Triangle Spirals and Analytical Geometry

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Triangles Spirals and Analytic Geometry

10 Grades from Jean Monnet High School – Bucharest, Romania and IIS Tommaso Salvini - Technological Pole–Rome, Italy

-eTwinning Project-2019-

Spirals in 10‌or more questions Spiral formed with equilateral triangles. Each equilateral triangle, starting with the second, has the side as the high as the previous one. Spiral Problem 1 E

C

ABC is equilateral triangle and A(1; 1), B(4;1). AD⊼BC , D∈BC, ADE equilateral triangle, etc. a) Determine the equation for the line BC. b) Determine the equation for the line AC. c) Find the coordinate for the point C. d) Determine the equilateral triangle height equation that passes through A. e) Calculates the AD height of the ABC triangle. f) Calculates the area of the ABC triangle. g) Determine the equation for the line AE. h) What is đ?’Ž(âˆĄđ?‘Ťđ?‘­đ?‘Ż)? i) What is the position of the line AH and AB? j) Calculates the area of the ADE triangle. SOLUTION: a) Let be d the equation for the line BC. ABC is equilateral triangle ⇒ đ?’Ž(âˆĄđ?‘¨đ?‘Šđ?‘Ş) = đ?&#x;”đ?&#x;Ž° ⇒ đ?’Ž(âˆĄđ?’…, đ?‘śđ?’™) = đ?&#x;?đ?&#x;?đ?&#x;Ž° ⇒ đ?’Žđ?‘Šđ?‘Ş = −√đ?&#x;‘ ⇒ đ?’š − đ?’šđ?‘Š = đ?’Žđ?’… (đ?’™ − đ?’™đ?‘Š ) ⇒ đ?’š = −√đ?&#x;‘đ?’™ + đ?&#x;’√đ?&#x;‘ + đ?&#x;? b) Let be e the equation for the line AC. ABC is equilateral triangle ⇒ đ?’Ž(âˆĄđ?‘Şđ?‘¨đ?‘Š) = đ?&#x;”đ?&#x;Ž° ⇒ đ?’Žđ?’† = đ?’•đ?’ˆđ?&#x;”đ?&#x;Ž° = √đ?&#x;‘. ⇒ đ?’š − đ?’šđ?‘¨ = đ?’Žđ?’† (đ?’™ − đ?’™đ?‘¨ ) ⇒ đ?’š = √đ?&#x;‘đ?’™ + √đ?&#x;‘ + đ?&#x;? đ?&#x;‘ đ?&#x;“√đ?&#x;‘ đ?&#x;“√đ?&#x;‘+đ?&#x;? đ?&#x;“√đ?&#x;‘ đ?’™=đ?&#x;? đ?’š= đ?&#x;? +đ?&#x;? đ?’š = −√đ?&#x;‘đ?’™ + đ?&#x;’√đ?&#x;‘ + đ?&#x;? đ?&#x;?đ?’š = đ?&#x;“√đ?&#x;‘ + đ?&#x;? đ?’š= đ?&#x;? đ?’š = đ?&#x;? +đ?&#x;? c) { ⇔{ ⇔{ ⇔{ ⇔{ ⇔{ ⇔ đ?&#x;“√đ?&#x;‘ đ?&#x;“√đ?&#x;‘ đ?’š = √đ?&#x;‘đ?’™ + √đ?&#x;‘ + đ?&#x;? đ?’š = √đ?&#x;‘đ?’™ + √đ?&#x;‘ + đ?&#x;? đ?’š = + đ?&#x;? đ?’š = √đ?&#x;‘đ?’™ + √đ?&#x;‘ + đ?&#x;? đ?’š = √đ?&#x;‘đ?’™ + √đ?&#x;‘ + đ?&#x;? = + √đ?&#x;‘đ?’™ √đ?&#x;‘ đ?&#x;? đ?&#x;? đ?&#x;‘ đ?&#x;“√đ?&#x;‘

đ?‘Ş (đ?&#x;? ;

đ?&#x;?

+ đ?&#x;?)

−đ?&#x;?

d) đ?’Žđ?‘¨đ?‘Ť ∙ đ?’Žđ?‘Šđ?‘Ş = −đ?&#x;? ⇒ đ?’Žđ?‘¨đ?‘Ť = đ?’Ž e) AB=3⇒ đ?‘¨đ?‘Ť = đ?’?đ?&#x;? √đ?&#x;‘

đ?’?√đ?&#x;‘ đ?&#x;?

=

đ?&#x;‘√đ?&#x;‘ đ?&#x;?

đ?‘Šđ?‘Ş

⇒ ��� =

⇒ �� =

đ?&#x;? √đ?&#x;‘

⇒ đ?’š − đ?’šđ?‘¨ = đ?’Žđ?‘¨đ?‘Ť (đ?’™ − đ?’™đ?‘¨ ) ⇒ đ?’š =

√đ?&#x;‘ đ?’™ đ?&#x;‘

+

√đ?&#x;‘ đ?&#x;‘

+đ?&#x;?

