Question 9:
In diagram below, ADB is a right-angled triangle and DBFE is a square. C is the midpoint of DB and CH = CD.
Calculate the area, in cm², of the coloured region.

Solution:
$\begin{array}{l}\text{Area of }△ABC\\ =\frac{1}{2}\times 6\times 8\\ =24{\text{ cm}}^2\\ \\ \text{Area of trapezium }BCHF\\ =\frac{1}{2}\times \left(12+6\right)\times 6\\ =54{\text{ cm}}^2\\ \\ \text{Area of }CDEFH\\ =\left(12\times 12\right)-54\\ =144-54\\ =90{\text{ cm}}^2\\ \\ \text{Area of coloured region}\\ =24+90\\ =114{\text{ cm}}^2\end{array}$

Question 10:
Diagram below shows a rectangle ACDE.

Calculate the area, in cm², of the coloured region.

Solution:
$\begin{array}{l}\text{Using Pythagoras}\text{'}\text{ theorem (Refer Form 2 Chapter 6)}\\ F{E}^2=D{F}^2-D{E}^2\\ ={13}^2-{12}^2\\ =169-144\\ =25\\ FE=\sqrt{25}=5\text{ cm}\\ \\ AF=8-5=3\text{ cm}\\ AB=12-8=4\text{ cm}\\ \\ \text{Area of rectangle }ACDE\\ =8\times 12\\ =96{\text{ cm}}^2\\ \\ \text{Area of }△ABF\\ =\frac{1}{2}\times 3\times 4\\ =6{\text{ cm}}^2\\ \\ \text{Area of }△DEF\\ =\frac{1}{2}\times 5\times 12\\ =30{\text{ cm}}^2\\ \\ \text{Area of coloured region}\\ =96-30-6\\ =60{\text{ cm}}^2\end{array}$

Question 11:
Diagram below shows a sketch of parallelogram shaped garden, PQRS that consists of flower beds and a playground.

Calculate the area, in m², of the flower beds.

Solution:
$\begin{array}{l}\text{Area flower bed}\\ =\left(12\times 14\right)-\left(\frac{1}{2}\times 12\times 7\right)\\ =168-42\\ =126{\text{ m}}^2\end{array}$

Question 5:
In diagram below, PQUV is a square, QRTU is a rectangle and RST is an equilateral triangle.


The perimeter of the whole diagram is 310 cm.
Calculate the length, in cm, of PV.

Solution:
$\begin{array}{l}PV=VU=TS=SR=QP\\ \text{Given perimeter of the whole diagram}=310\text{ cm}\\ \\ PV+VU+UT+TS+SR+RQ+QP=310\\ PV+PV+50+PV+PV+50+PV=310\\ 5PV+100=310\\ 5PV=210\\ PV=42\text{ cm}\end{array}$

Question 6:
In diagram below, ABCD and CGFE are rectangles. M, G, E and N are midpoints of AB, BC, CD and DA respectively.


Calculate the perimeter, in cm, of the coloured region.

Solution:


$\begin{array}{l}\text{Using Pythagoras}\text{'}\text{ theorem}\\ M{G}^2=M{F}^2+F{G}^2\\ =5^2+{12}^2\\ =25+144\\ =169\\ MG=13\text{ cm}\\ \\ \text{Perimeter of the coloured region}\\ =13+13+5+12+5+12\\ =60\text{ cm}\end{array}$

Question 7:
Diagram below shows a trapezium BCDE and a parallelogram ABEF. ABC and FED are straight lines.

The area of ABEF is 72 cm².
Calculate the area, in cm², of trapezium BCDE.

Solution:
$\begin{array}{l}\text{Base }\times \text{ Height }=\text{Area of }ABFE\\ 9\text{ cm }\times \text{ Height}=72{\text{ cm}}^2\\ \text{ Height}=\frac{72}{9}\\ =8\text{ cm}\\ CD=8\text{ cm}\\ BC=23-9\\ =14\text{ cm}\\ \\ \text{Area of trapezium }BCDE\\ =\frac{1}{2}\times \left(BC+ED\right)\times CD\\ =\frac{1}{2}\times \left(14+9\right)\times 8\\ =92{\text{ cm}}^2\end{array}$

Question 8:
In diagram below, ACEF is a trapezium and BCDG is a square.


