Theorem of Thales: answer key for math exercises in 3rd grade.

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Theorem of Thales and correction of the exercises in 3ème with the calculation of length and use of the cross product. Apply the direct and reciprocal part of Thales’ theorem.

Exercise 1:

figure 1: $\frac{AM}{AG}=\frac{AF}{AE}=\frac{MF}{GE}$

figure 2: $\frac{TS}{TG}=\frac{TA}{TB}=\frac{SA}{CB}$

Figure 3: $\frac{OB}{OD}=\frac{OG}{OE}=\frac{BG}{DE}$

Exercise 2:

Show using the direct part of Thales’ theorem that :

$CT=6,5\times \,\frac{5}{13}=\frac{32,5}{13}$

$AB=3\times \,\frac{13}{5}=\frac{39}{5}$

Exercise 3:

To consolidate a building, a wooden buttress was built.
In the drawing below, we give:

BS = 6 m; BN = 1.8 m; AM = 1.95 m; AB = 2.5m.

1. in the triangle ABS rectangular in B, according to the direct property of the Pythagorean theorem :

$AS=\sqrt{42,25}$

conclusion : AS = 6,5 meters.

2 Calculate the lengths SN and SM.
In the triangles SMN and SAB, the lines (MN) and (AB) are parallel.
$M\in(AS)$
$N\in(SB)$
According to the direct part of Thales’ theorem, we have the following equalities:

$\frac{SM}{SA}=\frac{SN}{SB}=\frac{MN}{AB}$

Let us substitute the known lengths.

$\frac{SM}{6,5}=\frac{6-1,8}{6}=\frac{MN}{2,5}$

$\frac{SM}{6,5}=\frac{4,2}{6}=\frac{MN}{2,5}$

Using the cross product, we obtain:

$SM=\frac{4,2\times \,6,5}{6}=\,4,55\,m$

$MN=\frac{4,2\times \,2,5}{6}=\,1,75\,m$

Exercise 5:

A funicular starts from D and goes to A along the line (DA) .
DM = 420m; DH = 1000m; MP = 252m.

The triangles DPM and DAH are respectively rectangles in P and H.

1) Calculate the distance DP in meters.

In the triangle DPM rectangular in P, according to the direct part of the Pythagorean theorem :

$DP=\sqrt{112896}$

${\color{DarkRed}\,DP=336\,m}$

2) a) Show that the lines (MP) and (HA) are parrallel.

We know that:

$\,\{\,(MP)\,\perp\,(DH)\\(AH)\perp\,(DH).$

Property: two lines perpendicular to a third are parallel to each other.

Conclusion: $(MP)\parallel\,(AH)$ .

b) Calculate the distance DA in meters and then in kilometers.

In the DPM and DHA triangles:

We know that:

$\,\{\,M\in\,(DA)\\P\in\,(DH)\,\\(MP)\parallel\,(AH).$

According to the direct part of Thales’ theorem, we have the following equalities:

$\frac{DM}{DA}=\frac{DP}{DH}=\frac{PM}{HA}$

$\frac{420}{DA}=\frac{336}{1000}=\frac{252}{HA}$

using the cross product, we obtain :

$DA=\frac{420\times \,1000}{336}=1250\,m=1,25\,km$

Exercise 6:

a. [SO] is a height of the pyramid so the triangle AOS is rectangular in O .

In the triangle AOS rectangular in O, according to the direct part of the Pythagorean theorem :

$OS=\sqrt{256}$

${\color{DarkRed}\,OS=16\,cm}$

b. The reduction coefficient is :

$k=\frac{SM}{SO}=\frac{2}{16}=\frac{1}{8}$

$SI=\frac{1}{8}SA=\frac{20}{8}=\frac{5}{2}=2,5\,cm$

$IA=SA-SI=20-2,5=17,5\,cm$

Exercise 7:

This is an application of Thales’ theorem.

$\frac{24}{200}=\frac{30}{x}$

$x=\frac{30\times \,200}{24}=250\,cm=2,5\,m$

Conclusion: He must place the screen at 2.5 meters from the light source.

Exercise 9:

We apply Thales’ theorem since we have an hourglass configuration.

We have the equality of the following ratios:

$\frac{L}{1}=\frac{20}{5}$

$L=\frac{20\times \,1}{5}=4\,pas$

The width of the river is 4 steps.

$65\times \,4=260\,cm$

The width of the river is 260 centimeters.

Exercise 10:

The lines (CA) and (DH) are perpendicular to the same line so they are parallel.

$H\in(AB)\,,\,D\in\,(CB)$

according to the direct part of Thales’ theorem, we have the following equality:

$\frac{BD}{BC}=\frac{DH}{CA}$

$\frac{BD}{1200}=\frac{150}{200}$

$BD=\frac{150\times \,1200}{200}=900\,m$

Conclusion: He has 900 meters to go.

Exercise 11:

The lines (BI) and (KA) are parallel, $S\in(KI)\,et\,S\in(AB)$

according to the direct part of Thales’ theorem :

$\frac{BI}{KA}=\frac{SI}{SK}$

$\frac{BI}{4,5}=\frac{4}{6}$

$BI=\frac{4\times \,4,5}{6}=3\,cm$

Conclusion: the length of the segment [BI] is 3 cm.