đ?&#x;‘√đ?&#x;‘ đ?&#x;?

.

đ?&#x;—√đ?&#x;‘

f) đ?‘¨đ?‘¨đ?‘Šđ?‘Ş = đ?&#x;’ ⇒ đ?‘¨đ?‘¨đ?‘Šđ?‘Ş = đ?&#x;’ . g) đ?‘¨đ?‘Ź ⊼ đ?‘¨đ?‘Š ⇒ đ?’š = đ?’šđ?‘¨ ⇒ đ?’š = đ?&#x;?. h) đ?‘š(âˆĄđ??ˇđ??šđ??ť) = 60° + 90° = 150°. i) đ?‘š(âˆĄđ??´đ??ťđ??ľ) = 60° + 30° = 90° ⇒ đ?‘¨đ?‘Ż ⊼ đ?‘¨đ?‘Š ⇒ đ?‘¨đ?‘¨đ?‘Ťđ?‘Ź =

đ?’?đ?&#x;? √đ?&#x;‘ đ?&#x;’

⇒ ���� =

đ?‘¨đ?‘Ťđ?&#x;? √đ?&#x;‘ đ?&#x;’

=

đ?&#x;?đ?&#x;•âˆšđ?&#x;‘ đ?&#x;?đ?&#x;”

Spiral Problem 2 ABC is equilateral triangle and A(6; 8), B(0;8). AD⊼BC , D∈BC, ADE equilateral triangle. a) Determine the points F, E, the distance AF and DE. b) Determine the Area and Perimeter of ABC. c) Determine the middle points D, F till to arrive to N. d) Determine the Areas of the triangles and the Perimeters. e) Determine the straight line beam with centre A (6,8).

SOLUTION: (đ?&#x;Ž+đ?&#x;”)

= đ?&#x;‘ √đ?&#x;‘ ⇒ CM=AD= height of sides of equilateral triangle= AE= đ?&#x;? đ?’” = đ?&#x;‘√đ?&#x;‘ đ?’š=đ?&#x;– đ??‚(đ?&#x;‘; đ?&#x;– − đ?&#x;‘√đ?&#x;‘) and đ??„(đ?&#x;”; đ?&#x;– − đ?&#x;‘√đ?&#x;‘) the y coordinate is determinat by the difference between yA = 8 distance from x axis and AE a) Let M midpoint [AB], MAB {

đ?’™=

đ?&#x;?

𝐀𝐅 = 𝐡𝐀𝐃𝐄 = H(𝟔; 𝟖 −

√𝟑 𝟐

𝟗

· 𝟑√𝟑 = 𝟐 ; 𝐀𝐇 = 𝐡𝐀𝐆𝐅 =

√𝟑 𝟐

∙ 𝑨𝑭 =

√𝟑 𝟗 ∙ 𝟐 𝟐

=

𝟗√𝟑 𝟒

⇒ 𝐀𝐇 =

𝟗

𝟗

𝟗

FH= 𝟐 : 𝟐 = 𝟒 H middle point of the segment FG, therefor F: { 𝒔∙

√𝟑 𝒔 𝟐

𝒙𝑭 = 𝟔 − 𝟒 = 𝒚𝑭 = 𝟖 −

√𝟑

√𝟑 𝟒

b) ABC equilateral triangle side= 6 Area ABC= = 𝒔𝟐 = 𝟐 𝟒 c) Asks intersections of all the heights with the sides till N 𝟎+𝟑 𝟑 𝒙𝑫 = 𝟐 = 𝟐 𝟑 𝟏𝟔−𝟑√𝟑 D midpoint of BC { ⇒ 𝑫 (𝟐 ; ) 𝟖+𝟖−𝟑√𝟑 𝟐 𝒚𝑫 = 𝟐 𝟏𝟓

F(𝟒 ; 𝟖− 𝟒

.

) the y coordinate is determinat by the difference between yA =8 distance from x axis and AH

𝟒

AF=FG=𝟐 side equilateral triangle

𝟗√𝟑

𝟒

𝟗√𝟑

𝟗

AH=

𝟗√𝟑

𝟏𝟓

𝟒 𝟗√𝟑

𝟏𝟓

⇒ 𝑭( 𝟒 ; 𝟖 −

𝟗√𝟑 𝟒

)

