**Question 1**:

In the diagram,

*ABCD*is a trapezium and

*ABEF*is a parallelogram.

Calculate the area, in cm

^{2}, of the coloured region.

**:**

SolutionSolution

$\begin{array}{l}\text{Areaoftrapezium}ABCD\\ =\frac{1}{2}\times \left(8+14\right)\times 10\\ =110{\text{cm}}^{2}\\ \\ \text{Areaofparallelogram}ABEF\\ =8\times 6\\ =48{\text{cm}}^{2}\\ \\ \text{Areaoftheshadedregion}\\ =110-48\\ =62{\text{cm}}^{2}\end{array}$

**Question 2**:

Diagram below shows a rectangle

*ABCD.*

Calculate the area, in cm

^{2}, of the coloured region.

**Solution**:$\begin{array}{l}\text{Theareaofthecolouredregion}\\ =\text{Areaofrectangle}-\text{Areaoftrapezium}\\ =\left(12\times 8\right)-\frac{1}{2}\times \left(4+6\right)\times 4\\ =96-20\\ =76{\text{cm}}^{2}\end{array}$

**Question 3**:

In diagram below,

*AEC*is a right-angled triangle with an area of 54 cm

^{2}and

*BCDF*is a rectangle.

Calculate

(a) the perimeter, in cm, of the coloured region.

(b) the area, in cm

^{2}, of the coloured region.

**Solution**:$\begin{array}{l}\text{(a)}\\ \text{Givenareaof}\u25b3\text{}ACE\\ \frac{1}{2}\times AC\times 9=54\\ \text{}AC=54\times \frac{2}{9}\\ \text{}AC=12\text{cm}\\ \\ \text{UsingPythagoras'theorem:}\\ AE=\sqrt{{9}^{2}+{12}^{2}}\\ \text{}=15\text{cm}\\ \\ \text{Perimeterofcolouredregion}\\ =6+4.5+6+4.5+15\\ =36\text{cm}\end{array}$

$\begin{array}{l}\text{(b)}\\ \text{Areaofthecolouredregion}\\ =\text{Areaof}\u25b3\text{}ACE-\text{Areaofrectangle}BCDF\\ =54-\left(6\times 4.5\right)\\ =54-27\\ =27{\text{cm}}^{2}\end{array}$

**Question 4**:

Diagram below shows a trapezium

*ABCDE. ABGF*is a square with an area of 36 cm

^{2}.

Calculate

(a) the perimeter, in cm, of the coloured region.

(b) the area, in cm

^{2}, of the coloured region.

**Solution**:$\begin{array}{l}\text{(a)}\\ \text{UsingPythagoras'theorem:}\\ \text{In}\u25b3\text{}CDH,\\ CD=\sqrt{{8}^{2}+{6}^{2}}\\ \text{}=10\text{cm}\\ \\ AB=BG=GF=FA=\sqrt{36}=6\text{cm}\\ \\ \text{Perimeterofcolouredregion}\\ =6+10+18+2+6+6\\ =48\text{cm}\end{array}$

$\begin{array}{l}\text{(b)}\\ \text{Areaofthecolouredregion}\\ =\text{Areaoftrapezium}ABCDE-\text{Areaofsquare}ABGF\\ =\left[\frac{1}{2}\left(12+18\right)\times 8\right]-36\\ =\left[\frac{1}{2}\times 30\times 8\right]-36\\ =120-36\\ =84{\text{cm}}^{2}\end{array}$