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Solving Grade 4 Math Exercise Book 2 Pages 13-14 VBT, Practice Session, Exercise 95 provides solutions and instructions for exercises 1, 2, 3, 4 according to the curriculum. All are meticulously prepared to closely align with the learning content, making it easy for students to refer to and complete exercises in the SBT.
Solving Grade 4 Math Exercise Book 2 Pages 13-14 VBT, Practice Session, Exercise 95
1. Solve problem 1 - Grade 4 Math Exercise Book 2 Page 13
Given:
Circle the letter before the correct answer:
The shape with the largest area is:
A. Shape (1)
B. Shape (2)
C. Shape (3)
Solving Method:
- Calculate the areas of the shapes and compare the results.
- Apply the formulas:
+ Area of a square = side x side.
+ Area of a rectangle = length x width.
Answer
The area of shape 1 is: 5 x 5 = 25 (cm2)
The area of shape 2 is: 20cm2
The area of shape 3 is: 4 x 6 = 24 (cm2)
So shape (1) has the largest area.
Choose A. Shape (1).
2. Solve problem 2 - Math workbook grade 4 volume 2 page 13
Question:
Fill in the blank (according to the model)
Perimeter: 20cm
Perimeter: ...
Perimeter: ...
Perimeter: ...
Solution Method
Let P be the perimeter of the parallelogram:
P = (a + b) x 2
where a, b are the lengths of the sides of the parallelogram measured in the same unit.
Answer
Let P represent the perimeter of the parallelogram:
P = (a + b) x 2
where a, b are the lengths of the sides of the parallelogram measured in the same unit.
Perimeter of parallelogram (1) is:
P = (6 + 4) x 2 = 20 (cm)
Perimeter of parallelogram (2) is:
P = (5 + 3) x 2 = 16 (cm)
Perimeter of parallelogram (3) is:
P = (4 + 4) x 2 = 16 (cm)
Perimeter of parallelogram (4) is:
P = (5 + 4) x 2 = 18 (cm)
3. Solving problem 3 - Workbook Grade 4 Mathematics Exercise 2 Page 14
Question:
Fill in the blank (according to the pattern)
Parallelogram
Base: 4cm, Height: 34cm, Area: 136cm2
Base: 14cm, Height: ..., Area: 182cm2
Base: ..., Height: 24cm, Area: 360cm2
Solution Method
Using the formula:
Area of parallelogram = base length x height
implies:
+ Base length (base) = area of parallelogram : height
+ Height = area of parallelogram : base length (base)
Answer
(2) Height: 13cm
(3) Base length: 15cm
Area of shape H = area of rectangle ABCD + parallelogram BEFC
Answer
Since ABCD is a rectangle, AB = BC = 4cm.
The area of rectangle ABCD is:
Area of shape H = area of rectangle ABCD + parallelogram BEFC
+ Area of rectangle = length x width
+ Area of parallelogram = base length x height
Answer
Since ABCD is a rectangle, AB = BC = 4cm.
The area of rectangle ABCD is:
4 x 3 = 12 (cm2)
The area of parallelogram BEFC is:
The area of rectangle is 12 square centimeters.
The area of shape (H) is:
12 + 12 = 24 square centimeters.
Answer: 24cm2.
You are currently viewing the guide for solving Grade 4 math problems on pages 13 and 14 of Exercise Book 2 Practice Exercise 95. You can revisit the guide for solving Grade 4 math problems on pages 12 and 13 of Exercise Book 2, or preview the guide for solving Grade 4 math problems on page 15 of Exercise Book 2 to gain a better understanding of the lesson.
Wishing you all success in your math studies.
