Neat Geometry | Can you find the length X? | (Math skills explained Step-by-Step) |

Learn how to find the length X of the right triangle in a rectangle. Important Geometry skills are also explained: Complementary angles; similar triangles. Step-by-step tutorial by PreMath.com
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Neat Geometry | Can you find the length X? | (Math skills explained Step-by-Step) | #math #maths
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[@wackojacko3962]

Soooooooo smoooooth! 🙂

[@PreMath]

Excellent! Thanks ❤️

[@engralsaffar]

I did it differently, I used pythagoras 3 times, x^2=25^2+h^2 - - - (1) y^2=20^2+h^2 - - - (2) x^2+y^2=45^2=25^2+h^2+20^2+h^2 2h^2=1000 h^2=500 Substitute in (1) x^2=625+500=1125 x=15sqrt(5)

[@kkulkulkan5472]

Exactly how I did it as well. Simple.

[@soli9mana-soli4953]

15√5 with Euclid's theorem on AEB

[@Nothingx303]

I have used a different method but still my answer is correct Let BE =y So in triangle AEB we have X² +y² = 2025 ______(1) Now extend E to G parallel to DP similarly extend E to H parallel to PC (not your pc😂) So AGHB is a Rectangle so let us the Pythagoras Theorem in the right triangle AGE GE² + AG² =X² [ GE =DP = 25 UNITS] AG² = X² - 625 similarly in triangle BEH we have BH² = y² -400 (since EH = PC = 20 units) Since AGHB is a Rectangle AG = BH So x² - y² = 225 ______(2) Now solve 1 and 2 we get X² =1125 So x≈ 33.54 units

[@thewolfdoctor761]

Draw line parallel to DC, going through E. Let EF = d and BE =a. There are 3 right triangles. Use pyth. theory on all 3. Eqtn. #1: 25^2 + d^2 = x^2, Eqtn. # 2: 20^2 + d^2 = a^2, Eqtn.#3: x^2 + a^2 =45^2. Subtract Eqtn. #2 from Eqtn. #1 to get rid of d^2. You get Eqtn. #4: x^2 - a^2 = 225. Add Eqtn.#4 and Eqtn. #3 and get 2x^2 = 225+45^2. Solve for x. Get 33.54

[@PreMath]

Thanks ❤️

[@flash24g]

x^2 = 25^2 + y^2 [1] z^2 = 20^2 + y^2 [2] x^2 + z^2 = 45^2 [3] Subtracting [2] from [1] and adding [3] gives 2 x^2 = 45^2 + 25^2 - 20^2 = 2250 But similarity is a more elegant method. Still, why overcomplicate with this stuff about complementary angles? Triangles AEB and AFE have an angle in common and separately from this a right angle each, so this fact is sufficient to show AA similarity. I've seen a proof of Pythagoras' theorem that works by the same construction involving similar triangles, but I can't remember the details at the moment. Maybe what I've done together with your solution can be rearranged into such a proof.

[@johnspathonis1078]

Much simpler method. Draw a circle through points A, B , E. Use intersecting chords to find EF. i.e. EF^2 = 20 x 25 =500. Then Pythag 500 + 625= X^2.

[@kennethstevenson976]

Using a for the height of the similar triangles with bases and marking alpha and beta angles to get the correct proportions you should get a/25 = 20/a . This produces a^2=500. Since x^2 = a^2 + 25^2 ; x^2 = 500 + 625 ; x^2 =1125 ; x = 33.54 units.

[@unknownidentity2846]

Let's find x: . .. ... .... ..... Since ABE is a right triangle, its height according to the base AB can be calculated by using Euclids formula: h²(AB) = DP*CP = 20*25 = 500 Now we can calculate x by using the Pythagorean theorem: x² = DP² + h²(AB) = 25² + 500 = 625 + 500 = 1125 ⇒ x = √1125 = 15√5 Best regards from Germany

[@PreMath]

Excellent! Thanks ❤️

[@marcgriselhubert3915]

Let's consider H the orthogonal projection of E on (AD) and K the orthogonal projection of E on (BC), then we have EH = PD =25 and EK = PC = 20. Let's consider now h =KB = HA the height of rectangle KBAH. In right triangle EKB we have EB^2 = EK^2 + KB^2 = 400 + h^2, and in right triangle EFA we have AE^2 = EF^2 + FA^2 = h^2 + 625. Now in right triangle AEB we have AB^2 = EB^2 + AE^2, so: 45^2 = (400 + h^2) + (h^2 + 625), or 2025 = 1025 + 2.(h^2), or h^2 = 500. Finally: x^2 = AE^2 = h^2 + 625 (already written), so x^2 = 500 + 625 = 1125, and x = sqrt(1125) = 15. sqrt(5).