Calculate the area, in cm², of the coloured region.

Solution:
$\begin{array}{l}\text{Area of trapezium }ACEF\\ =\frac{1}{2}\times \left(8+15\right)\times 10\\ =115{\text{ cm}}^2\\ \\ \text{Area of square }BCDG\\ =4\times 4\\ =16{\text{ cm}}^2\\ \\ \text{Area of trapezium }GDEF\\ =\frac{1}{2}\times \left(4+15\right)\times 6\\ =57{\text{ cm}}^2\\ \\ \text{Area of coloured region}\\ =115-16-57\\ =42{\text{ cm}}^2\end{array}$

Question 1:
In the diagram, ABCD is a trapezium and ABEF is a parallelogram.

Calculate the area, in cm², of the coloured region.

Solution:
$\begin{array}{l}\text{Area of trapezium }ABCD\\ =\frac{1}{2}\times \left(8+14\right)\times 10\\ =110{\text{ cm}}^2\\ \\ \text{Area of parallelogram }ABEF\\ =8\times 6\\ =48{\text{ cm}}^2\\ \\ \text{Area of the shaded region}\\ =110-48\\ =62{\text{ cm}}^2\end{array}$

Question 2:
Diagram below shows a rectangle ABCD.


Calculate the area, in cm², of the coloured region.

Solution:

$\begin{array}{l}\text{The area of the coloured region}\\ =\text{Area of rectangle}-\text{Area of trapezium}\\ =\left(12\times 8\right)-\frac{1}{2}\times \left(4+6\right)\times 4\\ =96-20\\ =76{\text{ cm}}^2\end{array}$

Question 3:
In diagram below, AEC is a right-angled triangle with an area of 54 cm² and BCDF is a rectangle.
Calculate
(a) the perimeter, in cm, of the coloured region.
(b) the area, in cm², of the coloured region.

Solution:
$\begin{array}{l}\text{(a)}\\ \text{Given area of }△ACE\\ \frac{1}{2}\times AC\times 9=54\\ AC=54\times \frac{2}{9}\\ AC=12\text{ cm}\\ \\ \text{Using Pythagoras' theorem:}\\ AE=\sqrt{9^2+{12}^2}\\ =15\text{ cm}\\ \\ \text{Perimeter of coloured region}\\ =6+4.5+6+4.5+15\\ =36\text{ cm}\end{array}$

$\begin{array}{l}\text{(b)}\\ \text{Area of the coloured region}\\ =\text{Area of }△ACE-\text{Area of rectangle }BCDF\\ =54-\left(6\times 4.5\right)\\ =54-27\\ =27{\text{ cm}}^2\end{array}$

Question 4:
Diagram below shows a trapezium ABCDE. ABGF is a square with an area of 36 cm².


Calculate
(a) the perimeter, in cm, of the coloured region.
(b) the area, in cm², of the coloured region.

Solution:

$\begin{array}{l}\text{(a)}\\ \text{Using Pythagoras' theorem:}\\ \text{In }△CDH,\\ CD=\sqrt{8^2+6^2}\\ =10\text{ cm}\\ \\ AB=BG=GF=FA=\sqrt{36}=6\text{ cm}\\ \\ \text{Perimeter of coloured region}\\ =6+10+18+2+6+6\\ =48\text{ cm}\end{array}$

$\begin{array}{l}\text{(b)}\\ \text{Area of the coloured region}\\ =\text{Area of trapezium }ABCDE-\text{Area of square }ABGF\\ =\left[\frac{1}{2}\left(12+18\right)\times 8\right]-36\\ =\left[\frac{1}{2}\times 30\times 8\right]-36\\ =120-36\\ =84{\text{ cm}}^2\end{array}$