Geometry through calculations

This e-book is a collection of works created by students within the etwinning project G.A.M.E. (Growing with Applied Mathematics Experiences) In the Geometry in real life theme, the students created problems and solved them, collaboratively. The result was the eBook: Geometry through calculations.

Coordinating Teachers: •

Mihaela Git, Jean Monnet High School, Romania

•

Ludmila Cojocari, Gaudeamus High School, R.Moldova

•

Benedicte Leduc, Broceliande High School, France

•

Tugba Varol, Baruthane Secondary School,Turkey

•

Vlatka Hizman Trzic, Technical School Virovitica, Croatia

•

Maria Roussaki, Othisi High School, Greece

•

María Bellver, IES Cristobal Perez Pastor, Spain

•

Cîtu Georgiana, Virgil Ierunca High School, Romania

Nr.

Problems

1

TERNARY TOWER IN SHANGHAI, CHINA

Solutions

1. Find a function that models this skyscraper.

2. Create your own design for a tower using other functions. 𝒇(𝒙) = 𝟎. 𝟔𝟑𝒔𝒊𝒏𝒙 with a rotation by 𝟗𝟎𝟎 around the point O. https://www.geogebra.org/m/qbrg8fvj Irina, Jean Monnet High School

Patrik, Technical School Virovitica, Croatia

2

FLUIDITY 2.0 HIGH-RISE

𝒇(𝒙) = −

𝟏 𝒙

is a Mega Luxury Hotel and Office Hybrid Tower in Swiss Alps

𝒈(𝒙) = 𝟕𝒙

2. - It has a twirling form and a spanning open-air spaces with the surrounding Swiss landscape.

-To extend the green and fresh 1. Insert the picture into a GeoGebra file. environment to the 2. Determine 2 functions that describe the upper levels of the silhouettes of this beautiful hotel.

Luxury Hotel and You can find several possible models.

Office.

Denisa, Jean Monnet High School, Romania

Malala Ramanantsoa and Lily-Rose Leseigneur, Brocéliande High School, France

3

THE LOUVRE PYRAMID

Solution: 1. The perimeter is a square, so we can assume that the perimeter is equal 4⋅side ⇔ 4⋅34 = 136.

The Louvre Pyramid have a perimeter of 136 meters.

2. We know that the base is a square, thus we can deduce the formula of the base surface area : side⋅side ⇔ 34⋅34 = 1156 m².

The Louvre Pyramid have a base surface area of 1156 square meters.

https://www.geogebra.org/m/jgnvkxdwf The Louvre Pyramid reaches the height of 21.6 metres and its square base has sides of 34 metres.

3. The Louvre is a pyramid, the formula for the area of a pyramid is : 𝟏 V = ⋅ 𝑨𝒃 ⋅ 𝒉 , 𝑨𝒃 =base area h=height 𝟑 Thus we can calculate : 𝟏 Vpyramid = ⋅ 𝟏𝟏𝟓𝟔 ⋅ 𝟐𝟏. 𝟔 = 𝟖𝟑𝟐𝟑. 𝟐 m3. 𝟑 The total area of the Louvre Pyramid is 8323.2 cubic meters.

Calculate: 1. The perimeter of the base 2. The base surface area 3. The total area of the pyramid.

Mariela D., Gaudeamus High School, Moldova

By Thaïs CLÉMENT Brocéliande High School, France

4.

THE EIFFEL TOWER

Solution:

𝒎 = 𝝆𝑽 ⟹ 𝑽 =

𝑽=

𝒎 𝟕𝟑𝟎𝟎𝟎𝟎𝟎 𝒌𝒈 ⟹𝑽= ⟹ 𝝆 𝟎. 𝟎𝟕𝟕 𝐤𝐠/𝟎. 𝟎𝟎𝟎𝟎𝟎𝟏𝒎𝟑

𝟕. 𝟑 𝒎𝟑 ⟹ 𝑽 ≈ 𝟗𝟒𝟖𝒎𝟑 𝟎. 𝟎𝟎𝟕𝟕

𝟒𝝅𝒓𝟑 𝟑 𝑽𝒔 = , 𝝅 ≈ 𝟑 ⟹ 𝟗𝟒𝟖 ≈ 𝟒𝒓𝟑 ⟹ 𝒓 ≈ √𝟐𝟑𝟕 ⟹ 𝒓 ≈ 𝟔. 𝟎𝟕 𝟑 ⟹ 𝐫≈𝟔𝐦 Some of the particulars to this construction are that it is comprised of 18,038 pieces of metal held together by some 2.5 million rivets. The iron from which it was fashioned has a mass of 7300 tonnes.

