Home / IB MYP Year 4-5: Standard Mathematics : Unit 3: Function -\(\rm{y = mx + c}\) MYP Style Questions

IB MYP Year 4-5: Standard Mathematics : Unit 3: Function -\(\rm{y = mx + c}\) MYP Style Questions

IB myp 4-5 MATHEMATICS – Practice Questions- All Topics

Topic :Function-y=mx+c

Topic :Function- Weightage : 21 % 

All Questions for Topic : Mappings,Function notation,Linear functions,y=mx+c,Parallel and Perpendicular lines,System of equations/simultaneous equations,Quadratic functions,Algorithms 

Question (33 marks)

The line $\rm{y=2 x+1}$ is shown on the coordinate axes below. Vertical line segments can be added between the line and the $\rm{x}$ axis, one unit apart horizontally. In this question, you will investigate areas of different trapeziums formed.

Question (a) : 2 marks

For stage 3 , show that the area of the trapezium is 6 units squared.

▶️Answer/Explanation

Ans:

To show that the area of the trapezium is 6 square units, we can use the formula for the area of a trapezium:

\[ \text{Area} = \frac{1}{2} (a + b) \times h \]

where \( a \) and \( b \) are the lengths of the parallel sides, and \( h \) is the distance between the parallel sides.

In this case, we are given that the lengths of the parallel sides are $5$ and $7$, and the distance between the parallel sides is $1$.

Substituting the given values into the formula, we have:

\[ \text{Area} = \frac{1}{2} (5 + 7) \times 1 \]

Simplifying the expression:

\[ \text{Area} = \frac{1}{2} \times 12 \times 1 \]

\[ \text{Area} = 6 \]

Therefore, the area of the trapezium is indeed 6 square units.

Question (b) : 1 marks

Write down the missing values in the table up to stage 6 .

▶️Answer/Explanation

Ans:

To fill in the missing values in the table up to stage 6:

To calculate the area of the trapezium at stage 5, we can observe that the area is increasing by 2 units at each stage. Thus, we can add 2 to the previous area (8) to find the area at stage 5, which gives us 10.

Similarly, to calculate the area of the trapezium at stage 6, we can add 2 to the previous area (10) to find the area at stage 6, which gives us 12.

Therefore, the missing values in the table are 10 for stage 5 and 12 for stage 6.

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