We are given a diagram with $PT \parallel SU$ and $QS \parallel TR$. We are also given that $SR = 6$ cm, $RU = 10$ cm, and the area of $\triangle TRU$ is 45 cm$^2$. We need to find the area of the trapezium $QTUS$.

1. Problem Description

We are given a diagram with $PT \parallel SU$ and $QS \parallel TR$. We are also given that $SR = 6$ cm, $RU = 10$ cm, and the area of $\triangle TRU$ is 45 cm$^2$. We need to find the area of the trapezium $QTUS$.

2. Solution Steps

First, we calculate the height of $\triangle TRU$. Let the height of $\triangle TRU$ be $h$.

The area of $\triangle TRU$ is given by

$\frac{1}{2} \times RU \times h = 45$

$\frac{1}{2} \times 10 \times h = 45$

$5h = 45$

$h = \frac{45}{5}$

$h = 9$ cm.

Since $QS \parallel TR$, $QTUS$ is a trapezium with parallel sides $QT$ and $SU$. The height of trapezium $QTUS$ is the same as the height of $\triangle TRU$, which is 9 cm.

Since $QS \parallel TR$, the height of parallelogram $QRST$ is equal to the height of $\triangle TRU$ which is 9 cm.

Let $h'$ be the height from $Q$ to $SR$. Since $QS \parallel TR$, the height from $Q$ to $SR$ is $h'$.

Since $QS \parallel TR$, $SR=6$. Area of parallelogram $QRST$ is $SR \times h' = 6 \times 9 = 54$ cm$^2$

Let the height from $Q$ to $SU$ be $h=9$. Then the area of $\triangle TRU$ is $\frac{1}{2} \times 10 \times 9 = 45$.

The area of trapezium $QTUS$ = Area of parallelogram $QRST$ + Area of $\triangle TRU$

Area($QTUS$) = $54 + 45 = 99$ cm$^2$.

Alternatively,

$SU = SR+RU = 6+10=16$ cm

Since $QT \parallel SU$, the height is $h=9$ cm.

We know that the triangles $\triangle PQS$ and $\triangle QRS$ has the same area because the have the same base and height ($QS$).

The area of trapezium $QTUS = \frac{1}{2} \times (QT+SU) \times h$

Since $QTUS$ is a trapezium with $QT$ parallel to $SU$ the area of parallelogram $QRST$ = $QS \times h = 54 cm^{2}$. But we don't know $QT$.

Let h be the height of the trapezium $QTUS$.

$h=9$ cm. Also $SU = 6+10 = 16$ cm

Since $QS \parallel TR$, $QS=TR$.

Area $TRU= 45$.

Area $\triangle TRU = \frac{1}{2} \times RU \times h = 45$

$RU = 10$, $h=9$

Also we know that $h$ is height between $QT$ and $SU$.

$QTUS = QRST + RTU$

Area of parallelogram $QRST=SR \times h = 6 \times 9 = 54$

Area of trapezium $QTUS= 54+45 = 99$ cm$^2$

3. Final Answer

The area of trapezium $QTUS$ is 99 cm$^2$.