The top of the tower (the sun-facing side) can shift up to 18 cm due to thermal expansion. The density of the wrought iron used to create this tower is 7.70 g/cm3 and was allegedly sourced from Reșița in Romania.

If one were to dismantle the Eiffel Tower and melt it to create an iron sphere, what would be its radius?

IF the top of the Eiffel Tower is seen from a distance of 600 meters at an angle of 30 degrees, find the tower height. https://opentextbc.ca/intermediatealgebraberg/chapter/chapte r-4-7-mathematics-in-life-the-eiffel-tower/

Afonschi Arina Gaudeamus High School, Moldova

𝒕𝒂𝒏𝟑𝟎° =

𝒙 𝟔𝟎𝟎√𝟑 ⟹ 𝒙 = 𝟔𝟎𝟎 ∙ 𝒕𝒂𝒏𝟑𝟎° = = 𝟐𝟎𝟎√𝟑 𝟔𝟎𝟎 𝟐 𝒙 ≈ 𝟐𝟎𝟎 ∙ 𝟏. 𝟕 = 𝟑𝟒𝟎

Miruna, Jean Monnet High School, Romania

5.

STATUE OF LIBERTY, NY, USA

The situation and the data The Statue of Liberty, erected in 1886, is 46.50 m high without its base and 93 m high with base.

A tourist in the front of a boat is looking at the statue. His eye is 5 m above the sea.

From what angle does the tourist see the Statue of Liberty?

The situation is represented on a GeoGebra worksheet as follows: - the statue and its base are assimilated to two vertical segments carried by the same line, - The observer is considered to be a vertical segment representing the height of his eye above sea level.

How does the angle of vision vary as the boat approaches the statue?

𝑡𝑎𝑛(∡𝐴𝐶𝐻) = 𝑡𝑎𝑛(∡𝑣) = 87 𝑥

𝐴𝐻 𝑥

, 𝐴𝐻 = 93 − 5 = 87 ⟹ 𝑡𝑎𝑛(∡𝑣) =

.

𝐼𝐻 41.5 = 𝑥 𝑥 𝒕𝒂𝒏(∡𝑎)– 𝒕𝒂𝒏(∡𝑏) 𝒕𝒂𝒏∡(𝒂 − 𝒃) = 𝟏 + 𝒕𝒂𝒏(∡𝑎) ∙ 𝒕𝒂𝒏(∡𝑏) 87 41.5 – 45.5 𝑥2 𝑥 𝑥 𝑡𝑎𝑛(∡𝑢) = = ∙ 2 ⟹ 87 41.5 𝑥 𝑥 + 3610.5 1+ ∙ 𝑥 𝑥 𝟒𝟓.𝟓𝒙 𝒕𝒂𝒏(∡𝒖) = 𝟐 . 𝒙 +𝟑𝟔𝟏𝟎.𝟓 For example, for 𝑥 = 100𝑚 ⟹ 𝑡𝑎𝑛(∡𝑢) = 0.3343 and the angle 𝑢 = 19°.

For, 𝑥 = 10𝑚 ⟹ 𝑡𝑎𝑛(∡𝑢) = 0.1226 and 𝑢 = 7°.

Alwena.BP, Brocéliande High School, France Daria, Jean Monnet High School, Romania 𝑡𝑎𝑛(∡𝐼𝐶𝐻) = 𝑡𝑎𝑛(∡𝑣) =

6.

THE BURJ AL ARAB HOTEL

𝐴𝐵 = 321: 4 = 80.25 𝐴𝐶 = 90 Pythagoras' Theorem ⇒ 𝑩𝑪𝟐 = 𝑨𝑪𝟐 + 𝑨𝑩𝟐 ⇒ 𝑩𝑪𝟐 = 𝟗𝟎𝟐 + 𝟖𝟎. 𝟐𝟓𝟐 ⇒ 𝑩𝑪𝟐 = 𝟖𝟏𝟎𝟎 + 𝟔𝟒𝟒𝟎. 𝟎𝟔𝟐𝟓 = 𝟏𝟒𝟓𝟒𝟎. 𝟎𝟔𝟐𝟓 ⇒ The Burj Al Arab hotel is 321 meters above the ground. Considering that the sides of each of the triangular poles in the body of the hotel are 90 meters, what is the length of the irons used as the hypotenuse?

Sude , Baruthane Secondary School

𝑩𝑪 ≃ 𝟏𝟐𝟎. 𝟓𝟖

Irina, Jean Monnet High School, Romania

7

EIFFEL TOWER

Asterix is on a visit to Paris. Today, he is at the Champs de Mars where he can see the Eiffel tower whose height BH is 324 m.