𝟒

· 𝟑𝟔 = 𝟗√𝟑 ⇒ 𝑨𝑨𝑩𝑪 = 𝟗√𝟑 𝐚𝐧𝐝 Perimeter ABC=𝟔 · 𝟑 = 𝟏𝟖 ⇒ 𝑷𝑨𝑩𝑪 = 𝟏𝟖

𝟗√𝟑 𝟒

) 𝐩𝐫𝐞𝐯𝐢𝐨𝐮𝐬𝐥𝐲 𝐟𝐨𝐮𝐧𝐝

𝐩𝐫𝐞𝐯𝐢𝐨𝐮𝐬𝐥𝐲 𝐟𝐨𝐮𝐧𝐝 𝒙𝑯 = 𝒙𝑨 = 𝟔

𝑯: { 𝟗√𝟑 𝒚𝑯 = 𝑫𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒑𝒐𝒊𝒏𝒕 𝑨 𝒇𝒓𝒐𝒎 𝒙 𝒂𝒙𝒊𝒔 − 𝑨𝑯 = 𝟖 − 𝟒 √𝟑

AK=AJ= 𝟐 ·

𝟗√𝟑 𝟒

=

Midpoint AK {𝒙 =

𝟐𝟕

; 𝒚𝑲 = 𝑨𝑩 + 𝑨𝑲 = 𝟔 +

𝟖 𝟔+

𝟕𝟓 𝟖

= 𝒚=𝟖 𝟐

𝟐𝟕 𝟖

⇒ K =(𝟔 +

𝟐𝟕 𝟖

𝟕𝟓

; 𝟖) ⇒ 𝑲 ( 𝟖 ; 𝟖)

𝟏𝟐𝟑 𝟏𝟔

𝒙𝑱 = 𝒙 𝒎𝒊𝒅𝒑𝒐𝒊𝒏𝒕 𝑨𝑲 = AL=J midpoint AK=

⇒𝑱={

𝒙𝑱 = 𝒙 𝒎𝒊𝒅𝒑𝒐𝒊𝒏𝒕 𝑨𝑲 = 𝒚𝑱 = 𝟖 − 𝑨𝑳 = 𝟖 −

√𝟑 𝟐

·

𝟐𝟕 𝟖

=

d) Asks areas of the triangles and the Perimeters. Equilateral triangle’s perimeter=side · 𝟑

𝟏𝟐𝟑 𝟏𝟔 𝟏𝟔−𝟐𝟕√𝟑 𝟏𝟔

and N= {

𝒙𝑵 = 𝑨𝑩 + 𝑨𝑵 = 𝟔 +

√𝟑 𝑨𝑳 𝟐

=𝟔+ 𝒚𝑱 = 𝟖

√𝟑 𝟐

·

𝟏𝟔−𝟐𝟕√𝟑 𝟏𝟔

=

𝟏𝟏𝟏+𝟏𝟔√𝟑 𝟑𝟐

𝐀𝐫𝐞𝐚 𝐭𝐫𝐢𝐚𝐧𝐠𝐥𝐞 =

𝒃𝒂𝒔𝒆∙𝒉𝒆𝒊𝒈𝒉𝒕

√𝟑

√𝟑

. Equilateral triangle: Base= side=s. Height= 𝟐 𝒔 ⇒ Area equilateral triangle =

𝟐

The next triangle has the side as the height of the previous one, therefor 𝒔𝟐 = 𝒉 = √𝟑

𝟑

Area2=( 𝟐 𝒔 ∙ 𝟒 𝒔) /𝟐 =

𝟑√𝟑 𝟐 𝒔 𝟏𝟔

⇒

√𝟑 𝒔; 𝟐

√𝟑

𝒉𝟐 = 𝟐 𝒔𝟐 =

𝒃𝒂𝒔𝒆∙𝒉𝒆𝒊𝒈𝒉𝒕 𝒔∙ 𝟐 𝒔 √𝟑 𝟐 = 𝟐 = 𝟒 𝒔 𝟐 𝟑 √𝟑 √𝟑 𝟐

∙

𝟐

𝒔 = 𝟒𝒔

√𝟑 𝟐 𝒔

𝟒 Ratio= 𝟑√𝟑 = 4/3 𝟏𝟔

𝒔𝟐

Determined in point b) ABC equilateral triangle side= 6 Area ABC= ADE equilateral triangle: 𝟒: 𝟑 = 𝟗√𝟑: 𝑨𝒓𝒆𝒂 𝑨𝑫𝑬 ⇒ 𝑨𝒓𝒆𝒂 𝑨𝑫𝑬 = 𝟐𝟕 AFG equilateral triangle: 𝟒: 𝟑 = 𝟐𝟕 √𝟑

𝒔∙

√𝟑 𝒔 𝟐

𝟐

=

√𝟑 𝟐 𝒔 𝟒

=

√𝟑 𝟒

· 𝟑𝟔 = 𝟗√𝟑. Perimeter ABC=𝟔 · 𝟑 = 𝟏𝟖

√𝟑 𝟒

√𝟑 : 𝑨𝒓𝒆𝒂 𝟒

√𝟑

𝑨𝑭𝑮 √𝟑

√𝟑

√𝟑

𝑨𝒓𝒆𝒂 𝑨𝑭𝑮 = 𝟖𝟏 𝟏𝟔; 𝑨𝒓𝒆𝒂 𝑨𝑯𝑰 = 𝟐𝟒𝟑 𝟔𝟒 ; 𝑨𝒓𝒆𝒂 𝑨𝑱𝑲 = 𝟕𝟐𝟗 𝟐𝟓𝟔; 𝑨𝒓𝒆𝒂 𝑨𝑳𝑴 = 𝟐𝟏𝟖𝟕 𝟏𝟎𝟐𝟒 ; 𝑨𝒓𝒆𝒂 𝑨𝑶𝑵 = 𝟔𝟓𝟔𝟏 𝟒𝟎𝟗𝟔 ; 𝑨𝒓𝒆𝒂 𝑨𝑸𝑹 = 𝟏𝟗𝟔𝟖𝟑

