Jie Min's answer to Amir's Primary 4 Maths Singapore question.
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
Hope this helps
Date Posted:
3 years ago
Thank you so much
Faster way without having to add up all the small individual lengths :
Realise that in the remaining piece of paper, the 5cm, 4cm section can be transferred leftwards to the dotted lines of the rectangle.
For the 6cm section, 5cm of it can be transferred downwards to the dotted lines of the rectangle, leaving a 1 cm part.
Mow there's another short part that is also 1cm, just directly below.
The perimeter of the remaining paper is therefore equal to the perimeter of the original rectangle, plus those two 1 cms
(20cm + 30cm + 1cm) x 2
= 51cm x 2
= 102cm
This method saves you the trouble of adding individual lengths (and lowering the risk of calculation errors / forgetting to include lengths, which many students do)
Realise that in the remaining piece of paper, the 5cm, 4cm section can be transferred leftwards to the dotted lines of the rectangle.
For the 6cm section, 5cm of it can be transferred downwards to the dotted lines of the rectangle, leaving a 1 cm part.
Mow there's another short part that is also 1cm, just directly below.
The perimeter of the remaining paper is therefore equal to the perimeter of the original rectangle, plus those two 1 cms
(20cm + 30cm + 1cm) x 2
= 51cm x 2
= 102cm
This method saves you the trouble of adding individual lengths (and lowering the risk of calculation errors / forgetting to include lengths, which many students do)
360 Tutoring Program