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

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[engralsaffar]

I did it differently, I used pythagoras 3 times,
x^2=25^2+h^2 - - - (1)
y^2=20^2+h^2 - - - (2)

2h^2=1000
h^2=500 Substitute in (1)
x^2=625+500=1125
x=15sqrt(5)

[Nothingx]

I have used a different method but still my answer is correct

Let BE =y

So in triangle AEB we have

X² +y² = 2025

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

Now solve 1 and 2 we get

X² =1125
So x≈ 33.54 units

[Mediterranean]

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

[unknownidentity]

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

[thewolfdoctor]

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

[flashg]

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.

[marcgriselhubert]

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).

[johnspathonis]

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.

[adhyanjaiswal]

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

[solimana-soli]

15√5 with Euclid's theorem on AEB

[MrPaulc]

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.

[jan-willemreens]

... 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

[sandanadurair]

X^2+BE^2 =45^2
X^2-25^2 = BE^2-20^2 =EF^2

X^2-BE^2 =

EQN 1 + EQN2 Gives

2* X^2 = 45^2+45*5
X^2 = 45 * 50 /2

X = 15* sqrt(3)

[NistkastenEppstein]

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

[michaelkouzmin]

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=
x= sqrt(25*45)=15sqrt(5).

[kennethstevenson]

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.

[montynorth]

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.

[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

[rebkh]

We can also solve this by establishing equation using Pythagoras theorem

[michaelstahl]

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 .