√𝟑 √𝟑 √𝟑 √𝟑 ; 𝑨𝒓𝒆𝒂 𝑨𝑺𝑻 = 𝟓𝟗𝟎𝟒𝟗 ; 𝑨𝒓𝒆𝒂 𝑨𝑼𝑽 = 𝟏𝟕𝟕𝟏𝟒𝟕 ; 𝑨𝒓𝒆𝒂 𝑨𝑾𝒁 = 𝟓𝟑𝟏𝟒𝟒𝟏 𝟏𝟔𝟑𝟖𝟒 𝟔𝟓𝟓𝟑𝟔 𝟐𝟔𝟐𝟏𝟒𝟒 𝟏𝟎𝟒𝟖𝟓𝟕𝟔

Ratio between perimeter of an equilateral triangle and the equilateral triangle that has the side as the high as the previous one. 𝟑𝒔 𝟏 Ratio=𝟑√𝟑 = √𝟑 𝟐

𝒔

𝟐

ADE equilateral triangle: √𝟑 𝟏: 𝟐

= 𝟏𝟖: 𝑷𝒆𝒓𝒊𝒎𝒆𝒕𝒆𝒓 𝑨𝑫𝑬. Perimeter ADE= 𝟗√𝟑 AFG equilateral triangle: 𝟏:

𝑷𝒆𝒓𝒊𝒎𝒆𝒕𝒆𝒓 𝑨𝑭𝑮 =

𝟐𝟕 𝟐

√𝟑 𝟐

= 𝟗√𝟑: 𝑷𝒆𝒓𝒊𝒎𝒆𝒕𝒆𝒓 𝑨𝑭𝑮

; 𝑷𝒆𝒓𝒊𝒎𝒆𝒕𝒆𝒓 𝑨𝑯𝑰 = 𝟐𝟕 √𝟑

√𝟑 ; 𝑷𝒆𝒓𝒊𝒎𝒆𝒕𝒆𝒓 𝟒

𝑨𝑱𝑲 =

𝟕𝟐𝟗

𝟖𝟏 𝟖

√𝟑

; 𝑷𝒆𝒓𝒊𝒎𝒆𝒕𝒆𝒓 𝑨𝑳𝑴 = 𝟖𝟏 𝟏𝟔 ; 𝑷𝒆𝒓𝒊𝒎𝒆𝒕𝒆𝒓 𝑨𝑶𝑵 = √𝟑

𝑷𝒆𝒓𝒊𝒎𝒆𝒕𝒆𝒓 𝑨𝑸𝑹 = 𝟐𝟒𝟑 𝟔𝟒 ; 𝑷𝒆𝒓𝒊𝒎𝒆𝒕𝒆𝒓 𝑨𝑺𝑻 = 𝟏𝟐𝟖 ; 𝑷𝒆𝒓𝒊𝒎𝒆𝒕𝒆𝒓 𝑨𝑼𝑽 = 𝟕𝟐𝟗 𝟐𝟓𝟔 ; 𝑷𝒆𝒓𝒊𝒎𝒆𝒕𝒆𝒓 𝑨𝑾𝒁 =

e) Asks straight line beam with centre A (6,8) y–8=m(x–6)

y=mx–6m+8

𝟐𝟏𝟖𝟕 𝟓𝟏𝟐

.

𝟐𝟒𝟑 𝟑𝟐

;