Exercise 12:

We can use Thales’ theorem only on figures b, d and e.

Exercise 14:

In the triangles ADE and ACB, according to the direct part of Thales’ theorem :

$\frac{AE}{AB}=\frac{DE}{BC}$

$BC=2BF=2\times \,1737=3474\,km$

$\frac{2,8}{AB}=\frac{0,026}{3474000}$

$AB=\frac{2,8\times \,3474000}{0,026}$

$AB=374123077\,m$

Exercise 15:

The lines (FC) and (DA) are parallel if $\frac{IA}{IC}=\frac{ID}{IF}$ .

$\frac{7x+5}{5x}=\frac{12}{7}$

$7(7x+5)=12\times \,5x$

$x=\frac{35}{11}$

Exercise 16:

The height of the stick is BD .

Note the height of the stick.

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

we have :

$x=\frac{-2250}{-30}$

${\color{DarkRed}\,x=75}$

The height of the stick is 75 cm.

Exercise 17:

Construct a triangle ABC such that AB=12cm,BC=16cm,AC=8cm .

1)Place the point E on (AB) such that AE=9cm then draw the parallel to (BC) passing through E.

It cuts (AC) at F.

Calculate AF.

$\,\{\,F\in(AC)\,\\E\in(AB)\\(FE)//(CB).$ according to the direct part of Thales’ theorem :

$\frac{AF}{AC}=\frac{AE}{AB}$

$\frac{AF}{8}=\frac{9}{12}$

$AF=\frac{8\times \,9}{12}$

${\color{DarkRed}\,AF=6\,cm}$

2)In the rest of the problem, the point E walks on [AB] and we pose AE=x .

a)Give a frame for x .

0<x<12

b) Calculate AF as a function of x .

$\frac{AF}{AC}=\frac{AE}{AB}$

$\frac{AF}{8}=\frac{x}{12}$

$AF=\frac{8}{12}x$

$AF=\frac{2}{3}x$

c) Deduce FC and also express EB as a function of x .

$FC=AC-AF=8-\frac{2}{3}x$

Exercise 18:

a) Calculate the angle Â .

The value of this angle is impossible to determine.

b) What is the nature of triangle AEF? Justify your answer.

The triangle AEF is any.

The points O,C,F; O,B,E and O,A,D are aligned.

(CB)//(FE) and (BA)//(ED).

Show that (CA)//(FD) .

Indication:

Use the equalities of the ratios in the OFE and OED triangles.

Then apply the reciprocal of Thales’ theorem to show that (CA)//(FD) .

Exercise 20:

1. in the triangles KMN and KLV, the segment [MN] is an enlargement of [LV]

and the magnification coefficient is $k=\frac{8}{5}$

$MN=\frac{8}{5}\times \,4=\frac{32}{5}$

2. The points D,C,M and F,C,N are aligned in the same order.

$\frac{CD}{CM}=\frac{45}{51}$ and $\frac{CF}{CN}=\frac{30}{40}$

or $45\times \,40=1800$ and $30\times \,51=1530$

so $\frac{CD}{CM}\neq\,\frac{CF}{CN}$ according to the contraposed of Thales’ theorem,

the lines (DF) and (MN) are not parallel.

Exercise 21:

According to the diagram we are well in a configuration of Thales where the points O,M,A,N are aligned (it was enough of 3 points but here I specify that these four are aligned). and the points 0, B and C are also aligned. And that there are several interesting pairwise triangles such as OAB and ONC or OBM and OCA.

The Thales theorem tells us that if the triangles OAB and ONC form a Thales configuration such that the points O,A,N on the one hand are aligned in this order, and O,B,C on the other hand are aligned in this order. And if the lines (AB) and (NC) are parallel then we have the following ratio equalities:

OC/OB = ON/OA = NC/AB (1)

The same is true for the triangles OBM and OCA which form a Thales configuration since the points O, B and C are aligned in this order and the points O, M and N are also aligned in this order. Since the lines (BM) and (AC) are parallel we have the following ratio equalities:

OC/OB = OA/OM = CA/BM (2)

Let us use (1) and notice that $\frac{OC}{OB}=\frac{ON}{OA}=\frac{NC}{AB}$ is equivalent to $\frac{OB}{OC}=\frac{OA}{ON}=\frac{AB}{NC}$

With (2) we have $\frac{OA}{OM}=\frac{OC}{OB}=\frac{1}{\frac{OB}{OC}}$ or in (1) $\frac{OB}{OC}=\frac{OA}{ON}$ so OA/OM = ON/OA since $\frac{1}{\frac{OA}{ON}}=\frac{ON}{OA}$

which by a cross product gives us OA² =OMxON

Exercise 22:
1. In the triangles ECD and EAB :

$\,(AB)\,//(CD)\,\\\,C\in\,(EB)\\\,D\in\,(AE)$

according to the direct part of Thales’ theorem, we have :

$\,\frac{EC}{EB}=\frac{ED}{EA}=\frac{CD}{AB}$

$\,\frac{EC}{16}=\frac{6}{10}=\frac{CD}{20}$

Let’s calculate CD :

$\,\frac{CD}{20}=\frac{6}{10}$

$\,10\times \,CD=6\times \,20$

$\,\frac{10\times \,CD}{10}=\frac{6\times \,20}{10}$

$\,\fbox{CD=12\,cm}$

2. In the BFG and BEA triangles:

$\,B,F,E\,\,B,G,A$ are aligned in the same order.