He places his camera on the ground at a distance AB = 600 m from the monument and programs it to take a photo.

1 What is the measure to the nearest degree of the angle HAB? 2 Knowing that Asterix is 1.70 m tall, at what distance AL must she stand to appear as tall as the Tower Eiffel in his photo?

Give an approximate value of the result to the nearest centimeter.

Jeanne, Brocéliande High School, France

To answer the first question, we can use the Pythagorean Theorem to find the length of the hypotenuse of the right triangle formed by the Eiffel Tower, the camera, and the ground. The hypotenuse is the line connecting the Eiffel Tower and the camera.

We have: 𝑩𝑯𝟐 + 𝑨𝑩𝟐 = 𝑪𝑯𝟐 , we get: 𝟑𝟐𝟒𝟐 + 𝟔𝟎𝟎𝟐 = 𝑪𝑯𝟐 Solving for CH, we find that CH = 675.6 meters.

To find the measure of the angle HAB, we can use the inverse cosine function. In this case, the side adjacent to the angle HAB is AB, and the hypotenuse is CH. Therefore, the cosine of the angle HAB is AB/CH. We can find the measure of the angle HAB by using the inverse cosine function on the value AB/CH: angle HAB = arccos(AB/CH) = arccos(600/675.6) Using a calculator, we find that this is approximately 31.7 degrees.

For the second question, we can use similar triangles to find the distance AL. Let's call the distance from the camera to Asterix's feet DL. Then we have the following similar triangles: [Eiffel Tower, camera, ground] ~ [Asterix, camera, DL] Since the ratios of corresponding sides of similar triangles are equal, we have: BH/AB = HA/AL Substituting the values given in the problem, we get: 324/600 = HA/AL. Solving for AL, we find that AL = 288 meters.

Since 1 meter is equal to 100 centimeters, the distance AL is approximately 28800 centimeters. To the nearest centimeter, this is 28800 centimeters.

Dino Špondreht, Technical School Virovitica Croatia

8.

TRIANGLES IN OP-ART This work in GeoGebra was inspired by an op-art painting made by Max Bill.

https://www.geogebra.org/m/n6wry2jg ABCD is a square with A(4; 3) and D(0; 0).

1. Find the equations of AD, AB, BC lines.

2. Calculate AD 3. If BF=DH 1.6 units, calculate the yellow area.

Luca D, Jean Monnet High School, Romania

Solution: 𝒚−𝒚𝑫

𝒙−𝒙

𝒚

𝒙

𝟑𝒙

𝟏

𝟏

𝑫 = ⇒ = ⇒𝒚= ⇒𝒚= 𝒙⇒𝒚= 𝒙 𝒚𝑨 −𝒚𝑫 𝒙𝑨 −𝒙𝑫 𝟑 𝟒 𝟏𝟐 𝟒 𝟒 𝒚 − 𝒚𝑩 𝒙 − 𝒙𝑩 𝒚−𝟕 𝒙−𝟏 𝑨𝑩: = ⇒ = ⇒ 𝒚𝑨 − 𝒚𝑩 𝒙𝑨 − 𝒙𝑩 −𝟒 𝟑 ⇒ 𝟑𝒚 − 𝟐𝟏 = −𝟒𝒙 + 𝟒 ⇒ −𝟒 𝟐𝟓 𝒚= 𝒙+ 𝟑 𝟑 𝒚 − 𝒚𝑩 𝒙 − 𝒙𝑩 𝒚−𝟕 𝒙−𝟏 𝑪𝑩: = ⇒ = ⇒ 𝒚𝑪 − 𝒚𝑩 𝒙𝑪 − 𝒙𝑩 −𝟑 −𝟒 𝟑 𝟐𝟓 ⇒ −𝟒𝒚 + 𝟐𝟖 = −𝟑𝒙 + 𝟑 ⇒ 𝒚 = 𝒙 + . 𝟒 𝟒 2. 𝑨𝑫𝟐 = 𝟑𝟐 + 𝟒𝟐 = 𝟐𝟓 ⇒ 𝑨𝑫 = 𝟓.