Spiral Problem 3 ABC is an equilateral triangle and A(6,10.38), D midpoint of AB on Yaxis. Determine the coordinates of B. a) Determine the equation for the line AB and call it r. b) Determine the equation for the line DC and call it s. c) Which are the characteristics of r and s? d) Which are the characteristics between A and B? e) Determine the coordinates of F ∈ đ?‘Ťđ?‘Ź. f) Determine the coordinates of H ∈ CE. g) Determine the equation for the line CE. h) Determine the coordinates of G. i) Which are the characteristics between E and G? j) Determine the equation for the line CG and call it t. k) Determine the equation for the line HI and call it u. l) Which are the characteristics of t and u? m) Compare r,s,t,u, which are the characteristics? SOLUTION: a) D midpoint of AB on Oy axis and ABC equilateral ⇒ đ??śđ??ˇ ⊼ đ??´đ??ľ ⇒ đ??´đ??ľ||đ?‘‚đ?‘Ľ ⇒ đ?‘Š(−đ?&#x;”; đ?&#x;?đ?&#x;Ž. đ?&#x;‘đ?&#x;–) Equation of AB is (r): đ?‘Ś = 10.38. b) C, D ∈ đ?‘‚đ?‘Ś ⇒ (đ?’”): đ?’™ = đ?&#x;Ž. c) đ?’” ⊼ đ?’“ d) A and B are symmetrical points vs.Oy. ⃗⃗⃗⃗⃗ = 4đ??šđ??´ ⃗⃗⃗⃗⃗ ⇒ đ?’™đ?‘¨ − đ?’™đ?‘Ş = đ?&#x;’(đ?’™đ?‘¨ − đ?’™đ?‘­ ) ⇒ 6 = 24 − 4đ?‘Ľđ??š ⇒ đ?’™đ?‘­ = đ?&#x;’. đ?&#x;“ ⇒ đ?‘Śđ??´ − đ?‘Śđ??ś = 4(đ?‘Śđ??´ − đ?‘Śđ??š ) ⇒ 10.38 = 41.52 − 4đ?‘Śđ??š ⇒ đ?’šđ?‘­ = đ?&#x;•. đ?&#x;•đ?&#x;–đ?&#x;“ ⇒ e) đ??śđ??´ đ?‘­(đ?&#x;’. đ?&#x;“; đ?&#x;•. đ?&#x;•đ?&#x;–đ?&#x;“) f) F midpoint of DE ⇒ 2đ?‘Ľđ??š = đ?‘Ľđ??ˇ + đ?‘Ľđ??¸ ⇒ đ?‘Ľđ??¸ = 9 − 0 = 9 ⇒ đ?’™đ?‘Ź = đ?&#x;— 2đ?‘Śđ??š = đ?‘Śđ??ˇ + đ?‘Śđ??¸ ⇒ đ?‘Śđ??¸ = 15.57 − 10.38 = 5.19 ⇒ đ?’šđ?‘Ź = đ?&#x;“. đ?&#x;?đ?&#x;— ⇒E(9; 5.19) ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ ⇒ đ?’™đ?‘Ź − đ?’™đ?‘Ş = đ?&#x;’(đ?’™đ?‘Ź − đ?’™đ?‘Ż ) ⇒ 9 = 36 − 4đ?‘Ľđ??ť ⇒ đ?’™đ?‘Ż = đ?&#x;”. đ?&#x;•đ?&#x;“ ⇒ đ?‘Śđ??¸ − đ?‘Śđ??ś = 4(đ?‘Śđ??¸ − đ?‘Śđ??ť ) ⇒ 5.19 = 20.76 − 4đ?‘Śđ??ť ⇒ đ?‘Śđ??ť = 15.57: 4 = đ??śđ??¸ = 4đ??ťđ??¸ 3.8925 ⇒ đ?‘Ż(đ?&#x;”. đ?&#x;•đ?&#x;“; đ?&#x;‘. đ?&#x;–đ?&#x;—đ?&#x;?đ?&#x;“) g) đ?‘Ś − đ?‘Śđ??ś = đ?‘š(đ?‘Ľ − đ?‘Ľđ??ś ); đ?‘š(âˆĄđ??¸đ??śđ??ş) = 30° ⇒ (đ?‘Źđ?‘Ş): đ?’š =

√đ?&#x;‘ đ?&#x;‘

đ?’™.

h) i) j) k) l)

FH: đ?‘Ś − đ?‘Śđ??š = đ?‘š(đ?‘Ľ − đ?‘Ľđ??š ) ⇒ đ?‘Ś − 7.785 = −√3(đ?‘Ľ − 4.5) ⇒ đ?’š = −√đ?&#x;‘đ?’™ + đ?&#x;?đ?&#x;“. đ?&#x;“đ?&#x;• ⇒ đ??šđ??ť ∊ đ?‘‚đ?‘Ľ: đ?’š = đ?&#x;Ž; đ?’™ = đ?&#x;?đ?&#x;“. đ?&#x;“đ?&#x;•: đ?&#x;?. đ?&#x;•đ?&#x;‘ = đ?&#x;— ⇒ đ?‘Ž(đ?&#x;—; đ?&#x;Ž) đ?’™đ?‘Ž = đ?’™đ?‘Ź CG: y=0 (t) HI||Oy⇒ đ?’™ = đ?’™đ?‘Ż ⇒ đ?’™ = đ?&#x;”. đ?&#x;•đ?&#x;“ (u) đ?‘Ą ⊼ đ?‘˘ ; r||t.

