11.2.3 Perimeter and Area, PT3 Practice


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:
Area of  ABC = 1 2 ×6×8 =24  cm 2 Area of trapezium BCHF = 1 2 ×( 12+6 )×6 =54  cm 2 Area of CDEFH =( 12×12 )54 =14454 =90  cm 2 Area of coloured region =24+90 =114  cm 2


Question 10:
Diagram below shows a rectangle ACDE.

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

Solution:
Using Pythagoras' theorem (Refer Form 2 Chapter 6) F E 2 =D F 2 D E 2         = 13 2 12 2         =169144         =25 FE= 25 =5 cm AF=85=3 cm AB=128=4 cm Area of rectangle ACDE =8×12 =96  cm 2 Area of  ABF = 1 2 ×3×4 =6  cm 2 Area of  DEF = 1 2 ×5×12 =30  cm 2 Area of coloured region =96306 =60  cm 2


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:
Area flower bed =( 12×14 )( 1 2 ×12×7 ) =16842 =126  m 2



11.2.2 Perimeter and Area, PT3 Practice


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:
PV=VU=TS=SR=QP Given perimeter of the whole diagram=310 cm PV+VU+UT+TS+SR+RQ+QP=310 PV+PV+50+PV+PV+50+PV=310 5PV+100=310  5PV=210    PV=42 cm


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:


Using Pythagoras' theorem M G 2 =M F 2 +F G 2 = 5 2 + 12 2 =25+144 =169 MG=13 cm Perimeter of the coloured region =13+13+5+12+5+12 =60 cm


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:
Base × Height =Area of ABFE 9 cm × Height=72  cm 2  Height= 72 9 =8 cm CD=8 cm BC=239  =14 cm Area of trapezium BCDE = 1 2 ×( BC+ED )×CD = 1 2 ×( 14+9 )×8 =92  cm 2


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


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

Solution:
Area of trapezium ACEF = 1 2 ×( 8+15 )×10 =115  cm 2 Area of square BCDG =4×4 =16  cm 2 Area of trapezium GDEF = 1 2 ×( 4+15 )×6 =57  cm 2 Area of coloured region =1151657 =42  cm 2

11.2.1 Perimeter and Area, PT3 Practice


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

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

Solution
:

Area of trapezium ABCD = 1 2 ×( 8+14 )×10 =110  cm 2 Area of parallelogram ABEF =8×6 =48  cm 2 Area of the shaded region =11048 =62  cm 2


Question 2:
Diagram below shows a rectangle ABCD.


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

Solution:

The area of the coloured region =Area of rectangleArea of trapezium =( 12×8 ) 1 2 ×( 4+6 )×4 =9620 =76  cm 2


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:
(a) Given area of  ACE 1 2 ×AC×9=54    AC=54× 2 9    AC=12 cm Using Pythagoras' theorem: AE= 9 2 + 12 2   =15 cm Perimeter of coloured region =6+4.5+6+4.5+15 =36 cm

(b) Area of the coloured region =Area of  ACEArea of rectangle BCDF =54( 6×4.5 ) =5427 =27  cm 2


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:

(a) Using Pythagoras' theorem: In  CDH, CD= 8 2 + 6 2  =10 cm AB=BG=GF=FA= 36 =6 cm Perimeter of coloured region =6+10+18+2+6+6 =48 cm

(b) Area of the coloured region =Area of trapezium ABCDEArea of square ABGF =[ 1 2 ( 12+18 )×8 ]36 =[ 1 2 ×30×8 ]36 =12036 =84  cm 2