3. 𝑪𝑭 = 𝑪𝑩 − 𝑩𝑭 = 𝟓 − 𝟏. 𝟔 = 𝟑. 𝟒 𝑪𝑯 = 𝑪𝑫 − 𝑫𝑯 = 𝟓 − 𝟏. 𝟔 = 𝟑. 𝟒 𝑨𝑩 ⋅ 𝑩𝑭 𝟓 ⋅ 𝟏. 𝟔 𝟖 𝑨𝑩𝒍𝒖𝒆 = = = = 𝟒 = 𝑨𝑹𝒆𝒅 𝟐 𝟐 𝟐 𝑪𝑭 ⋅ 𝑪𝑯 𝟑. 𝟒 ⋅ 𝟑. 𝟒 𝟏𝟏. 𝟓𝟔 𝑨𝑮𝒓𝒆𝒆𝒏 = = = = 𝟓. 𝟕𝟖 𝟐 𝟐 𝟐 𝑨𝑨𝑩𝑪𝑫 = 𝒍𝟐 = 𝟐𝟓 𝑨𝒀𝒆𝒍𝒍𝒐𝒘 = 𝟐𝟓 − (𝟑. 𝟒 + 𝟑. 𝟒 + 𝟓. 𝟕𝟖) = 𝟐𝟓 − 𝟏𝟐. 𝟓𝟖 𝑨𝒀𝒆𝒍𝒍𝒐𝒘 = 𝟏𝟐. 𝟒𝟐

1. 𝑨𝑫:

Vlad, Jean Monnet High School, Romania

9.

TRIANGLES IN OP-ART ABC is a equilateral triangle with AB=8.

Each triangle, starting with the second, has a side 0.9 from the side of the previous triangle https://www.geogebra.org/m/yhey8pkq

Solution: 𝟖𝟐 √𝟑

𝟔𝟒√𝟑

1. 𝑨 = = = 𝟏𝟔√𝟑 𝟒 𝟒 2. 𝒍𝟐 = 𝟖 ⋅ 𝟎, 𝟗 = 𝟕, 𝟐 𝒍𝟑 = 𝟕, 𝟐 ⋅ 𝟎, 𝟗 = 𝟔, 𝟒𝟖 𝒍𝟒 = 𝟔, 𝟒𝟖 ⋅ 𝟎, 𝟗 = 𝟓, 𝟖𝟑𝟐 𝒍𝟓 = 𝟓, 𝟖𝟑𝟐 ⋅ 𝟎, 𝟗 = 𝟓, 𝟐𝟒𝟖𝟖 𝑷𝟏 = 𝟑 ⋅ 𝟖 = 𝟐𝟒 𝑷𝟐 = 𝟑 ⋅ 𝟕, 𝟐 = 𝟐𝟏, 𝟖 𝑷𝟑 = 𝟑 ⋅ 𝟔, 𝟒𝟖 = 𝟏𝟗, 𝟒𝟒 𝑷𝟒 = 𝟑 ⋅ 𝟓, 𝟖𝟑𝟐 = 𝟏𝟕, 𝟒𝟗𝟔 𝑷𝟓 = 𝟑 ⋅ 𝟓, 𝟐𝟒𝟖𝟖 = 𝟏𝟓, 𝟕𝟒𝟔𝟒 𝑷𝟏 + 𝑷𝟐 + 𝑷𝟑 + 𝑷𝟒 + 𝑷𝟓 = 𝟐𝟒 + 𝟐𝟏, 𝟖 + 𝟏𝟗, 𝟒𝟒 + 𝟏𝟕, 𝟒𝟗𝟔 + 𝟏𝟓, 𝟕𝟒𝟔𝟒 = 𝟗𝟖, 𝟒𝟖𝟐𝟒 3. 𝑨𝟐 =

1. Find the area of ABC triangle.

2. Calculate the sum of the perimeters for the first 5 equilateral triangles.

3. Calculate the sum of the areas for the first 5 equilateral triangles.

Florentina, Jean Monnet High School, Romania

𝟕,𝟐𝟐 √𝟑 𝟒

= 𝟏𝟐, 𝟗𝟔√𝟑

𝟔, 𝟒𝟖𝟐 √𝟑 𝑨𝟑 = = 𝟏𝟎, 𝟒𝟗𝟕𝟔√𝟑 𝟒 𝟓, 𝟖𝟑𝟐𝟐 √𝟑 𝑨𝟒 = = 𝟖, 𝟓𝟎𝟑𝟎𝟓𝟔√𝟑 𝟒 𝟓, 𝟐𝟒𝟖𝟖𝟐 √𝟑 𝑨𝟓 = = 𝟔, 𝟖𝟖𝟕𝟒𝟕𝟓𝟑𝟔√𝟑 𝟒 𝑨𝟏 + 𝑨𝟐 + 𝑨𝟑 + 𝑨𝟒 + 𝑨𝟓 = 𝟏𝟔√𝟑 + 𝟏𝟐, 𝟗𝟔√𝟑 + 𝟏𝟎, 𝟒𝟗𝟕𝟔√𝟑 + 𝟖, 𝟓𝟎𝟑𝟎𝟓𝟔√𝟑 + +𝟔, 𝟖𝟖𝟕𝟒𝟕𝟓𝟑𝟔√𝟑 = 𝟓𝟒, 𝟖𝟒𝟖𝟏𝟑𝟏𝟑𝟔√𝟑 Buștiuc Arina, Gaudeamus High-School, Moldova

10.