Spiral Problem 4 a) Determine the straight line beam with center A(0;0). b) Analyze the slopes of the straight lines ∈ to the sides of the equilateral triangles. c) Proof triangles AEM and CEM are right angle triangles. d) Proof triangle AEG and DGE are right angle triangles. e) Calculate the ratios between areas ACB and ADM. f) Calculate the ratios between areas ADM and AEF and so on where each equilateral triangle, with the one that has the side as the high as the previous one. g) Analyze the ratios. SOLUTION: a) y=mx đ?’š −đ?’š

b) m=đ?’•đ?’‚đ?’? đ?œś = đ?’™đ?&#x;? −đ?’™đ?&#x;? đ?&#x;?

đ?&#x;?

√đ?&#x;‘

m=tan0°=0 for the line AB; m=tan30° = đ?&#x;‘ for the line AM; m=tan60°=√đ?&#x;‘ for the line AC; m=tan 90°= ∄ for the line AD, being the equation line x=0; m= tan 120°= - √đ?&#x;‘ for the line AF; m=tan 150°= -

√đ?&#x;‘ đ?&#x;‘

for the line AH; m=tan 180°=0 for

√đ?&#x;‘

line AJ, same line AB, being the equation line x=0; m= tan 210°== đ?&#x;‘ for line AL, same line AM; tan 240°= √đ?&#x;‘ for the line AO, same line AC; tan 270°=°= ∄ for the line AQ, same line AD; tan 300°= - √đ?&#x;‘ for the line AS, same line AF; tan 330°= 0°=0 c)CB=√(đ?’™đ?’ƒ − đ?’™đ?’„ )đ?&#x;? + (đ?’šđ?’ƒ − đ?’šđ?’„ )đ?&#x;?= √đ?&#x;’ + đ?&#x;?đ?&#x;?= √đ?&#x;?đ?&#x;”=4

√đ?&#x;‘ đ?&#x;‘

for the line AS, same line AH; tan 360°=tan

CA=√(đ?’™đ?’„ − đ?’™đ?’‚ )đ?&#x;? + (đ?’šđ?’„ − đ?’šđ?’‚ )đ?&#x;? = √đ?&#x;’ + đ?&#x;?đ?&#x;?= √đ?&#x;?đ?&#x;”=4 AM=√(đ?’™đ?’Ž − đ?’™đ?’‚ )đ?&#x;? + (đ?’šđ?’Ž − đ?’šđ?’‚ )đ?&#x;? = √đ?&#x;— + đ?&#x;‘= √đ?&#x;?đ?&#x;?=2√đ?&#x;‘ EM=(MC∙AM)/CA=(2∙2√đ?&#x;‘)/4=√đ?&#x;‘ EA=√(đ?‘¨đ?‘´đ?&#x;? − đ?‘´đ?‘Źđ?&#x;? )=√đ?&#x;?đ?&#x;? − đ?&#x;‘=√đ?&#x;—=3 CE=4-3= đ?&#x;? According to Euclid's theorem, in a right-angled triangle, the square built on the height relative to the hypotenuse is equivalent to the rectangle whose dimensions are the projections of the cathetus on the hypotenuse. ME=√đ?‘Şđ?‘Ź ∙ đ?‘¨đ?‘Ź=√đ?&#x;? ∙ đ?&#x;‘=√đ?&#x;‘ Another way is using the perpendicular condition of slopes m=tan60°=√đ?&#x;‘ for the line AC, M(3; √đ?&#x;‘), D=(0; 2√đ?&#x;‘)

đ?’š −đ?’š

m=đ?’™đ?&#x;? −đ?’™đ?&#x;? = đ?&#x;?

đ?&#x;?

đ?&#x;?√đ?&#x;‘−√đ?&#x;‘ đ?&#x;Žâˆ’đ?&#x;‘

=-

√đ?&#x;‘ for đ?&#x;‘

the line

MD; đ?’Ž ∙ đ?’ŽâŠĽ = −đ?&#x;?. d) Equation line passing through AD is x=0, through FE y=3. Parallel lines of x axes are perpendicular to y axes therefor AEG and DGE are right angle triangles e)Area ACB=(CB∙ đ?‘¨đ?‘´)/2=(4∙ đ?&#x;?√đ?&#x;‘)/đ?&#x;? = đ?&#x;’√đ?&#x;‘ Area ADM=(AD∙ đ?‘¨đ?‘Ź)/2=(đ?&#x;?√đ?&#x;‘ ∙ đ?&#x;‘)/đ?&#x;? = đ?&#x;‘√đ?&#x;‘ Ratio1= (đ?&#x;’√đ?&#x;‘)/( đ?&#x;‘√đ?&#x;‘)=4/3 đ?&#x;‘

đ?&#x;—

f) Area ADM=(AD∙ đ?‘¨đ?‘Ź)/2=(đ?&#x;?√đ?&#x;‘ ∙ đ?&#x;‘)/đ?&#x;? = đ?&#x;‘√đ?&#x;‘ . Area AEF==(EF∙ đ?‘¨đ?‘Ž)/2=(đ?&#x;‘ ∙ đ?&#x;? √đ?&#x;‘)/đ?&#x;? = đ?&#x;’ √đ?&#x;‘ đ?&#x;—

Ratio2=(đ?&#x;‘√đ?&#x;‘)/( đ?&#x;’ √đ?&#x;‘)=4/3 g)Area triangle=

đ?’ƒđ?’‚đ?’”đ?’†âˆ™đ?’‰đ?’†đ?’Šđ?’ˆđ?’‰đ?’• đ?&#x;?