[@Copernicusfreud]

That is how I did it. I drew a line parallel to CD and AB that had an intersect point at E. I also used the line segment of EF. EF = h. AE = x. BE = y. Using the Pythagorean Theorem for three different interior triangles, you can generate 3 equations with three unknowns with one of the unknowns as x. Doing the plug-and-chug, x = (15)*(√5). The solution in the video is much prettier.

[@PreMath]

Excellent! Thanks ❤️

[@MrPaulc222]

Draw a horizontal joining the two sides and crossing point E. Call BE, y, and the verticals down from A or B as far as the horizontal with E, a. x^2 - 625 = a^2 y^2 - 400 = a^2 x^2 - y^2 = 225 (equation 1) x^2 + y^2 = 2025 (equation 2) (because AB is 45). Add the two equations for 2x^2 = 2250. Therefore, x^2 = 1125 x = sqrt(1125) x = sqrt(45)sqrt(25) = 5*sqrt(45) 5*3*sqrt(5) x = 15*sqrt(5) = 33.54 (rounded) Unusually, I thought my way was a bit cleaner.

[@adhyanjaiswal8367]

draw //to dc and take EB as Y then using pythagoras in triABE we get X^2+Y^2=45^2 and then take angle BE as ○ so angle AE is 90-○ so take Xcos(90-○)= 25 =>Xsin○=25 and Ycos○=20 now put values of sin○ and cos ○ in sin^2(○)+cos^2(○)=1 we get y^2=(20)^2 X^2/X^2-(25)^2 now put y^2 in x^2+y^2=45^2 and sq. of (x^2-1125)^2=0

[@mathlover6169]

Love the explanation

[@jamestalbott4499]

Thank you!

[@rebkh8061]

We can also solve this by establishing equation using Pythagoras theorem

[@PreMath]

Thanks ❤️

[@salimahmad7414]

Draw a circle on BE as diameter x² = 25×(25+20) = 25×45 = 25×9×5 x = 5×3×√5 = 15√5

[@Mediterranean81]

Let h be the height of the rectangle and a = EP Area of trapezoid (h+a)10 Area of trapezoid (h+a)12.5 Their ratios (ADEP/BECP) are equal to 0.8 or 4/5 Let's focus on the right triangle AE is a scaled up BD So We have BE = 4/5 AE = 4/5 X AB is 25+20=45 So let's use the Pythagorean theorem 2025 = X^2+16/25 X^2 2025= 41/25 X^2 25*2025/41= X^2 5*45 sqrt41/41 = X So X=225 sqrt 41/41

[@NistkastenEppstein]

by Euclid's law: Height of triangle ABE: EF^2 = AF x BF From Pythagoras x = sqrt(EF^2 + AF^2)

[@raya.pawley3563]

Thank you

[@himo3485]

AD'E♾️E'CB AD'=BC'=y 25/y=y/20 y=√500 x=√[25^2+(√500)^2]=√1125=15√5

[@montynorth3009]

Draw horizontal from point E to meet line AD at point Q. Then triangles AQE and AEB are similar. X/25 = 45/X. Cross multiplying. X^2 = 25 x 45 = 1125. X = 33.541.

[@PreMath]

Excellent! Thanks ❤️

[@ybodoN]

If we translate DPC so that P = E, we get three similar triangles. x is the major cathetus of the one whose hypotenuse is 45 units. x is the hypotenuse of the one whose major cathetus is 25 units. Therefore, x / 25 = 45 / x ⇒ x² = 1125 ⇒ x = 15√5 units.

[@PreMath]

Excellent! Thanks ❤️

[@denisweimer8791]

This was my solution as well. MUCH quicker this way.

[@michaelstahl1515]

Nice Video Sir , but I used Euklid`s theorem : p = 25 , q = 20 , p*q = h^2 . so h = Squareroot ( 25 * 20 ) = 22,36.. . Then x ² = 25² + 22,36 ² = 1125 . Squareroot ( 1125 ) = x = 33,54 units .

