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 will learn how to deal with Math Modeling, Problem Sum from Lower Primary School to Upper Primary School level. This posting is an upper primary school math question on Perimater and Area.

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

Read also the posting on **Sarin learns Shapes and Solids in school (Math Concept)** to know about shapes

**Upper primary school mathematics question UPQ238**

Three semi-circles and a right-angled triangle ABC lie inside Figure EFGH. Triangle ABC has a perimeter of 48 cm. If AC = 20cm, BC = 12cm, find

(a) the length of AB

(b) the area of Figure EFGH

(c) the area of the Pink shaded figure ( take π = 3.14)

Solution:

AB = 48 – 20 – 12 = 16 cm

By extending the right-angle triangle ABC to a rectangle as blue dotted lines shown

You can reduce that J, K and L are the centre points of the three semicircles respectively

So,

JA = JC = JM = JN = 20 ÷ 2 = 10 cm

LA = LB = LO = JK = 16 ÷ 2 = 8 cm

KC = KB = KP = JL = 12 ÷ 2 = 6 cm

Length EF = 10 + 8 + 6 = 24 cm

Breadth FG = 10 + 8 + 6 = 24 cm

The area of EFGH = 24 × 24 = 576 cm^{2}

The area of semi-circle = 1/2 × π × r × r = 1.57 × r × r

The area of the three semi-circles = 1.57 × (10 × 10 + 8 × 8 + × 6 × 6) = 314 cm^{2}

The area of Triangle ABC = ½ × 16 × 12 = 96 cm^{2}

The area of pink shaded figure = 576 – 314 – 96 = 166 cm^{2}

Note : EF = FG, EFGH is a square