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primary 6 | Maths | Fractions
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Maple Chan
Maple Chan

primary 6 chevron_right Maths chevron_right Fractions chevron_right Singapore

Help pl

Date Posted: 3 years ago
Views: 446
J
J
3 years ago
Summary explanation :

Difference between their perimeters
= TQ + UR = ​8cm

Since TQ = UR, each of them = 8cm ÷ 2 = 4cm

① PQRS and TQRU have common side QR, and PS = TU

Since 10% of PQRS is shaded, TQRU has 10% of PQRS area and because of point ①,

TQ and UR are 10% (or 1/10) of PQ and RS.
PQ and RS are 10 times of TQ and UR.

PT = PQ - TQ
= 10 × 4cm - 4cm
= 40 cm - 4cm
= 36cm

Or

PT = 9 times of TQ
= 9 × 4cm
= 36cm

See 1 Answer

Explanation :
For the perimeters of PQRS and PTUS,
①PS is common to both rectangles (shared side)
②TU has the same length as QR

The net difference is: (PQ + SR) - (PT + SU)
Since 10% of the rectangle PQRS is shaded, rectangle TQRU has 10% or 1/10 of PQRS's area.
Since their sides QR are the same, then this means the breadths TQ and UR must be 1/10 of the lengths PQ and SR
We can let TQ and UR be 1u each and PQ and SR be 10u each
Then this means that PT and SU are 9u each.
The overall difference is 8cm.
(10u + 10u) - (9u + 9u) = 8cm
20u - 18u = 8cm
2u = 8cm
1u = 8cm ÷ 2 = 4cm
(Notice that the difference is equal to the lengths of TQ + UR)
Finally, PT = 9u = 9 × 4cm = 36cm
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J
J's answer
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Maple Chan
Maple Chan
3 years ago
Thank you so much. Understand now.