[@jan-willemreens9010]

... Angle(EAB) and Angle(ABE) are complementary angles ... assuming Angle(EAB) = a so Angle(ABE) = 90 - a ... Angle(CBE) = 90 - (90 - a) = a ... now drawing a perpendicular line segment on BC (naming point F on BC) from obtuse Angle(BEP), creating right angled triangle (FEB) ... applying SIN(X) = OPP/HYP on Triangle(EAB) and Triangle(BEF) as follows ... (1) SIN(Angle(EAB)) = I BE I / I AB I = I BE I / 45 & (2) SIN(Angle(CBE) = I EF I / I BE I = 20 / I BE I ... setting (1) = (2) ... I BE I / 45 = 20 / I BE I .... I BE I^2 = 900 .... I BE I = 30 ... finally SQRT( I AB I^2 - I BE I^2 ) = I AE I = X = SQRT( 45^2 - 30^2 ) = approx. 33.541 ( exactly 15 * SQRT(5) ) .... thank you for your clear explanation .... happy Sunday, Jan-W

[@PreMath]

Excellent! You are very welcome! Thanks ❤️

[@sandanadurair5862]

X^2+BE^2 =45^2 .....1 X^2-25^2 = BE^2-20^2 =EF^2 X^2-BE^2 = 25^2-20^2=45*5.....2 EQN 1 + EQN2 Gives 2* X^2 = 45^2+45*5 X^2 = 45 * 50 /2 X = 15* sqrt(3)

[@Alishbafamilyvlogs-bm4ip]

Nice ❤

[@michaelkouzmin281]

another solution, yours is better. We can use Pithagorean theorem thrice: 1. Let h=FE, y= BE; 2. x^2= =h^2+25^2 (1) x^2+y^2=45^2 (2) y^2= h^2+20^2 (3) 2. from (1) h^2 = x^2 - 25^2 (4) 3. Let us put (3) into (2): x^2+h^2+20^2 =45^2 (5) 4. Let us put (4) into (5) x^2+ (x^2-25^2)+20^2 = 45^2 2x^2= 45^2+25^2-20^2; x^2= ((45-20)(45+20)+25^2)/2=25*90/2=24*45; x= sqrt(25*45)=15sqrt(5).

[@PreMath]

Excellent! Thanks ❤️

[@ankitgodara1565]

Thank You so Such for Questions. This is simple one. Give some hard question. Make a video of questio and a separate video for answers. 🎉🎉🎉🎉🎉🎉

[@PreMath]

Okay sure Thanks ❤️

[@ankitgodara1565]

@@PreMath Welcome

[@pralhadraochavan5179]

Good morning sir

[@murdock5537]

a = BC - EP → tan⁡(ϑ) = 20/a = a/25 → a = 10√5 → x = 15√5

[@PreMath]

Excellent! Thanks ❤️

[@GetMatheFit]

Mit dem Kathetensatz geht es auch mega schnell. Aber wie immer eine sehr schöne Lösung. Kathetensatz: x² = c * p x² = 45 * 25 x² = 3*3*5 * 5*5 x = 3*5*√5 x=15*√5 LG Gerald

[@Copernicusfreud]

Yay! I solved the problem.

[@PreMath]

Excellent! Glad to hear that! Thanks ❤️

[@mikenorman2525]

The problem is unaltered if you move the line DC up so that E is coincident with P and then with the same point F drawn we end up with ED being the same as AF (25) and EC the same as FB (20). Then let EB = y and let AD = z and you get three right angled triangles such that x^2 + y^2 = 45^2, 20^2 + z^2 = y^2 and 25^2 + z^2 = x^2. This is three equations in three variables which can be reduced by substituting y and z to get x^2 = 1125 = 25*9*5 so x = 5*3*sqrt(5) = 15*sqrt(5)

[@PreMath]

Thanks ❤️

[@phungpham1725]

Let EF be h. We have sqh= 25 . 20= 500----> sq x = sq 25 +sq h-----> x =33.54 units

[@PreMath]

Excellent! Thanks ❤️

[@WhiteGandalfs]

Again an interesting instance of "many ways to Rome" :D I just took the height FE as helper unit and pythagoras.

