A cuboid has a volume of 250 cm^3, correct to the nearest cubic centimetre. The height of the cuboid is 8.4 cm, correct to 1 decimal place. Calculate the greatest possible base area of the cuboid. Answer ... cm^2 In the figure, AC and PQ are parallel lines. P lies on AB such that AP : PB = 1 : 5 and R lies on AC such that AR : RC = 3 : 2. Explain why triangles ABR and PBQ are similar. Answer ... Show that the ratio of area of triangle PBQ to the area of trapezium APQR is 25 : 11. Answer If the area of the trapezium APQR is 22 cm^2, calculate the area of triangle ABC. Answer ... cm^2

1. Find the upper bound for the volume and the lower bound for the height.

Volume is $250\text{ cm}^3$, correct to the nearest cubic centimetre.

$249.5 \leq V < 250.5$

So the greatest possible volume is $250.5\text{ cm}^3$.

Height is $8.4\text{ cm}$, correct to 1 decimal place.

$8.35 \leq h < 8.45$

So the least possible height is $8.35\text{ cm}$.

2. Calculate the greatest possible base area using the appropriate bounds.

For a cuboid,

$\text{Volume}=\text{base area}\times\text{height}$

Hence

$\text{base area}=\frac{\text{Volume}}{\text{height}}$

Greatest possible base area:

$\frac{250.5}{8.35}=30$

Thus the greatest possible base area is $30\text{ cm}^2$.

1. Identify one pair of corresponding equal angles using the parallel lines.

Since $AC \parallel PQ$ and $AB$ is a transversal,
$\angle BAR = \angle BPQ$.

These are corresponding angles.

2. Identify a second pair of equal angles and conclude similarity.

Since $A,P,B$ are collinear and $R,Q,B$ are collinear,
$\angle ABR = \angle PBQ$.

Therefore, triangles $ABR$ and $PBQ$ are similar by AA similarity.

1. Use the side ratio of the similar triangles to find their area ratio.

Given $AP:PB=1:5$,
$AB:PB=(1+5):5=6:5.$
Since $\triangle ABR$ and $\triangle PBQ$ are similar,
$\text{Area of }\triangle ABR: \text{Area of }\triangle PBQ=6^2:5^2=36:25.$

2. Find the trapezium area as the difference between the two triangle areas.

Trapezium $APQR$ is the part of $\triangle ABR$ left after removing $\triangle PBQ$.

So
$\text{Area of trapezium }APQR=36-25=11\text{ parts}.$
Hence
$\text{Area of }\triangle PBQ : \text{Area of trapezium }APQR=25:11.$

1. Use the given trapezium area and the ratio from part (ii) to find the area of triangle $PBQ$.

From part (ii),
$\text{Area of }\triangle PBQ : \text{Area of trapezium }APQR = 25:11.$
Given $\text{Area of trapezium }APQR=22\text{ cm}^2$,
$\text{Area of }\triangle PBQ=\frac{25}{11}\times 22=50\text{ cm}^2.$

2. Find the area of triangle $ABR$ using the area of trapezium $APQR$.

Since trapezium $APQR$ is the part of $\triangle ABR$ outside $\triangle PBQ$,
$\text{Area of }\triangle ABR=\text{Area of trapezium }APQR+\text{Area of }\triangle PBQ.$
$\text{Area of }\triangle ABR=22+50=72\text{ cm}^2.$

3. Use $AR:RC=3:2$ to find the area of triangle $ABC$.

Triangles $ABR$ and $ABC$ have the same base $AB$, and their heights to $AB$ are proportional to $AR$ and $AC$ since $R$ lies on $AC$.

Given $AR:RC=3:2$, so
$AR:AC=3:5.$
Hence
$\text{Area of }\triangle ABR : \text{Area of }\triangle ABC=3:5.$
Therefore
$\text{Area of }\triangle ABC=\frac{5}{3}\times 72=120\text{ cm}^2.$