Volumes and sections: corrected math exercises in 3rd grade.

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Answers to the math exercises in 3ème on the calculation of volumes and the study of solid sections in space. To know by heart the formulas of volumes (right block, cube, prism, cone of revolution, ball, cylinder and pyramid) and to study sections of solids in third grade.

Exercise 10:
The volume of the right prism is given by :

$V=Base\times \,hauteur=\frac{BA\times \,BC}{2}\times \,BF=\frac{5\times \,5}{2}\times \,5=62,5\,cm^3$

Exercise 11:

The volume of a right prism is given by :

$V=\frac{base\times \,hauteur}{3}=\frac{\frac{CB\times \,AC}{2}\times \,AD}{3}$

$V=\frac{\frac{4\times \,5}{2}\times \,7}{3}$

$V=\frac{70}{3}$

$V=23,33\,\,mm^2$

Exercise 12:

The volume of a pyramid is given by :

$V=\frac{base\times \,hauteur}{3}$

$V=\frac{8^2\times \,11}{3}=\frac{64\times \,11}{3}=\frac{704}{3}$

${\color{DarkRed}\,V=234,67\,\,cm^2}$

Exercise 13:

The volume of a cylinder is given by :

$V=Base\times \,hauteur=\pi\times \,R^2\times \,h=\pi\times \,3^2\times \,5=45\pi$

Exercise 14:

The volume of a cone of revolution is given by :

$V=\frac{Base\times \,hauteur}{3}=\frac{\pi\times \,R^2\times \,hauteur}{3}$

$V=\frac{\pi\times \,6^2\times \,8}{3}=36\pi$

$V=113,1\,\,mm^2$

Exercise 15:

The volume is given by

$V=L\times \,l\times \,h=4\times \,2\times \,1,5=12\,\,\,cm^3$

Exercise 16:

1/ a. Express in two different ways, SM as a function of h.

SM = h-OM or h -SM = OM and SM/h = EF/AB =3/7 according to Thales (hard to schematize but it is what it is).

Proof: S,E,A aligned, S,F,B aligned and (EF)//(AB) we have the equality of the ratios SE/SA = EF/AB = 3/7; Moreover S,O,M aligned and S,E and A aligned with (EM)//(AO). it comes that the ratio SE/SA which is equal to 3/7 is also equal according to Thales to SM/SO or SO=h.

So SM = 3/7 h and SM = h-60

b. Deduce an equation for which h is a solution.

h-OM=3/7 h <==> 4/7h =OM

c. Solve this equation to find the value of h.

h = 7/4*OM = 7/4*60=7*60/4=7*15=105cm

d. Calculate the volume of this flower box.

Volume of a pyramid = 1/3 Base * height

Volume of the flower box = Volume of the complete pyramid – Volume of the top of the pyramid whose base is EFGH

Volume of the complete pyramid = 1/3 * AB²*h = (1/3)x70²x105 = 171 500 $cm^{3}$

Volume of the top pyramid =1/3 * EF² * SM with SM=h -OM = 105-60 =45 cm

=(1/3)x30²x45 =13500 $cm^{3}$

Volume of the flower box = 171 500 -13 500 = 158 000 $cm^{3}$

2/Here is how the Hindu mathematician Bhaskara calculated the volume of a pyramid trunk in the 12th century:

The sum of the area of the bases and the area of a rectangle of width the sum of the width of the bases and of length the sum of the length of the bases, being divided by six then multiply by the depth gives the volume.

Apply this method to calculate the volume of the flower box above:

Let’s go back to the terms we use:

“The sum of the base areas (the base itself squared) = AB²+EF²

“A rectangle of width is the sum of the width of the base (AB+EF) since it is a square the width is the length of the side.

“and of length the sum of the lengths of the bases” (AB+EF) always for the same reasons.

“The area of this rectangle” is (AB+EF)²

So we take the base areas + that of the hypothetical rectangle = (AB²+EF² +(AB+EF)²)

This is divided by 6 and then multiplied by the depth: ((AB²+EF² +(AB+EF)²)/6)*60 and we are supposed to get the volume.

This gives (70²+30²+100²)*10 since 60/6 = 10

The volume according to Bhaskara would be : 105 800 $cm^{3}$

Exercise 17:

We give: AB =6 m, AE = 5m, AD = 1.80m, BC = 0.80m .

On the diagram above, the dimensions are not respected.

