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Solving Grade 5 Math Problems on Pages 15 and 16 of VBT Volume 2, Focus Practice, Exercise 99 is a resource to help students excel in Math with exercises from exercise 99. Through guided solutions and explanations for exercises 1, 2, 3..., students can easily complete the exercises and reinforce their knowledge related to the area and circumference of circles.
Solving Grade 5 Math Problems on Pages 15 and 16 of VBT Volume 2, Focus Practice, Exercise 99
1. Solve Exercise 1 - Grade 5 Math Exercise Book Volume 2 Page 15
Problem:
Calculate the length of the steel wire used to bend into a flower as shown in the diagram.
Solution Method
The length of the steel wire used to bend into a flower as shown in the diagram is four times half the circumference of a circle with a diameter of 9cm.
Answer
The circumference of a circle with a diameter of 9cm is:
9 x 3.14 = 28.26 (cm)
Half the circumference of a circle is:
14.13 (cm)
The length of the steel wire is four times half the circumference of the circle:
4 x 14.13 = 56.52 (cm)
Answer: 56.52cm
2. Solve Exercise 2 - Math Workbook Grade 5 Page 15
Problem:
Two circles with the same center O as shown in the figure. The small circle has a radius of 5m. The circumference of the large circle is 40.82m. What is the difference in meters between the radius of the large circle and the radius of the small circle?
Solution method:
- Find the diameter of the large circle = circumference of the large circle : 3.14 : 2.
- The difference between the two radii = radius of the large circle - radius of the small circle.
Answer
The diameter of the large circle is:
Diameter of the large circle is:
The radius of the large circle is:
Difference between the two radii is:
So the radius of the large circle is longer than the radius of the small circle by 1.5 meters.
Difference between the two radii is:
The radius of the large circle is longer than the radius of the small circle by 1.5 meters.
3. Solve Problem 3 - Math Workbook Grade 5 Unit 2 Page 16
Problem:
The area of the figure is:
The figure above is formed by a semicircle and a triangle.
The area of the shaded part is:
A. 46.26cm2
B. 50.13cm2
C. 28.26cm2
D. 32.13cm2
Solution:
- Area of the drawing = area of the semicircle + area of the triangle.
- Area of the circle = radius x radius x 3.14.
- Area of the triangle = base length x height : 2.
Answer
The radius of the circle is:
3 (cm)
The area of the circle is:
28.26 (cm2)
Area of the semicircle is:
14.13 (cm2)
Area of the triangle is:
18 (cm2)
Area of the figure is:
32.13 (cm2)
Choose answer D
4. Solve problem 4 - Exercise book for grade 5 lesson 2 page 16
Problem:
Circle the letter before the correct answer:
The area of the shaded part of the square is:
A. 243cm2
B. 126cm2
C. 314cm2
D. 86cm2
Solution:
- The area of 2 semicircles equals the area of 1 circle with a diameter of 20cm.
- The shaded area = area of a square with a side length of 20cm - area of 2 semicircles.
Answer:
The area of the square is:
The area of the square is:
The radius of the circle is:
10 (cm)
The area of two semicircles is:
The area of the shaded part is:
The area of the square is:
86 (cm2)
Choose answer D
You are currently viewing the guide for solving math problems for Grade 5 on pages 15 and 16 of VBT Exercise Book 2, Concentrated Practice, Exercise 99. You can review the guide for solving math problems for Grade 5 on page 14 of VBT Exercise Book 2, Concentrated Practice, or preview the guide for solving math problems for Grade 5 on pages 16 and 17 of VBT Exercise Book 2, Introduction to Pie Charts to understand more about the lesson.
Wishing you all the best in studying math.
