48cm
Area of rectangle is 128cm²
So use algebra to solve (remember area of rectangle = LxW)
2x * x = 128 (where 2x = length and x = width. Use 2x given length is twice width)
3x=128
2x²=128
x²=64
x=8
So perimeter = (8*4 for the length + 8*2 for the width) = 48cm
72cm
(15 * 4) + (3*4) = 60 + 12
C
The answer would be the highest factor of both 8 & 20 from the options since each tile is square. The answer to this is 4
Area = 200cm²
- If 2x long as wide, then need to calculate each side
- Double length v width means length is 20cm and width is 10
- Area = 200cm²
a) 160cm
b) 168cm²
a) 0.33cm²
Overlap is 1/3rd of a triangle. So area if full would be 1×1 = 1cm. So answer is 1/3rd of this
b) 12.25cm²
2 full shaded squares. In addition, 1.5 other shaded squares. So 3.5 x 3.5
462cm
Large Triangle = 16cm * 3 (all sides same as its equilateral). So 48cm
Middle Triangle = 7cm * 3 triangles * 3 sides (7 small triangles of 1cm each on the horizontal base tells us its a 7x7x7 triangle). So 63cm
Next size Triangle of which there are 3 = 3cm * 3 triangles * 3 sides = 27cm. There are 3 of them so 27cm
Smallest triangle = 36 small triangles all of 1cm per side. So 3cm *36 triangles * 3 sides = 324cm
Total of all of these is 48 + 63 + 27 + 324
64cm
Perimeter = 14+3+4+6+7+2+17+11
a) 30°
180 – Australia – Africa = 180-90-60
b) 1/8
45/360 = 9/72 = 1/8
c) 30
180/1 * 60/360 = 180*1/6 = 30
d) 1000
Degrees for Europe = 180 – USA = 180-45 = 135
1600 * 225/360. Simplified this is 1600 * 45/72 = 15/24
1600/1 * 15/24
1/4
4 parts of 16 shaded = 4/16 = 1/4
a) 1200mm²
Area of rectangle = LxW = 60mm x 20mm
b) 40 triangles
2 Triangles will joined together have dimensions of LxW = 12mmx5mm. So they form a rectangle
Therefore, in 60mm across, you get 5 “blocks” of 2 triangles. And you can go to a height of 4 blocks
So in total 5 * 4 blocks = 20 blocks. 2 triangles per block so 40 triangles
c) 1200mm
Perimeter of 1 triangle = 12+5+13=30mm
Total number of triangles = 40
Total Perimeter =1200mm
d) 1.2m
10mm = 1cm
100cm = 1m
1000mm = 1m
Area of 1 rectangle = 15 x 5 = 75cm²
30cm
- If Length is 11cm more than width, need possible combinations to give multiple of 60
- So this could be 30×2 or 15×4. Or several others
- 15×4 is the only combination that works and this gives a length 11cm more than the width
- So perimeter is 11cm + 4cm + 11cm + 4cm
20cm²
Its 10cm length x 2cm height. So 20cm²
80cm
- 9 identical squares with area of 144 means square has an area of 16cm
- Each square must be 4×4
- Number of sides showing on the shape above if you count all the way around as a perimeter is 20
- So perimeter is 80cm
96 tiles
3metres long and 4m wide = 12m² area
You will get 12 tiles in the width from 25cm tiles (3metres = 300cm. 300cm/25 = 12 tiles)
Across you will get 8 tiles in the length from 50cm tiles ( 4metres = 400cm/50cm – 8 tiles)
12 x 8 = 96 tiles
a) 64cm²
Perimeter = 32. Therefore, each side is 8cm
Area = 8 x 8 = 64
b) 48cm²
20 students
- Work on basis 6 students = 30°
- Therefore, 2 students = 10°
- So, 100° = 20 students
Approximately 40°
An acute angle
65°
- Triangle = 180°
- 180 – 50 – 130
- 130/2 = 65
8m
- Area is LxW
- So working backwards, 72=9xW
- Rearrange. W=8
72 stickers
- 80/10 = 8
- 45/5 = 9
- 8 * 9 = 72