1. Show that the volume of this pool is 39 m ^{3 .}

$V_{ABCD}=\frac{(AD+BC)\times \,AB}{2}=\frac{(1,8+0,8)\times \,6}{2}=7,8m^2$

$V_{piscine}=V_{ABCD}\times \,AE=7,8\times \,5=39\,m^3$

2. At the end of the summer, Mr. Dujardin empties his pool with a pump whose flow rate is 5m ³ per hour. Calculate the number of ^m3 remaining in the pool after 5 hours.

In 5 hours he will have emptied 25 , there will be 14 .

Exercise 18:

A parallelepipedic tank is shown here to measure the height of water falling in a garden during a rain shower (see below)

1. We assimilate the drops of water to balls of diameter 4mm.

Calculate the volume of a drop of water. Give their exact value.

The volume of a ball or a sphere is: 4/3 $\pi$ $R^{3}$ where R is the radius of the sphere, 2 mm

it becomes obvious that the drop has a volume of 32 $\pi$ /3 mm cubic.

2. The height of water that fell during this shower is equal to 8cm.

Calculate the number of drops of water in the tank. We will give the approximate value by default.

To do this we must first count the volume of water collected in the container.

4cm x 4cm x8cm = 16×8 cm cube= 128 $cm^{3}$ which is equivalent to 128 000 $mm^{3}$ of water in the container

Then we just divide by the volume of a drop to find the number of drops.

128 000 /( 32 $\pi$ /3) = (3 * 128 000)/ (32 $\pi$ )= 384 000/ (32 $\pi$ ) $\approx$ 3819 drops, by default approximated to the unit

After the rain the container contains 3819 drops of water.

Exercise 19:

A SABCD pyramid with a rectangular base by a plane parallel to the base at 5 cm from the vertex. AB=4.8cm; BC=4.2cm and SO =8cm.

a. Calculate the coefficient of K of reduction between the pyramids SABCD and SA’B’C’D’ .

$k=\frac{5}{8}$

b. Calculate the volume of the pyramid SABCD .

$V=\frac{base\times \,hauteur}{3}$

$V=\frac{4,8\times \,4,2\,\times \,8}{3}$

${\color{DarkRed}\,V\simeq\,53,75\,cm^3}$

c. Deduce the volume of the pyramid SA’B’C’D’ .

The volume will be multiplied by

$k^3=(\frac{5}{8})^3=\frac{5^3}{8^3}=\frac{125}{512}$

$V_{SA'B'C'D'}=\frac{125}{512}V_{SABCD}$

$V_{SA'B'C'D'}=\frac{125}{512}\times \,53,75$

$V_{SA'B'C'D'}\simeq\,13,12\,cm^3$

Exercise 20:

a brass ball measures 10cm in diameter.

Brass is an alloy of 40% zinc and 60% copper.

1) Calculate the volume of this ball (round to the nearest 1/10 cm3)

$V=\frac{4}{3}\pi\times \,R^3$

$V=\frac{4}{3}\pi\times \,5^3$

$V=\frac{500\pi}{3}\,cm^3$

${\color{DarkRed}\,V=523,6\,cm^3}$

2) We want to cover this ball with gold paint.

a)Calculate the area of the surface of the ball. Give the exact value.

$A=4\pi\times \,R^2=4\pi\times 5^2=100\pi$

b) How much paint is needed if 1dl covers 0.1m²?

$A\simeq\,314,16\,cm^2$

$A\simeq\,0,031416\,m^2$

$1dLrightarrow\,0,1\,m^2$

$xrightarrow\,0,031416\,m^2$

$x=\frac{1\times \,0,031416}{0,1}$

$x=0,31\,dL$

3) the ball is sawn along a plane located at 3cm from its center.

a) calculate the radius of the sectional circle, the length of the sectional circle and the area of the sectional disk.

Give the exact values then the values rounded to the nearest cm and cm².

In the triangle ABC rectangle in A , according to the direct part of the Pythagorean theorem :

$L=2\pi\,R=2\pi\times \,5=10\pi\simeq\,32\,cm$

The length of the circle is 32 cm .

$A=\pi\times \,R^2=\pi\times \,5^2=25\pi\simeq\,79\,cm$

The area of the sectional disk is 79 cm .

Exercise 21:

$V=\frac{base\times \,hauteur}{3}=\frac{\pi\times \,R^2\times \,SO}{3}$

$V=\frac{\pi\times \,7^2\times \,12}{3}$

${\color{DarkRed}\,V=196\pi}$

The reduction coefficient is $k=\frac{3}{12}=\frac{1}{4}$ .

The volume will be multiplied by $k^3=(\frac{1}{4})^3=\frac{1}{64}$ .