√đ?&#x;‘

. Equilateral triangle: Base= side=s. Height= đ?&#x;? đ?’” √đ?&#x;‘

đ?’ƒđ?’‚đ?’”đ?’†âˆ™đ?’‰đ?’†đ?’Šđ?’ˆđ?’‰đ?’• đ?’”∙ đ?&#x;? đ?’” √đ?&#x;‘ đ?&#x;? = = đ?&#x;? = đ?&#x;’ đ?’” đ?&#x;?

Area equilateral triangle The next triangle has the side as the height of the previous one, therefor √đ?&#x;‘ đ?’”đ?&#x;? = đ?’‰ = đ?’” đ?&#x;? đ?&#x;‘ đ?&#x;‘ đ?&#x;‘√đ?&#x;‘ √đ?&#x;‘ √đ?&#x;‘ √đ?&#x;‘ √đ?&#x;‘ đ?’‰đ?&#x;? = đ?&#x;? đ?’”đ?&#x;? = đ?&#x;? ∙ đ?&#x;? đ?’” = đ?&#x;’ đ?’” ⇒ Area2=( đ?&#x;? đ?’” ∙ đ?&#x;’ đ?’”)/đ?&#x;? = đ?&#x;?đ?&#x;” đ?’”đ?&#x;? √3 2 đ?‘

4 Ratio= 3√3 = 4/3 ⇒Ratio = 4/3 16

đ?‘ 2

Spiral Problem 5 ABC is equilateral triangle and A(4; 2), B(0;6). AD⊼BC , D∈BC, ADE equilateral triangle, etc. a) Determine the equation for the line AB. b) Determine the equation for the line CD. c) Determine the equilateral triangle height equation that passes through C. d) Find the coordinate for the point C. e) Determine the perimeter of triangle ABC f) Determine the perimeter of triangle CHF g) Determine the area of triangle ABC h) Determine the area of red triangle. i) Determine the area of pink triangle. j) Determine the area of purple triangle.

SOLUTION: đ?‘Śđ??ľ −đ?‘Śđ??´

a) đ?‘šđ??´đ??ľ =

đ?‘Ľđ??ľ −đ?‘Ľđ??´

=

6−2 −4

= −1 ⇒ đ?‘Ś − đ?‘Śđ??ľ = đ?‘šđ??´đ??ľ (đ?‘Ľ − đ?‘Ľđ??ľ ) ⇒ đ?‘Ś − 6 = −đ?‘Ľ

b) D middle of [AB] ⇒ đ?‘Ľđ??ˇ =

đ?‘Ľđ??´ +đ?‘Ľđ??ľ 2

= 2;

đ??śđ??ˇ ⊼ đ??´đ??ľ ⇒ đ?‘šđ??śđ??ˇ =

−1 đ?‘šđ??´đ??ľ

đ?‘Śđ??ˇ =

đ?‘Śđ??´ +đ?‘Śđ??ľ 2

⇒ đ?‘¨đ?‘Š: đ?’š = −đ?’™ + đ?&#x;”

= 4 ⇒ đ?‘Ť(đ?&#x;?; đ?&#x;’)

= 1 ⇒ đ?‘šâˆĄ(đ??śđ??ˇ, đ?‘‚đ?‘Ľ) = 45° ⇒ đ?‘Ś − đ?‘Śđ??ˇ = đ?‘šđ??śđ??ˇ (đ?‘Ľ − đ?‘Ľđ??ˇ ) ⇒ đ?‘Ś − 4 = đ?‘Ľ − 2 ⇒ CD: đ?’š = đ?’™ + đ?&#x;?

c) AB2 = (đ?‘Ľđ??ľ − đ?‘Ľđ??´ )2 + (đ?‘Śđ??ľ − đ?‘Śđ??´ )2 = 32 ⇒ đ?‘¨đ?‘Š = đ?&#x;’√đ?&#x;? ⇒ đ??śđ??ˇ =

đ?’?√đ?&#x;‘ đ?&#x;?