SQUARES&TRIANGLES IN OP-ART This drawing was made in GeoGebra, after a painting by Max Bill.

Solution: 1. DIJ=45=DCA (corresp.angles)⇒ 𝑰𝑱||𝑨𝑪 ACB=45=LKB (corresp.angles) ⇒ 𝑲𝑳||𝑨𝑪 2. ΔAPJ is an isosceles right triangle with hypotenuse equal to half of the side of the square ΔAPL, isosceles right triangle with hypotenuse equal to half of the side of the square Δ KOC, isosceles right triangle with hypotenuse equal to half of the side of the square ΔAPJ≡ΔAPLΔ ≡KOC 3. The red surface has an area equal to the area of two squares with side: √𝟐

𝒍

𝟐

𝒍√𝟐

√𝟐

𝑷𝑱 = 𝑨𝑱 ⋅ = ⋅ = ⇒ 𝑨𝑹𝒆𝒅 = (𝟐√𝟐) = 𝟖. 𝟐 𝟐 𝟐 𝟒 The blue surface has an area equal to the area of two squares with side 𝑴𝑶 = 𝑷𝑱 = 𝑨𝑱 ⋅

With the notations in the drawing, ABCD is a square with side 8 units and L, K, I, J are the means of sides AB, BC, CD, respectively DA. 1. Prove that the lines aC, KL, IJ are parallel.

2. Show that ΔAPJ, ΔAPL, Δ KOC are congruent triangles.

3. Prove that the surface colored in red is equal to the surface colored in blue.

4. Calculate the ratio between the area of the orange and the area of the square Teofil, Jean Monnet High School, Romania

√𝟐 𝟐

𝒍

√𝟐

𝟐

𝟐

= ⋅

=

𝒍√𝟐 𝟒

⇒ 𝑨𝑩𝒍𝒖𝒆 = 𝟖.

⇒ 𝑨𝑹𝒆𝒅 = 𝑨𝑩𝒍𝒖𝒆 = 𝟖 4. The orange surface has an area equal to the area of two squares with side 𝒍√𝟐 𝑷𝑴 = 𝑷𝑱 = = 𝟐√𝟐 ⇒ 𝑨𝑶𝒓𝒂𝒏𝒈𝒆 = 𝟖 𝟒 𝑨𝑶𝒓𝒂𝒏𝒈𝒆 𝟖 𝟏 = = 𝑨𝑨𝑩𝑪𝑫 𝟔𝟒 𝟖 Miruna, Jean Monnet High School, Romania

11.

Model inspired by a carpet from SINAIA - Romania

Solution: 𝒅𝒊𝒂𝒈𝒐𝒏𝒂𝒍𝟏 ⋅𝒅𝒊𝒂𝒈𝒐𝒏𝒂𝒍𝟐

𝑨𝑪 ⋅ 𝑩𝑫

𝟖𝟎

⇒ 𝑨𝑨𝑩𝑪𝑫 = = = 𝟒𝟎 ⇒ In the attached image, ABCD is a rhombus with 𝑨𝒓𝒉𝒐𝒎𝒃𝒖𝒔 = 𝟐 𝟐 𝟐 diagonals of 10u and 8u. Each rhombus, starting with 𝑨𝑨′𝑩′𝑪′𝑫′ = 𝒌𝟐 ⋅ 𝑨𝑨𝑩𝑪𝑫 ⇒ 𝑨𝑨′𝑩′𝑪′𝑫′ = 𝟎. 𝟔𝟒 ⋅ 𝟒𝟎 = 𝟐𝟓. 𝟔 the second, has a side 0.8 of the side of the previous one. = 𝑨𝑮𝒓𝒆𝒆𝒏 𝑨𝑾𝒉𝒊𝒕𝒆𝑹𝒉𝒐𝒎𝒃𝒖𝒔 = 𝟎. 𝟔𝟒 ⋅ 𝟐𝟓. 𝟔 = 𝟏𝟔. 𝟑𝟖𝟒 ⇒ 𝑨𝑹𝒆𝒅𝑹𝒉𝒐𝒎𝒃𝒖𝒔 = 𝟎. 𝟔𝟒 ⋅ 𝟏𝟔. 𝟑𝟖𝟒 = 𝟏𝟎. 𝟒𝟖𝟓𝟕𝟔 ⇒ (𝑻𝒉𝒆 𝒂𝒓𝒆𝒂 𝒐𝒇 𝒕𝒉𝒆 𝒔𝒎𝒂𝒍𝒍 𝒓𝒆𝒅 𝒓𝒉𝒐𝒎𝒃𝒖𝒔) 𝑨𝑭𝑮𝑯𝑰 = 𝟎. 𝟔𝟒 ⋅ 𝟏𝟎. 𝟒𝟖𝟓𝟕𝟔 = 𝟔, 𝟕𝟏𝟎𝟖𝟖𝟔𝟒 𝑨𝑭𝑮𝑯𝑰 = 𝟔. 𝟕𝟏𝟎𝟖𝟖𝟔𝟒