$V'=\frac{1}{64}\times \,k^3=\frac{196}{64}\pi$

${\color{DarkRed}\,V'=\frac{49}{16}\pi}$

${\color{DarkRed}\,V'\simeq\,9,6\,cm^3}$

Exercise 22:

a) the reduction coefficient is :

$\frac{SE}{SA}=\frac{3}{12}=\frac{1}{4}$

so $EF=\frac{1}{4}AB=\frac{9}{4}=2,25\,cm$ .

b) In the triangle SAB right-angled at A, according to the direct part

of the Pythagorean theorem, we have :

$SB=\sqrt{225}$

${\color{DarkRed}\,SB=15\,cm}$

$V_{SABCD}=\frac{AB^2\times \,SA}{3}=\frac{81\times \,12}{3}=324\,cm^3$

b) It is of $\frac{1}{4}.$

$V_{SEFGH}=\,(\frac{1}{4}\,\,)^3\times \,V_{SABCD}=\frac{324}{64}\simeq\,5\,cm^3$

Exercise 23:

1. In the right-angled triangle DAB, according to the direct part

of the Pythagorean theorem:

$DA=\sqrt{16}$

${\color{DarkRed}\,DA=4\,cm}$

2. $V_{SABCD}=\frac{1}{3}\times \,base\times \,hauteur$

$V_{SABCD}=\frac{1}{3}\times \,AD\times \,AB\times \,SO$

$V_{SABCD}=\frac{1}{3}\times \,3\times \,4\times \,6$

$V_{SABCD}=24\,cm^3$

3. a.The section is still a rectangle.

b. O ‘ is the middle of [SO] so the ratio of the reduction is $k=\frac{1}{2}$ .

The volume will be multiplied by $(\frac{1}{2})^3$

$V_{SA'B'C'D'}=(\frac{1}{2})^3V_{SABCD}$

$V_{SA'B'C'D'}=\frac{1}{2^3}V_{SABCD}$

$V_{SA'B'C'D'}=\frac{24}{8}$

${\color{DarkRed}\,V_{SA'B'C'D'}=3^\,cm^3}$ .

Exercise 24:

$V_1=\frac{1}{3}\pi\times \,R^2\times \,hauteur$

$V_1=\frac{1}{3}\pi\times \,5^2\times \,9$

$V_1=75\pi\,cm^3$

The reduction coefficient is :

$k=\frac{3}{9}=\frac{1}{3}$

$V_2=(\frac{1}{3})^3V_1=\frac{1}{27}V_1=\frac{75\pi}{27}$

$V_2=\frac{25\pi}{9}\,cm^3$

Exercise 25:

A cylindrical box contains 3 tennis balls of radius 3.4 cm. a) Make a figure, if the box has minimal dimensions.

b) What are the minimum dimensions of this box (height and radius)?

height = 3x2x3,4= 20,4 cm

radius = 3.4 cm

c) Calculate the volume of the box and the volume of the three balls.

$V_{boite}=\pi\times \,3,4^2\times \,20,4\simeq\,741\,cm^3$

$V_{3\,balles}=3\times \,\frac{4}{3}\pi\times \,3,4^3=4\times \,\pi\times \,3,4^3\simeq\,494\,cm^3$

d) Calculate the percentage of “empty” in this box containing the 3 balls.

$\frac{494}{741}\times \,100\simeq66,7%$ ball occupancy.

The void occupies about 33.3% or $\frac{1}{3}$ of the box.

Exercise 26:

In a conical glass of height 8cm and radius 6 cm,

I put 3 scoops of ice cream of radius 3cm each.

I don’t have time to eat them! too many copies to correct.

The 3balls are melting!

Is the ice going to overflow? if so, how many cL of ice did I lose?

Let’s calculate the volume of the glass and then the volume of the three balls.

$V_{verre}=\frac{4}{3}\pi\times \,6^2\times \,8=\frac{4}{3}\pi\times \,36\times \,8=\frac{4}{3}\pi\times 3\times \,12\times \,8=4\pi\times \,12\times \,8=384\pi$

$V_{3_,boules}=3\times \,\frac{4}{3}\pi\times \,3^3=4\pi\times \,3^3=108\pi$

Conclusion:

The volume of the glass is greater than the volume of the three balls

so the glass will not overflow.

Exercise 27:

For his show, a magician wants to stick swords in a box in which a spectator would be locked.

The box is a cube of 1m side.

For his project, the magician must have swords made.

He needs swords that are all the same size so that wherever he thrusts the sword, it can protrude at least 10 cm.

