Solutions — 5H Worksheet
Year 10 Mathematics Core · Cambridge Ch 5, §5H · Mr Wong
ANSWER KEY
Part A — Rectangle dimensions
Q1. Area $40\ \text{m}^2$, length $3$ m more than width:
· $x(x+3)=40 \Rightarrow x^2+3x-40=0 \Rightarrow (x+8)(x-5)=0$; reject $x=-8$.
Width $5$ m, length $8$ m.
Q2. Area $24\ \text{m}^2$, length $5$ m more than width:
· $x(x+5)=24 \Rightarrow (x+8)(x-3)=0$; reject $x=-8$.
Width $3$ m, length $8$ m.
Q3. Area $63\ \text{m}^2$, length $2$ m less than width:
· $x(x-2)=63 \Rightarrow (x-9)(x+7)=0$; reject $x=-7$.
Width $9$ m, length $7$ m.
Part B — Triangles
Q4. Area $10\ \text{cm}^2$, base $1$ cm more than height (height $=x$):
· $\tfrac12\,x(x+1)=10 \Rightarrow x^2+x-20=0 \Rightarrow (x+5)(x-4)=0$; reject $x=-5$.
Height $4$ cm, base $5$ cm.
Part C — Number problems
Q5. A positive number $12$ less than its square:
· $x=x^2-12 \Rightarrow x^2-x-12=0 \Rightarrow (x-4)(x+3)=0$; reject $x=-3$.
The number is $4$.
Q6. Two consecutive even integers, product $168$:
· $x(x+2)=168 \Rightarrow (x-12)(x+14)=0$; positive case $x=12$.
The integers are $12$ and $14$.
Part D — Pythagoras
Q7. Legs $x$ and $x+7$, hypotenuse $13$:
· $x^2+(x+7)^2=13^2 \Rightarrow 2x^2+14x-120=0 \Rightarrow x^2+7x-60=0 \Rightarrow (x-5)(x+12)=0$;
reject $x=-12$. $x=5$ (legs $5$ cm and $12$ cm).
Challenge
Q8. Garden bed $x$ by $(x+6)$, area $91\ \text{m}^2$:
· $x(x+6)=91 \Rightarrow x^2+6x-91=0 \Rightarrow (x-7)(x+13)=0$; reject $x=-13$.
Dimensions $7$ m by $13$ m.