# Sarin learns Perimeter, Area and Volume in school part 70 (Math Question)

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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 acquire  the skill of dealing with Math Modeling, Math Problem Solving from Lower Primary School to Upper Primary School level after reading the blog postings. This posting is an upper primary school math question on Shapes 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

Challenge yourself with the question before look out for the given solution‼!

Upper primary school mathematics question UPQ265

The figure below is made up of Squares EFGH and PQRS, a circle and two identical semicircles. J and K are the midpoints of PQ and RQ respectively. P, Q, R and S are the midpoints of EF, FG, GH and EH respectively. Use the calculator value of π to find the total area of the shaded parts, correct to 2 decimal places.

Solution:

Shifted portions of the coloured figure and recoloured to 2 portions

The area of the yellow triangle = ¼ × 14 × 14 = 49 cm2

The area of the blue arc = ¼ × π × 14 × 14 – ½ × 14 × 14 = (49π – 98) cm2

The area of the coloured figure = 49 + 49π – 98 = 49 (π – 1) = 104.94 cm2

Alternative solution:

Shifted the coloured portions to form three different coloured portions as shown below

The area of the yellow arc = ¼ × π × 7 × 7 – ½ × 7 × 7 = 49 (1/4 π – ½)

The area of the blue arc = ¼ × π × 14 × 14 – ½ × 14 × 14 = 196 (1/4 π – ½)

The area of the orange portion = 1/4 × 14 × 14 – 49(1/4 π – ½) = 49 (1 1/2 – ¼ π)

The area of the coloured figure = 49 (1 1/2 – 1/4 π) + 196 (1/4 π – ½) + 49 (1/4 π – ½) = 49π – 49 = 104.94 cm2