1. Calculate the area of the rhombus FGHI. 2. If the dimensions of the black rectangle are 12u and 10u, find the length of the green polygonal line, formed by segments that describe isosceles right triangles.

Carlos Sanz Ene, Jean Monnet High School, Romania

𝒙𝟐 + 𝒙𝟐 = 𝟒 ⇒ 𝟐𝒙𝟐 = 𝟒 ⇒ 𝒙𝟐 = 𝟐 ⇒ 𝒙 = √𝟐 𝑷 = 𝟒𝟎𝒙 ⇒ 𝑷 = 𝟒𝟎√𝟐 Teofil, Jean Monnet High School, Romania Ellie, Othisi High School, Greece

12.

SPIRALS

GeoGebra shapes:

You may get the shape on the left by cutting this one in half and sliding.

Which is larger? The red or the yellow area?

You may draw this design by joining 10 equally spaced dots with semicircles!

Ellie and Marianna, Othisi High School

Since everything except the center is symmetric, the larger area is red!

Matei, Jean Monnet High School, Romania

13

CHRISTMAS TREE

Solution: 𝑨𝑩𝑯𝑭𝑯𝑱𝑳 =

𝟑𝒍𝟐 √𝟑 𝟐

=

𝟑⋅𝟔𝟒√𝟑 𝟐

= 𝟗𝟔√𝟑.

𝑨𝑰𝑲𝑯 = √𝒑(𝒑 − 𝒂)(𝒑 − 𝒃)(𝒑 − 𝒄), 𝒑 =

𝑨𝑰𝑲𝑯 = √

𝟐𝟏 𝟐𝟏 𝟐

(

𝟐

𝟐𝟏

− 𝟖)(

𝟐

− 𝟔. 𝟓)(

√𝟏𝟎𝟓 ⋅ 𝟒 ⇒ 𝑨𝑰𝑲𝑯 = 𝟐√𝟏𝟎𝟓

𝑽= https://www.geogebra.org/m/uddqhtrv This tree is built as a sequence of 5 regular pyramids in which each pyramid, starting with the second one, has the elements 0.5 of the previous one.

The base is a star polygon with 12 sides.

IJ=JK=KL=...+HI=6.5 and BHFHJL is a regular hexagon with BD=8. The great high is 16.

Calculate the volume of the third pyramid.

𝟐𝟏 𝟐

𝒂+𝒃+𝒄 𝟐𝟏 ⇒𝒑= 𝟐 𝟐

− 𝟔. 𝟓) = √

𝟐𝟏 𝟐

𝟓

⋅ ⋅𝟒⋅𝟒= 𝟐

⇒ 𝑨𝒃 = 𝟗𝟔√𝟑 + 𝟏𝟐√𝟏𝟎𝟓

𝑨𝒃 ⋅ 𝒉 (𝟗𝟔√𝟑 + 𝟏𝟐√𝟏𝟎𝟓)𝟏𝟔 = = (𝟑𝟐√𝟑 + 𝟒√𝟏𝟎𝟓) ⋅ 𝟏𝟔 𝟑 𝟑 = 𝟔𝟒(𝟖√𝟑 + √𝟏𝟎𝟓) 𝑽𝟏 =

𝟏 ⋅ 𝟔𝟒(𝟖√𝟑 + √𝟏𝟎𝟓) = 𝟐(𝟖√𝟑 + √𝟏𝟎𝟓) 𝟖

𝑽𝟐 =

𝟏 𝟖√𝟑 + √𝟏𝟎𝟓 ⋅ 𝟐(𝟖√𝟑 + √𝟏𝟎𝟓) = 𝟖 𝟒 𝑽𝟐 =

√𝟑(𝟖 + √𝟑𝟓) 𝟒

Ana, Jean Monnet High School, Romania Eymen, Baruthane Secondary School, Turkey

14

Solution

MONTREAL BIOSPHERE

𝑽𝒄𝒖𝒃𝒆 = 𝒍𝟑

https://www.geogebra.org/3d/dgp27q5w What is the volume of the cube with the largest side length that can fit inside this sphere with a diameter of 78 meters?