$\,\frac{BF}{BE}=\frac{12,8}{16}=\frac{128}{160}$

$\,\frac{BG}{BA}=\frac{16}{20}=\frac{128}{160}$

Conclusion:

$\,\frac{BF}{BE}=\frac{BG}{BA}$

therefore according to the reciprocal part of Thales’ theorem, the lines (FG) and (AE) are parallel.

Exercise 23:
1. In triangles ABC and AMN :

$\,(BC)\,//(MN)\,\\\,M\in\,(AB)\\\,N\in\,(AC)$

according to the direct part of Thales’ theorem, we have :

$\,\frac{AB}{AM}=\frac{AC}{AN}=\frac{BC}{MN}$

$\,\frac{2,4}{AM}=\frac{5,2}{7,8}=\frac{BC}{4,5}$

Let’s calculate AM :

$\,\frac{2,4}{AM}=\frac{5,2}{7,8}$

$\,5,2\times \,AM=2,4\times \,7,8$

$\,\frac{5,2\times \,AM}{5,2}=\frac{2,4\times \,7,8}{5,2}$

$\,AM=\frac{2,4\times \,7,8}{5,2}=3,6$

$\,\fbox{Am=3,6\,cm}$

Let’s calculate BC:

$\,\frac{BC}{4,5}=\frac{5,2}{7,8}$

$\,7,8\times \,BC=4,5\times \,5,2$

$\,\frac{7,8\times \,BC}{7,8}=\frac{4,5\times \,5,2}{7,8}$

$\,BC=\frac{4,5\times \,5,2}{7,8}=3$

$\,\fbox{Am=3\,cm}$

2. In the triangles ABC and APR:

$\,P,A,C\,\,R,A,B$ are aligned in the same order.

$\,\frac{AP}{AC}=\frac{2,6}{5,2}=\frac{26}{52}=0,5$

$\,\frac{AR}{AB}=\frac{1,2}{2,4}=\frac{12}{24}=0,5$

Conclusion:

$\,\frac{AP}{AC}=\frac{AR}{AB}$

therefore according to the reciprocal part of Thales’ theorem, the lines (PR) and (BC) are parallel.

Exercise 24:
1. In the triangles OAC and OBE :

$\,A,O,B\,\,C,O,E$ are aligned in the same order.

$\,\frac{OA}{OB}=\frac{60}{72}=\frac{5}{6}$

$\,\frac{OC}{OE}=\frac{50}{60}=\frac{5}{6}$

Conclusion:

$\,\frac{OA}{OB}=\frac{OC}{OE}$

therefore according to the reciprocal part of Thales’ theorem, the lines (AC) and (EB) are parallel.

2. In the triangles OAC and OBE :

$\,(AC)\,//(EB)\,\\\,C\in\,(OE)\\\,A\in\,(OB)$

according to the direct part of Thales’ theorem, we have :

$\,\frac{OA}{OB}=\frac{OC}{OE}=\frac{AC}{EB}$

$\,\frac{60}{72}=\frac{50}{60}=\frac{100}{EB}$

Let’s calculate EB:

$\,\frac{100}{EB}=\frac{50}{60}$

$\,50\times \,EB=100\times \,60$

$\,\frac{50\times \,EB}{50}=\frac{100\times \,60}{50}$

$\,\fbox{CD=120\,cm}$

Exercise 28:

Show that O is the midpoint of [ST].

Assumptions:

The lines (OM) and (RT) are parallel (from the previous question).

M middle of [SR] according to the coding.

Ownership:

If a line passes through the middle of one side of a triangle and is parallel to a second side

then this line cuts the third side in its middle.

Conclusion:

O is the middle of [ST]

Exercise 29:

1. Show that the lines (IO) and (BD) are parallel.

Assumptions:

I medium of [AB] and O medium of [AD] (according to the coding)

Ownership:

If a line passes through the middle of two sides of a triangle then

this line is parallel to the third side.

Conclusion:

(IO) and (BD) are parallel

2. Show that J is the middle of [AC].

Assumptions:

The lines (OJ) and (DC) are parallel (from the previous question).

O middle of [AD] by assumptions

Ownership:

If a line passes through the middle of one side of a triangle and is parallel to a second side

then this line cuts the third side in its middle.

Conclusion:

J is the middle of [AC]

The answer key to the math exercises on Thales’ theorem in 3rd grade.

After consulting the answers to these math exercises on Thales’ theorem, you can go back to the exercises in third grade.

The exercises in the third grade.

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