Sarin learns Perimeter, Area and Volume in school part 78 (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 and Problem Sum from Lower Primary School to Upper Primary School level after reading the blog postings. This posting is an upper primary school math question on FractionArea and Problem Sum.

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

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

Upper primary school mathematics question UPQ333

The figure below is made up of a rectangle and a square. The overlapped portion of the two shapes is marked in orange. The area of the square is 2/5 the area of the rectangle. The area of the unoverlapped portion of the square is 1/6 the area of the unoverlapped portion of the rectangle. Given that the length of the square is 6 cm, find the area of the figure below.

UPQ333-1

Solution:

            The area of the square is 2/5 fraction of the area of the triangle

            The area of the square = 6 × 6 = 36cm2

UPQ333-2

            From the model

            2 units = 36

            1 unit = 18cm2

            The area of the rectangle = 18 × 5 = 90cm2

            The area of square plus rectangle = 36 + 90 = 126cm2

            The area of unoverlapped square is 1/6 fraction of the area of the unoverlapped rectangle

            The orange portion is same in area

            The updated model

UPQ333-3

            Per the model

            5 units = 90 – 36 = 54cm2

            1 unit = 10.8cm2

            The area of orange portion of the figure = 36 – 12.8 = 23.2cm2

            The area of the figure = 36 + 6 × 10.8 = 100.8cm2 

 

More Questions on Area. Click here…. 

More Questions on Fraction. Click here…. 

More Questions on Problem Sum. Click here….

 

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