A trapezium $PQRS$ is given, with $PQ$ parallel to $RS$. The perpendicular distance from $P$ to $RS$ is 40 cm. We are given $|PQ| = 20$ cm, $|SP| = 50$ cm, and $|SR| = 60$ cm. We need to find the area of the trapezium and the angle $QRS$, correct to 2 significant figures.

1. Problem Description

A trapezium $PQRS$ is given, with $PQ$ parallel to $RS$. The perpendicular distance from $P$ to $RS$ is 40 cm. We are given $|PQ| = 20$ cm, $|SP| = 50$ cm, and $|SR| = 60$ cm. We need to find the area of the trapezium and the angle $QRS$, correct to 2 significant figures.

2. Solution Steps

(a) Area of the trapezium:

Let $h$ be the height of the trapezium, which is given as 40 cm. Let $a$ and $b$ be the lengths of the parallel sides, where $a = |PQ| = 20$ cm and $b = |SR| = 60$ cm.

The area of a trapezium is given by:

$Area = \frac{1}{2} (a+b)h$

$Area = \frac{1}{2} (20+60)(40)$

$Area = \frac{1}{2} (80)(40)$

$Area = 40 \times 40$

$Area = 1600$ cm$^2$.

Rounding to two significant figures, the area is 1600 cm$^2$.

(b) Angle $QRS$:

Let $X$ be the foot of the perpendicular from $P$ to $RS$. Then $PX = 40$ cm.

Let $Y$ be the foot of the perpendicular from $Q$ to $RS$. Since $PQRS$ is a trapezium, $QY = PX = 40$ cm.

Also, $XY = PQ = 20$ cm.

We can consider the right-angled triangle $PXS$. $SX^2 + PX^2 = SP^2$.

So $SX^2 + 40^2 = 50^2$.

$SX^2 = 2500 - 1600 = 900$.

$SX = \sqrt{900} = 30$ cm.

Now consider the length $YR$. Since $SR = 60$ cm, $SX = 30$ cm, and $XY = 20$ cm,

$YR = SR - (SX + XY) = 60 - (30+0) = 60 - (SX + XY) = 60 - (SX + XY) = 60 - XY - XR$

We have $XY = 20$.

$SR = SX + XY + YR$. So $YR = SR - SX - XY$.

We need to find $YR$. We have $SR=60$, $SX=30$ and $XY = PQ = 20$

Then $YR = SR - SX - XY=60 - 30 - 0=60-20-XS = 10$

$XR$

We have $XY=20$. So $YR = SR -XY - XS$. $XR=SR-(PQ+XS)=SX - x$

So YR =

1

0. Thus $YR = 60 - 30 - 20 = 10$ cm.

In right-angled triangle $QYR$, $\tan(\angle QRS) = \frac{QY}{YR} = \frac{40}{10} = 4$.

$\angle QRS = \arctan(4) \approx 75.96^{\circ}$.

Rounding to two significant figures, $\angle QRS = 76^{\circ}$.

3. Final Answer

(a) The area of the trapezium is $1.6 \times 10^3$ cm$^2$ or 1600 cm$^2$.

(b) $\angle QRS = 76^{\circ}$.