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Question
primary 6 | Maths
| Fractions
2 Answers Below
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Help plss
Date Posted: 1 year ago
Views: 296
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Hope this helps
Date Posted:
1 year ago
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What I did:
I rearranged the squares to find a pattern (on the top left of the picture)
I found the pattern to be
Figure 1 = 3 3 1
Figure 2 = 5 5 1 1
where Figure X = 2[3 + 2(x-1)] + x
Using this idea, I was able to solve for a & b.
Since 3 + 2(x-1) + x = no of squares
for (c), 142 = 2[3 + 2(x-1)] + x
I rearranged the squares to find a pattern (on the top left of the picture)
I found the pattern to be
Figure 1 = 3 3 1
Figure 2 = 5 5 1 1
where Figure X = 2[3 + 2(x-1)] + x
Using this idea, I was able to solve for a & b.
Since 3 + 2(x-1) + x = no of squares
for (c), 142 = 2[3 + 2(x-1)] + x
Date Posted:
1 year ago
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