⇒ đ??śđ??ˇ =

4√6 2

⇒ đ?‘Şđ?‘Ť = đ?&#x;?√đ?&#x;” .

d) đ??ś ∈ đ??śđ??ˇ ⇒ đ?‘Śđ??ś = đ?‘Ľđ??ś − 2 24 = (đ?‘Ľđ??ˇ − đ?‘Ľđ??ś )2 + (đ?‘Śđ??ˇ − đ?‘Śđ??ś )2 ⇒ (2 − đ?‘Ľđ??ś )2 + (2 − đ?‘Ľđ??ś )2 = 24 ⇒ 2đ?‘Ľđ??ś2 − 8đ?‘Ľđ??ś − 16 = 0 ⇒ đ?‘Ľđ??ś2 − 4đ?‘Ľđ??ś − 8 = 0 đ?‘Ľđ??ś =

4Âąâˆš48 2

, đ?‘Ľđ??ś < 0 ⇒ đ?’™đ?‘Ş = đ?&#x;? − đ?&#x;?√đ?&#x;‘ ⇒ đ?‘Śđ??ś = 2 − 2√3 + 2 ⇒ đ?’šđ?‘Ş = đ?&#x;’ − đ?&#x;?√đ?&#x;‘ ⇒ đ?’™đ?‘Ş â‰ˆ −đ?&#x;?. đ?&#x;’đ?&#x;”; đ?’šđ?‘Ş â‰ˆ đ?&#x;Ž. đ?&#x;“đ?&#x;’ đ?‘ˇđ?‘¨đ?‘Šđ?‘Ş = đ?&#x;‘đ?‘¨đ?‘Š = đ?&#x;?đ?&#x;?√đ?&#x;?

Spiral Problem 6 Theodorus ‘ spiral a) b) c) d) e) f) g)

ΔABC, m(ďƒ?ABC)=90â—Ś, A(0; 0), B(1; 0), C(1; 1) Determine the equation for the line AB. Determine the equation for the line CB. Determine the equation for the line CA. Determine the equation for the line CD. Find the coordinate for the point D. Determine the equation for the line AD. Determine the height for the triangle ACD.

SOLUTION: a) A(0; 0), C(1: 1) ⇒ AC: y = x đ??śđ??ˇ ⊼ đ??´đ??ś ⇒ đ?‘šđ??śđ??ˇ ∙ đ?‘šđ??´đ??ś = −đ?&#x;? ⇒ đ?‘šđ??śđ??ˇ = −1 ⇒ đ?‘Ś − đ?‘Śđ??ś = đ?‘šđ??śđ??ˇ (đ?‘Ľ − đ?‘Ľđ??ś ) ⇒ đ?’š − đ?&#x;? = −(đ?’™ − đ?&#x;?) ⇒ đ?‘Şđ?‘Ť: đ?’š = −đ?’™ + đ?&#x;? b) đ??ˇ ∈ đ??śđ??ˇ ⇒ đ?‘Śđ??ˇ = −đ?‘Ľđ??ˇ + 2 AD2 = (đ?‘Ľđ??ˇ − đ?‘Ľđ??´ )2 + (đ?‘Śđ??ˇ − đ?‘Śđ??´ )2 ⇒ 3 = (đ?‘Ľđ??ˇ − 0)2 + (đ?‘Śđ??ˇ − 0)2 ⇒ 3 = đ?‘Ľđ??ˇ2 + đ?‘Śđ??ˇ2 ⇒ đ?‘Ľđ??ˇ2 + (−đ?‘Ľđ??ˇ + 2)2 = 3 ⇒ 2đ?‘Ľđ??ˇ2 − 4đ?‘Ľđ??ˇ + 1 = 0 ⇒ đ?‘Ľđ??ˇ = c)

4Âąâˆš8 4

đ?‘Śâˆ’đ?‘Śđ??´ đ?‘Śđ??ˇ −đ?‘Śđ??´

d) â„Ž =

⇒ đ?‘Ľđ??ˇ = đ?‘Ľâˆ’đ?‘Ľđ??´

=đ?‘Ľ

đ??śđ??ˇ ∙đ??śđ??´ đ??´đ??ˇ

đ??ˇ −đ?‘Ľđ??´

2Âąâˆš2 2 đ?‘Ś

⇒ đ?‘Śđ??ˇ = 2 − đ?‘Ľ

⇒� =� ⇒�=

⇒ℎ=

đ??ˇ

1∙√2 √3

đ??ˇ

⇒�=

2Âąâˆš2

2+√2 2

2

⇒ đ?‘Śđ??ˇ = 2

2∓√2

. đ?‘Ľđ??ˇ 2 2+√2

đ?&#x;?−√đ?&#x;?

< đ?‘Śđ??ˇ ⇒ đ?‘Ť (

đ?‘Ľ ∙ 2−√2 ⇒ đ?‘Ś = 2−√2 đ?‘Ľ ⇒ đ?‘Ś =

6+4√2 2

đ?&#x;?

;

đ?&#x;?+√đ?&#x;? đ?&#x;?

).

đ?‘Ľ ⇒ đ?‘¨đ?‘Ť: đ?’š = (đ?&#x;‘ + đ?&#x;?√đ?&#x;?)đ?’™

∙√đ?&#x;” đ?&#x;‘

Teachers coordinating: Mihaela Git and Normalisa Neiman

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