𝑩𝑯 = 𝟐𝑹 = 𝟕𝟖 𝑩𝑯 = 𝒍√𝟑, 𝒂𝒔 𝒅𝒊𝒂𝒈𝒐𝒏𝒂𝒍 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒖𝒃𝒆, 𝒔𝒐 𝒍√𝟑 = 𝟕𝟖 ⇒ 𝟕𝟖 𝟕𝟖√𝟑 𝒍= = = 𝟐𝟔√𝟑 𝟑 √𝟑 𝟑

𝑽 = (𝟐𝟔√𝟑) = 𝟏𝟕𝟓𝟕𝟔 ⋅ 𝟑√𝟑 = 𝟓𝟐𝟕𝟐𝟖√𝟑 ⇒ Beyza, Baruthane Secondary School, Turkey

𝑽 = 𝟓𝟐𝟕𝟐𝟖√𝟑 Miruna, Jean Monnet High School, Romania

15.

WHITE HOUSE Solution: The length of the white house is 25 meters and the We use the formulas for the perimeter and area of the rectangle: width is 52 meters, so what is the perimeter and the area of the white house?

P=2(L+l); A=L ⋅ l P=2∙(25+52)=2∙77=154⇒ P=154 m A=25 ⋅ 52 = 1300 𝑚2 A= 1300 𝒎𝟐

Çağan Aras, Baruthane Secondary School, Turkey 16.

CUBS

Mara, Jean Monnet High School, Romania The large cube that Natalija made has 27 cubes and Diana

Natalija and Dijana make a large cube from small only has 20 in her work, therefore she needs to add 7 more cubes. Natalija finished her cube like which we see in cubes (Answer:C.7). the left picture.

How many more small cubes must Diana add to her work that she gets a cube equal to Natalia's?

Mariela D., Gaudeamus High School, Moldova

Patrik Živković, Technical School Virovitica, Croatia

17.

SQUARES

Solution: We know that the side of the square is 2√3 cm.

So the area of the purple square is 2√3 ∙ 2√3 which is equal to 12cm².

We know that the lenght of the base is of 3 cm and that it’s an isosceles triangle. The other two sides measure twice the size of the base: 3∙2 so 6 cm. The perimeter is the sum of all sides: 3+3+6= 12 cm

Jeanne, Brocéliande High School, France

What is the area of the purple square, knowing that a side of the square is 2√3 cm.

Calculate the perimeter of the blue triangle, knowing that it is an isosceles triangle, whose base is 3 cm, and the other two are congruent and are twice the size of the base.

Arina B., Gaudeamus High School, Moldova

18.

TOWER IN PISA The tower of Pisa is 56.294 meters high. If the diameter of the leaning tower of Pisa is 15.484 meters, what is volume of the tower in Pisa?

Solution:

We use the formula for the volume of the cylinder: 𝑉 = 𝜋 ⋅ 𝑟2 ⋅ ℎ 1 15,484 𝑟= 𝑑= = 7,742 (𝑚) 2 2 according to the formula, the volume of the pisa tower is: 𝑉 = 𝜋 ⋅ 7.72 ⋅ 56.29 = 59.29 ⋅ 56.29𝜋 = 3337.4341𝜋 𝑉 = 3337.4341𝜋 𝑚3 ≈ 10479.54 𝑚3

Aparatu Maria Magdalena, Gaudeamus High School, Moldova

Petar Petrinović.Elementary School of Nikola Andrić.

Vukovar/Croația

19.

Solution:

THE CAKE

https://www.geogebra.org/m/mvvdbphm 𝒍𝟐 √𝟑 𝑽 = 𝑨𝑨𝑩𝑪 ⋅ 𝒉, 𝑨𝑩𝑪 − 𝒆𝒒𝒖𝒊𝒍𝒂𝒕𝒆𝒓𝒂𝒍 𝒕𝒓𝒊𝒂𝒏𝒈𝒍𝒆 ⇒ 𝑨𝑨𝑩𝑪 = 𝟒 Before eating this cake, we are going to calculate its volume. The length AB equals 8 cm and the length EB equals 2 cm. ABC is an equilateral triangle. CBG is a rectangle triangle in G. The point G intersects the

⇒ 𝒍𝟐 √𝟑 𝟔𝟒√𝟑 𝑽= ⋅ 𝑩𝑬 ⇒ 𝑽 = ⋅ 𝟐 = 𝟏𝟔√𝟑 ⋅ 𝟐 = 𝟑𝟐√𝟑 𝟒 𝟒

𝑽 = 𝟑𝟐√𝟑

line AB in two.

