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If you're eager to see the solutions for exercises 1, 2, 3, and 4 in Concentrated Practice, exercise 93 on pages 7 and 8 of VBT Exercise Book 2, then the solution to 5th-grade math on pages 7 and 8 of VBT Exercise Book 2, Concentrated Practice, exercise 93, will be a helpful hint for you. Let's take a look together to complete the homework accurately and easily.
Solving 5th-grade math on pages 7 and 8 of VBT Exercise Book 2, Concentrated Practice, exercise 93
1. Solve exercise 1 - Math exercise book for 5th grade Exercise 2, page 7
Problem Statement:
Among the four figures below, identify a figure with an area different from the area of the remaining three:
Solution Method
Calculate the areas of the figures using the formulas below, then compare the results:
- Area of a square = side x side.
- Area of a rectangle = length x width.
- Area of a triangle = base x height : 2.
- Area of a rhombus = length of first diagonal x length of second diagonal : 2.
Answers
The area of figure A is:
4.5 x 4.5 = 20.25 (cm2)
The area of figure B is:
9 x 6.3 = 56.7 (cm2)
The area of figure C is:
9 x 12.6 : 2 = 56.7 (cm2)
The area of figure D is:
13.5 x 8.4 : 2 = 56.7 (cm2)
Therefore, figure A has a different area compared to the remaining figures.
2. Solve exercise 2 - Math exercise book for 5th grade Exercise 2, page 8
Problem Statement:
Calculate the area of a triangle given:
a. Base length 10cm, height 8cm is: ...
b. Base length 2.2dm, height 9.3cm is: ...
c. Base length 4/5 m, height 5/8 m is: ...
Solution Method
To calculate the area of a triangle, multiply the base length by the height (using the same unit of measurement), then divide by 2.
Answers
The area of the triangle is:
a. Base length 10cm, height 8cm is:
The area of the triangle is:
10 ⨯ 8 : 2 = 40 (cm2)
b. Base length 2.2dm, height 9.3cm is:
2.2dm = 22cm
The area of the triangle is:
22 ⨯ 9.3 : 2 = 102.3 (cm2)
c. Base length 4/5 m, height 5/8 m is:
The area of the triangle is:
4/5 x 5/8 : 2 = 1/4 (m2)
3. Solve exercise 3 - Math exercise book for 5th grade Exercise 2, page 8
Problem Statement:
How much greater is the area of trapezoid ABCD compared to the area of triangle MDC (see the figure on the side)?
Solution Method
- Area of a triangle = base length x height : 2.
- Area of a trapezoid = (large base + small base) x height : 2.
- The difference of the two areas = Area of trapezoid ABCD - Area of triangle MDC.
Answers
Area of triangle MDC:
6.8 ⨯ 2.5 : 2 = 8.5 (cm2)
Area of trapezoid ABCD:
(3.2 + 6.8) x 2.5/2 = 12.5 (cm2)
The trapezoid's area greater than the triangle's area is:
12.5 - 8.5 = 4 (cm2)
Answer: 4 (cm2)
4. Solve exercise 4 - Math exercise book for 5th grade Exercise 2, page 8
Problem Statement:
A rectangle has a length of 16m and a width of 10m. If the length increases by 4m, what percentage will the rectangle's area increase?
Solution Method
- Calculate the area of the original and new rectangles using the formula:
Area = length x width.
- To find the percentage increase in area, calculate the ratio of the new rectangle's area to the original rectangle's area, then multiply the resulting ratio by 100 and add the % symbol to the right.
- To find the percentage increase, calculate the percentage ratio between the area of the new rectangle and the original rectangle, then subtract 100%.
Answers
The area of the original rectangle is:
16 ⨯ 10 = 160 (m2)
After increasing by 4m, the new length is:
16 + 4 = 20 (m)
The area of the new rectangle is:
20 ⨯ 10 = 200 (m2)
The percentage ratio between the area of the new rectangle and the original rectangle is:
200 : 160 = 1.25 = 125%
The increase in the area of the new rectangle is:
125% - 100% = 25%
Answer: 25%
Above is the guide for solving 5th-grade math page 7,8 VBT exercise 2, Concentrated practice, exercise 93. Readers are invited to continue with the instructions for Solving 5th-grade math page 10 VBT exercise 2, Circle, Circular line, exercise 94, or review the Solving 5th-grade math page 6, 7 VBT exercise 2, Practice, exercise 92 to reinforce and grasp the concepts.
