The blog postings are about the Singapore Math. The readers can learn from the postings about Solving Singapore Primary School Mathematics. The blog presenting the Math Concept, Math Questions with solutions that teaches in Singapore Primary Schools. You or the kids can learn the skills of dealing with Math Modeling, Problem Sum from Lower Primary School to Upper Primary School level after reading the postings. This posting is an upper primary school math question on Perimeter, Area and Problem Solving.

You can read the posting on **Sarin’s family and Ramadan** and **Sarin’s family celebrating Hari Raya Puasa** in Welcome page about story of Sarin’s family during Ramadan and celebrating Hari Raya Puasa.

You shuold read the posting on **Sarin learns Perimeter, Area and Volume in school(Math Concept) **to learn the concept of Area.

*Challenge yourself to solve the question before look out for the given solution‼*

**Upper primary school mathematics question UPQ259**

After tidying up the house, Fatimah’s mother sews a piece of carpet to use during Hari Raya Puasa. The carpet has a brown outskirt with 10cm wide and fills with yellow and purple circles with identical size. A green cloth is used as the centre piece of the full carpet. The figure below showed a quarter of the carpet design. If there are 38 circles in total in a quarter of the carpet,

a) Find the outskirt area that is not covered by the circles.

b) Find the perimeter and area of the green cloth.

Solution:

Base on a quarter of the carpet.

The length of a quarter of brown outskirt = 10 × 38 ÷ 2 ÷ 2 = 95cm

The area of a quarter of the brown outskirt = 95 × 10 = 950cm^{2}

The radius of each circle = 10 ÷ 4 =2.5cm

The area of 38 circles = 38 × π × 2.5 × 2.5 = 38 × 3.14 × 2.5 × 2.5 = 745.75cm^{2}

The area of a quarter of the outskirt not cover with circles = 950 – 745.75 = 204.25cm^{2}

The area of the outskirt not cover with circles = 4 × 204.25 = 817cm^{2}

Base on a quarter of the carpet,

Consider only the inner ring circles,

Set the number of inner ring circles as X and Y respectively

X + Y = (38 – 4) ÷ 2 = 17

The length of (X + Y) = 17 × 10 ÷ 2 = 85cm

The perimeter of the green cloth = 4 × 85 = 340cm

Assume maximum size for the carpet

17 ÷ 2 = 8R1

X = 8 and Y = 9

The length of X = 8 × 10 ÷ 2 = 40cm

The length of Y = 9 × 10 ÷ 2 = 45cm

The area of a quarter of the green cloth = 40 × 45 = 1800cm^{2}

The area of the green cloth = 4 × 1800 = 7200cm^{2}