Louise Wester, Brocéliande High School, France

Matei, Jean Monnet High School, Romania

20.

THE ARC OF TRIOMPHE, PARIS, FRANCE

Solution: 1 𝑉1 = 𝑉𝑝𝑎𝑟𝑎𝑙𝑒𝑙𝑖𝑝1 + ∙ 𝑉𝑐𝑦𝑙1 2

𝐿 = 18.68 𝑙 = 22 ℎ = 18.68 𝑉𝑝𝑎𝑟𝑎𝑙𝑒𝑙𝑖𝑝1 = 18.68 ∙ 14.62 ∙ 22 = 𝟔𝟎𝟎𝟖. 𝟐𝟑𝟓𝟐 14.62 𝑟1 = = 7.31 2 𝑉𝑐𝑦𝑙1 = 22 ∙ 𝜋 ∙ 7.312 = 1175.5942π ≈ 𝟑𝟔𝟗𝟏. 𝟑𝟔𝟓𝟕 1 3691 𝑉1 = 𝑉𝑝𝑎𝑟𝑎𝑙𝑒𝑙𝑖𝑝1 + ∙ 𝑉𝑐𝑦𝑙1 = 6008 + ⇒ 2 2 𝑉1 = 6008 + 1845 = 𝟕𝟖𝟓𝟑 45– 14.62 = 15.19 2 𝑙 = 8.44 𝐻 = 18.68– 4.22 = 14.46 𝑉𝑝𝑎𝑟𝑎𝑙𝑒𝑙𝑖𝑝2 = 15.19 ∙ 18.44 ∙ 14.46 = 𝟒𝟎𝟓𝟎. 𝟐𝟗𝟖𝟎 8.44 𝑟2 = = 4.22 2 𝑉𝑐𝑦𝑙2 = 15.19 ∙ 𝜋 ∙ 4.222 = 270.5𝜋 ≈ 𝟖𝟒𝟗. 𝟒 𝐿= Listenbourg students are on a trip to Paris, Ella, a student wants to find the volume of the Arc de Triomphe.

The main volume is 50∙45∙22 = 49500m3.

But Ella can't find the total volume ( without corbels, baseboards, sculptures)...

Task: Find the volume of the main and secondary arch It's up to you to help Ella and find it !

Manon L, Brocéliande High School, France

1 7853 𝑉2 = 𝑉𝑝𝑎𝑟𝑎𝑙𝑒𝑙𝑖𝑝2 + ∙ 𝑉𝑐𝑦𝑙2 = 4050 + ⇒ 2 2 𝑉2 = 4050 + 3926.5 = 𝟕𝟗𝟕𝟔. 𝟓 𝑉3 = 49500– 7853– 2 ∙ 7976.5 = 41647– 15953 = 𝟐𝟓𝟔𝟗𝟒 Denisa, Jean Monnet High School, Romania

21

PLAYING CARDS

Solution:

Ivan built houses from playing cards. The picture To know how many cards are needed for a 4 story house we shows the houses that Ivan built, a one-story house, need to use this formula: n∙(3∙n+1)/2 two-story and three-story. How many cards does he For example: need to build a four-story house? 𝟏

𝟒

𝟐

𝟐

First story: 𝟑 ⋅ 𝟏 + = = 𝟐 cards Second story: 2(3⋅2+1)/2 = 2⋅7/2 = 14/2 = 7 cards Third story: 3(3⋅3+1)/2 = 3⋅10/2 = 30/2 = 15 cards And now for the fourth story: 4(3⋅4+1)/2 = 4⋅13/2 = 52/2 = 26 cards one-story=2 cards story=15 cards

two-story=7 cards

three-

Explain your solution !

Marko Zalović, Technical School Virovitica, Croatia

Mariela D., Gaudeamus High School, Moldova

22.

Solution: CAMP NOU STADIUM A = L⋅l= 220m ⋅ 250m= 55000mp Andreea / Virgil Ierunca High School Romania

What is the area of the Camp Nou Stadium, which is 220 meters wide and 250 meters long?

Eymen,Baruthane Secondary School, Turkey