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primary 6 | Maths | Measurement
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H.
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primary 6 chevron_right Maths chevron_right Measurement chevron_right Singapore

Please help for (b)

Date Posted: 3 years ago
Views: 363

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Let L be the length of one side of the small square.

Since Perimeter of big square is 4 times larger, length of one side of big square = 4L

Area of small square = L x L = L²

Area of big square = 4L x 4L = 16L²

Hence the area of the small square is 1/15 that of the area of the big square.

For part B, it is important to recognise that the shaded region is 1/2 of the diagram, therefore the unshaded region is 1/2 of the total area.

Difference in area = 16L² – L² = 15L² = 885

L² = 885/15 = 59
L = √59 = 7.68

Unshaded area = half the area of small square and big square

Unshaded area = 1/2 x (17 x 59) = 501.5
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Ghazi Emir Jennings
Ghazi Emir Jennings's answer
16 answers (Tutor Details)
1st
Ghazi Emir Jennings
Ghazi Emir Jennings
3 years ago
To clarify the last step,
Unshaded area = half of total area
Total Area = 16L² + L² = 17L² = 17 x 59 = 1003cm²
Unshaded area = 1/2 x 1003 = 501.5 cm²
H.
H.
3 years ago
How to get 17?
Ghazi Emir Jennings
Ghazi Emir Jennings
3 years ago
It’s the sum of the areas of
the small square (L x L) + big square (4L x 4L)