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primary 5 | Maths | Percentage
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Thks

Date Posted: 2 years ago
Views: 491

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AC Lim
Ac Lim's answer
12418 answers (A Helpful Person)
1st
Hope this helps
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2 years ago
Thks
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2 years ago
Thks
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MisterXYZ
Misterxyz's answer
37 answers (A Helpful Person)
Another method that is different from the answe4 that was provided, But i hope this aids in your understanding.
J
J
2 years ago
Your method , although arrives at the same final answer, is incorrect.

Notice that the perpendicular height of △APR is shorter than BQ (18cm), so to write △APR'S area as ½ × 12cm × 18cm is incorrect.


This means that area of PBQR is not 108cm²


The final answer was obtained correctly only because the two area PBQRs cancelled out when you did 108cm² - 36cm² = 72cm²

i.e

difference

= Area of △APR - Area of △RQC

= 108cm² - 36cm²

= (Area of △ABQ - Area of PBQR) - (Area of △PBC - Area of PBQR)

= (216cm² - 108cm²) - (144cm² - 108cm²)

= 216cm² - 108cm² - 144cm² + 108cm²

= 216cm² - 144cm²

= 72cm²

(216cm² - 144cm² would be the correct method which AC Lim used)


We could let area of PBQR be any value (eg. 50, 100,2) and you will still get the same difference of 72cm²
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HiderApplecross
Hiderapplecross's answer
1783 answers (Tutor Details)
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