What is the minimum length of sword blade he should order from the blacksmith?

The maximum length of a cube is its diagonal.

Using Pytahgore’s theorem twice:

the length of the diagonal of a face is :

$a=\sqrt{1^2+1^2}=\sqrt{2}$

the length of the diagonal of the cube is :

$c=\sqrt{a^2+1^2}=\sqrt{\sqrt{2}^2+1^2}=\sqrt{3}$

The sword must protrude at least 10 cm, so the minimum length of the blade is ${\color{DarkRed}\,\sqrt{3}+10}$ .

Exercise 33:

A dovecote is composed of a rectangular parallelepiped ABCDEFGH and a pyramid SEFGH whose height [SO] measures 3.1 m.

We know that AB = 3 m, BC = 3.5 m and AE = 4 m.

Calculate the length BD and deduce the length of BH.

In the triangle ABD rectangular in A, according to the direct part of the Pythagorean theorem, we have the following equality:

$BD=\sqrt{21,25}$

Approximate values of these results will be given to the nearest $10^{-1}$ .

In the triangle BDH rectangular in D, according to the direct part of the Pythagorean theorem, we have the following equality:

$BH=\sqrt{37,25}$

${\color{DarkRed}\,BH\simeq\,6,1\,m}$

2. Calculate in the volume of this dovecote.

$V_1\simeq\,3\times \,3,5\times \,4+\frac{3\times \,3,5\times \,3,1}{3}$

$V_1\simeq\,42+10,85$

${\color{DarkRed}\,V_1\simeq\,52,85\,m^3}$

3. A model maker wants to build a scale model of this pigeon house $\frac{1}{24}$ .

Calculate in the volume of the model.

$V_2=\,(\frac{1}{24}\,\,)^3\times \,V_1$

$V_2=\frac{1}{13824}\,\times \,V_1$

We will give an approximate value of this result to the nearest $10^{-3}$ .

$V_2=\frac{1}{13824}\,\times \,52,85\simeq\,0,004\,m^3$

Exercise 32:

ABCDEFGH is a right block with a square base. We give AD = 3 cm and DC =2cm and CG = 4 cm.

1.calculate the volume in ^cm3 of the pyramid with vertex G and base ABCD.

$V=\frac{1}{3}\times \,3\times \,2\times \,4=8\,cm^3$

Calculate DG.

In the triangle DCG rectangle in C, according to the direct part of the Pythagorean theorem :

$DG=\sqrt{20}$

$DG=\sqrt{4\times \,5}$

$DG=2\sqrt{5}$

Exercise 33:

A dovecote is composed of a rectangular parallelepiped ABCDEFGH and a pyramid SEFGH whose height [SO] measures 3.1 m.

We know that AB = 3 m, BC = 3.5 m and AE = 4 m.

Calculate the length BD and deduce the length of BH. Approximate values of these results to the nearest ^10-1 will be given.

In the triangle ABD rectangular in A, according to the direct part of the Pythagorean theorem,

we have :

$BD=\sqrt{AB^2+AD^2}$

$BD=\sqrt{3^2+3,5^2}$

$BD=\sqrt{9+12,25}$

$BD=\sqrt{21,25}$

${\color{DarkRed}\,BD\simeq,4,6\,\,m}$

In the triangle BDH rectangular in D , according to the direct part of the Pythagorean theorem,

we have :

$BH=\sqrt{BD^2+DH^2}$

$BH=\sqrt{21,25+4^2}$

$BH=\sqrt{21,25+16}$

$BH=\sqrt{37,25}$

${\color{DarkRed}\,BH\simeq,6,1\,\,m}$

2. Calculate in^m3 the volume _V1 of this loft.

$V_1=3\times ,3,5\times 4+\frac{1}{3}\times 3\times 3,5\times 3,1$

$V_1=42+\frac{1}{3}\times 3\times 3,5\times 3,1$

${\color{DarkRed}\,V_1=52,85\,\,m^3}$

3. A model maker wants to build a scale model of this pigeon house $\frac{1}{24}\,$ .

Calculate in ^dm3 the volume _V2 of the model.

$V_2=\frac{V_1}{24}$

$V_2=\frac{52,85}{24}$

$V_2\simeq\,2,2021$

$V_2\simeq\,2,2021\,\,m^3$

${\color{DarkRed}V_2\simeq\,2202,1\,\,dm^3}$

The exercises in the third grade

After having consulted the answers to these exercises on the calculation of volumes and the study of sections of solids in 3rd grade, you can return to the exercises in 3rd grade.

The exercises in the third grade.

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