[@giuseppemalaguti435]

x^2=1125..posto h=CB-FP..25:h=h:20..h^2=500...x^2=25^2+500=1125

[@PreMath]

Thanks ❤️

[@batavuskoga]

I've drawn a horizontal line through point E parallel to AB with exactly the same calculation and result

[@PreMath]

Thanks ❤️

[@LuisdeBritoCamacho]

Being AE = x Being BE = y x^2 + y^2 = 2.025 Being DP = 25 and CP = 20, dividing AB in 2 parts. Let's say AE' = 25 and BE' = 20 EE' divide the Triangle [ABE] in two Triangles: [AEE'] and [BEE'] Solving this System of Equations with two unknowns (x ; y): x^2 - EE'^2 = 625 EE'^2 = 625 - x^2 y^2 - EE'^2 = 400 EE'^2 = 400 - y^2 So: 625 - x^2 = 400 - y^2 625 - 400 = x^2 - y^2 225 = x^2 - y^2 Knowing that : x^2 + y^2 = 2.025 The only possible Integer Solutions : (15*sqrt(5) ; 30) Answer: x = 15*sqrt(5) linear units or x ~ 33,5410 linear units

[@PreMath]

Excellent! Thanks ❤️

[@nemesiochupaca5024]

También podía resolverse con puro Pitágoras Sea x el lado rojo e y el azul Entonces x^2+y^2=45^2 (la base del rectángulo) Por otro lado, haciendo K la altura del rectángulo (que no nos va a importar) nos dará que 25^2+K^2=x^2 ; 20^2+K^2=y^2 Anulamos K y nos da que x^2-y^2=225 Sumamos ambas ecuaciones y nos da que 2x^2=2250 de lo que se deduce x=15 √5

[@PreMath]

Excellent! Thanks ❤️

[@Kamini-Singh]

By using Pythagoras theorem I got still a correct answer

[@alster724]

After seeing the x²= (45)(25) I was prepared for it

[@prossvay8744]

x=15√5=33.54 units. ❤❤❤ Thanks

[@PreMath]

Excellent! You are very welcome! Thanks ❤️

[@ИванПоташов-о8ю]

x=√(AB*AF)=√(25*45)=15√3

[@PreMath]

Thanks ❤️

[@Waldlaeufer70]

Let's try that one: x = red line, y = blue line, h = height of triangle ABE 1) x² = h² + 25² 2) y² = h² + 20² 3) x² + y² = 45² 1) and 2) into 3) h² + 25² + h² + 20² = 45² 2h² = 45² - 25² - 20² 2h² = (45 + 25) (45 - 25) - 20² 2h² = 70 * 20 - 20² 2h² = 1400 - 400 = 1000 h² = 500 x² = h² + 25² x² = 500 + 625 = 1125 = 225 * 5 x = 15√5 ≈ 33.54 units

[@PreMath]

Excellent! Thanks ❤️

[@jan-willemreens9010]

... Good day, SQRT(45^2 - 30^2) = approx. 33.541 is X .... I'll explain later .... thanks, Jan-W

[@PreMath]

Thanks ❤️

[@Train_tv_22]

Sach pochy mujy math bilkul b achi ni lgti thi main kehti thi math bilkul b na ho kisi b class😅main

[@PreMath]

Thanks for being honest! No one is perfect. Apne bachon ko math zaroor sikhana 😀

[@hcgreier6037]

Challenge: What is the height of rectangle ABCD? 🤣

[@GetMatheFit]

😂😂😂 Geniale Aussage. Ich werde es mal berechnen 😜 Werde aber unendlich lang damit beschäftigt sein 😝 Eigentlich versteckt sich hier nur der Kathetensatz. Kathetensatz: x² = c * p x² = 45 * 25 x² = 3*3*5 * 5*5 x = 3*5*√5 x = 15*√5 LG Gerald

[@MeksinShira]

𝙽𝚒𝚌𝚎 𝚜𝚒𝚛,, 𝚔𝚎𝚎𝚙 𝚒𝚝 𝚞𝚙 𝚑𝚘𝚠 𝚌𝚊𝚗 𝙸 𝚙𝚛𝚘𝚋𝚕𝚎𝚖 𝚝𝚘 𝚢𝚘𝚞.?

[@PreMath]

Thanks ❤️ My email: [e-mail]

[@MeksinShira]

@@PreMath 𝚝𝚑𝚊𝚗𝚔 𝚢𝚘𝚞 𝚜𝚒𝚛.