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**Solution of Exercises Lesson 37: Triangular vertical prism and quadrilateral standing prism (Chapter 10 Math 7 Connecting)**

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### Solution 10.11 page 98 Math 7 Textbook Connecting knowledge volume 2 – KNTT

Observe and name the bottom, side, bottom, and side faces of the triangular prism in Figure 10.31.

Detailed explanation guide

**Solution method**

– The bottom two sides are parallel to each other.

The side faces are rectangles.

– The sides are parallel and equal.

**Detailed explanation**

+ 2 bottom faces: ABC, MNP

+ 3 side panels: ACPM, BAMN, BCPN

+ Bottom edge: NM, MP, NP, AB, BC, CA

+ Side: AM, BN, CP

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### Solve problem 10.12 page 99 Math textbook 7 Connecting knowledge volume 2 – KNTT

Observe Figure 10.12 and tell which of the edges (1), (2), (3) joins side AB to have a vertical prism

Detailed instructions for solving problems 10.12

**Solution method**

– The bottom two sides are parallel to each other.

The side faces are rectangles.

– The sides are parallel and equal.

**Detailed explanation**

Side number (1) joins side AB to get a vertical prism.

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### Solve problem 10.13 page 99 Math textbook 7 Connecting knowledge volume 2 – KNTT

Calculate the surrounding area and volume of the vertical prism shown in Figure 10.33.

Detailed instructions for solving problems 10.13

**Solution method**

Surrounding area of triangular vertical prism, tetragonal vertical prism:

\({S_{xq}} = Ch\)

where \({S_{xq}}\): The surrounding area of the prism,

C: Perimeter of one base of the prism,

h: The height of the prism.

**Detailed explanation**

The area around the vertical prism is:

(6 + 10 + 8) .15 = 360 (m^{2} )

The area of one base of the prism is:

$\frac{1}{2}$.6.8 = 24 (m^{2} )

The volume of the vertical prism is

24.15 = 360 ( m^{3})

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### Solve problem 10.14 page 99 Math textbook 7 Connecting knowledge volume 2 – KNTT

The barrel of an agricultural machine has the shape of a prism standing quadrilateral as shown in Figure 10.34. The bottom of this vertical prism (the side of the barrel) is a square trapezoid with a large base length of 3 m, a small base of 1.5 m. How many cubic meters does the barrel have?

Detailed instructions for solving problems 10.14

**Solution method**

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Surrounding area of triangular vertical prism, tetragonal vertical prism:

\({S_{xq}} = Ch\)

where \({S_{xq}}\): The surrounding area of the prism,

C: Perimeter of one base of the prism,

h: The height of the prism.

**Detailed explanation**

The area of the base surface of a vertical prism is:

\(\frac{1}{2}\) (3 + 1.5).1.5 = 3,375 (m .)^{2})

The volume of a vertical prism is:

3,375.2 = 6.75 (m^{3})

So the capacity of the barrel will be 6.75 (m^{3})

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### Solve problem 10.15 page 99 Math textbook 7 Connecting knowledge volume 2 – KNTT

A figure consisting of two vertical prisms combined with the dimensions shown in Figure 10.35. Calculate the volume of the puzzle.

Detailed instructions for solving problems 10.15

**Solution method**

Volumes of a triangular prism and a tetragonal vertical prism:

\(V = {S_{day}}.h\)

where V is the volume of the vertical prism,

Base: Area of one base of a vertical prism,

h: Height of the vertical prism.

**Detailed explanation**

The volume of a triangular prism is:

\(\frac{1}{2}.3.5.8\) = 60 (cm^{3})

The volume of a rectangular prism is:

10. 5. 8 = 400 (cm^{3})

The volume of the jigsaw is:

60 + 400 = 460 (cm^{3})

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### Solve problem 10.16 page 99 Math textbook 7 Connecting knowledge volume 2 – KNTT

A medical mask box is made of cardboard in the shape of a rectangular box, the size is as shown in Figure 10.36.

a) Calculate the volume of the box.

b) Calculate the area of cardboard used to make the box (ignoring the gluing edge).

Detailed instructions for solving problems 10.16

**Solution method**

Area of rectangle \({S_{xq}} = 2\left( {a + b} \right).c\)

**Detailed explanation**

a) The volume of the box is:

20. 10. 8 = 1600 (cm^{3})

b) The area of cardboard used to make the box is the surrounding area and the area of the two bottoms of the box.

The area around the box is:

2.( 20 + 10 ).8 + 2. 20.10 = 880 (cm^{3})

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