{2.5^2 - 2^2} = \sqrt{6.25-4} = \sqrt{2.25} = 1.5\text{ m}$. (2) Volume $= \tfrac13\pi r^2 h_c = \tfrac13\pi(4)(1.5) = 2\pi\text{ m}^3$. (2)

(b) Curved cylinder $= 2\pi r h = 2\pi(2)(5) = 20\pi$. Curved cone $= \pi r l = \pi(2)(2.5) = 5\pi$. Base circle $= \pi r^2 = \pi(4) = 4\pi$. (3) Total $= 20\pi + 5\pi + 4\pi = 29\pi\text{ m}^2$. Numeric $= 29(3.14159) = 91.11\text{ m}^2$ (2 d.p.). (3)

(c) Total interior volume $=$ cylinder $+$ cone $= \pi r^2 h + 2\pi = \pi(4)(5) + 2\pi = 20\pi + 2\pi = 22\pi\text{ m}^3$. (2) $V = 22(3.14159) = 69.115\text{ m}^3$. Mass $= \rho V = 750 \times 69.115 = 51836\text{ kg}$. Weight $= mg = 51836 \times 9.81 = 508513\text{ N} \approx 508{,}500\text{ N}$ (nearest 100 N). (3)

(d)

import math
def silo_volume(r, h, l):
    hc = math.sqrt(l**2 - r**2)        # cone height via Pythagoras
    cyl = math.pi * r**2 * h           # cylinder volume
    cone = (1/3) * math.pi * r**2 * hc # cone volume
    return cyl + cone

Marks: Pythagoras height (2), cylinder term (2), cone term (2), correct return/structure (1). silo_volume(2,5,2.5) returns $22\pi \approx 69.115\text{ m}^3$. (value: 1 of the 7)

Question 3

(a) Diagonals are $PR$ (from $(0,4)$ to $(0,-4)$) and $QS$ (from $(3,0)$ to $(-3,0)$).

• $PR$ is vertical (line $x=0$); $QS$ is horizontal (line $y=0$). Vertical ⟂ horizontal, so perpendicular. (2)
• Midpoint of $PR = (0,0)$; midpoint of $QS = (0,0)$. They share midpoint = origin, so each bisects the other. (2)
• Both diagonals pass through and are bisected at $(0,0)$, and meet at right angles → perpendicular bisectors. ∎ (2)

(b) Order of rotational symmetry $= 2$ (rotations of $180^\circ$ and $360^\circ$ map $PQRS$ to itself; $P\to R$, $Q\to S$ under $180^\circ$). (2) Lines of symmetry $= 2$: the two diagonals $x=0$ and $y=0$. (The sides are unequal in direction so no other axes.) (2)

(c) Reflection in $y=x$ maps $(x,y)\to(y,x)$: $P'(4,0),\ Q'(0,3),\ R'(-4,0),\ S'(0,-3)$. (2) Rotation $90^\circ$ anticlockwise maps $(x,y)\to(-y,x)$: $P\to(-4,0),\ Q\to(0,3),\ R\to(4,0),\ S\to(0,-3)$. (2) Self-mapping transformation: rotation of $180^\circ$ about the origin (or reflection in the $x$-axis or $y$-axis). (1)

(d) Diagonal lengths: $PR = 8$, $QS = 6$. Area $= \tfrac12 d_1 d_2 = \tfrac12(8)(6) = 24$. (2) Four right triangles each with legs $4$ and $3$: area $= 4\times\tfrac12(4)(3) = 4\times 6 = 24$. Confirmed equal. (1)

[
  {"claim":"Q1a: 8x-20=180 gives x=25 and acute angle 85","code":"x=solve(Eq((3*symbols('x')+10)+(5*symbols('x')-30),180),symbols('x'))[0]; result=(x==25 and 3*x+10==85)"},
  {"claim":"Q1c: remaining area = 120 - 0.5*pi*9 approx 105.86","code":"val=120-0.5*3.14159*9; result=abs(val-105.86)<0.01"},
  {"claim":"Q2b: total surface area = 29*pi approx 91.11","code":"val=29*3.14159; result=abs(val-91.11)<0.01"},
  {"claim":"Q2c: weight approx 508500 N (nearest 100)","code":"V=22*3.14159; m=750*V; w=m*9.81; result=abs(round(w/100)*100-508500)<=200"},
  {"claim":"Q3d: rhombus area from diagonals equals 24","code":"result=(Rational(1,2)*8*6==24 and 4*(Rational(1,2)*4*3)==24)"}
]