Can you calculate the value of X? | (Isosceles Trapezoid) |
Вставка
- Опубліковано 8 лют 2025
- Learn how to find the value of X. Important Geometry and Algebra skills are also explained: Isosceles Trapezoid; Isosceles Trapezium; Thales' theorem; Pythagorean Theorem; similar triangles. Step-by-step tutorial by PreMath.com
Today I will teach you tips and tricks to solve the given olympiad math question in a simple and easy way. Learn how to prepare for Math Olympiad fast!
Step-by-step tutorial by PreMath.com
• Can you calculate the ...
Need help with solving this Math Olympiad Question? You're in the right place!
I have over 20 years of experience teaching Mathematics at American schools, colleges, and universities. Learn more about me at
/ premath
Can you calculate the value of X? | (Isosceles Trapezoid) | #math #maths | #geometry
Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
#FindX #IsoscelesTrapezoid #SemiCircle #GeometryMath #PythagoreanTheorem #ThalesTheorem #SimilarTriangles
#MathOlympiad #IntersectingChordsTheorem #RightTriangle #RightTriangles
#PreMath #PreMath.com #MathOlympics #HowToThinkOutsideTheBox #ThinkOutsideTheBox #HowToThinkOutsideTheBox? #FillInTheBoxes #GeometryMath #Geometry #RightTriangles
#OlympiadMathematicalQuestion #HowToSolveOlympiadQuestion #MathOlympiadQuestion #MathOlympiadQuestions #OlympiadQuestion #Olympiad #AlgebraReview #Algebra #Mathematics #Math #Maths #MathOlympiad #HarvardAdmissionQuestion
#MathOlympiadPreparation #LearntipstosolveOlympiadMathQuestionfast #OlympiadMathematicsCompetition #MathOlympics #CollegeEntranceExam
#blackpenredpen #MathOlympiadTraining #Olympiad Question #GeometrySkills #GeometryFormulas #Angles #Height #ComplementaryAngles
#MathematicalOlympiad #OlympiadMathematics #CompetitiveExams #CompetitiveExam
How to solve Olympiad Mathematical Question
How to prepare for Math Olympiad
How to Solve Olympiad Question
How to Solve international math olympiad questions
international math olympiad questions and solutions
international math olympiad questions and answers
olympiad mathematics competition
blackpenredpen
Andy Math
math olympics
olympiad exam
olympiad exam sample papers
math olympiad sample questions
math olympiada
British Math Olympiad
olympics math
olympics mathematics
olympics math activities
olympics math competition
Math Olympiad Training
How to win the International Math Olympiad | Po-Shen Loh and Lex Fridman
Po-Shen Loh and Lex Fridman
Number Theory
There is a ridiculously easy way to solve this Olympiad qualifier problem
This U.S. Olympiad Coach Has a Unique Approach to Math
The Map of Mathematics
mathcounts
math at work
Pre Math
Olympiad Mathematics
Two Methods to Solve System of Exponential of Equations
Olympiad Question
Find Area of the Shaded Triangle in a Rectangle
Geometry
Geometry math
Geometry skills
Right triangles
imo
Competitive Exams
Competitive Exam
Calculate the length AB
Pythagorean Theorem
Right triangles
Intersecting Chords Theorem
coolmath
my maths
mathpapa
mymaths
cymath
sumdog
multiplication
ixl math
deltamath
reflex math
math genie
math way
math for fun
Subscribe Now as the ultimate shots of Math doses are on their way to fill your minds with the knowledge and wisdom once again.
One of my degrees is in Math. If I had teachers as excellent as you have proven to be, I would have done better. I loved the way you not only went step by step, you also applied theorems, rules and observations before moving on. You made learning easier, thank you.
That’s so kind of you to say, I’m glad you found it helpful!
You are very welcome!😀
Thanks for the feedback ❤️🙏
I didn't need Pythagorus, because there is no need to know the length of BC or of CE. Triangles ABC and ACE are similar, so AC/AB=3/9=AE/AC which means AE = AC/3 = 1.
I totally agree. Pythagorus is kind of an unnecessary step. One third of 9 is 3. One third of three is one. 1 x 2 = 2. 9 - 2 = 7. Done.
Thanks for the feedback ❤️🙏
excellant very quick @@scottdort7197
I hate to be pedantic, but there's no "u" in Pythagoras.
Ce where is the letter E?
Let H be the perpendicular projection of point C on line AB. We have AH=(9-x)/2. In right triangle ABC, let AH*AB=AC². From this, ((9-x)/2)*9=9. Therefore, x=7.
Excellent!
Thanks for sharing ❤️🙏
You can use Ptolemy's Theorem and everything becomes very simple:
AD*BC = AC*BD+AB*CD
Being AD*BC = AD² = AB²-DB², we have
9²-3² = 3*3 + 9*x;
72 = 9 + 9x;
63 = 9x
x = 7
Excellent!
Thanks for sharing ❤️🙏
Smart guy, Ptolemy.
Too bad that he taught a geocentric solar system.
@@calvinmasters6159 You are partially right.
If you want to describe the motions of celestial bodies as seen from the Earth's surface, it is more convenient to reason like Ptolemy rather than Copernicus!
@@Claudio_Bruzzone Haha, you have a point. We still say that the sun rises and sets.
Even so, I'm not letting him off the hook that easy. He published meticulous diagrams that stood for over a thousand years as textbook science til Copernicus.
Epic as always
Glad to hear that!
Thanks for the feedback ❤️🙏
sinα= ½c/R= (½3)/(½9)= 1/3
α = 19,4712° ; 2α= 38,9424°
x = 2 (R cos2α) = 2*½9*cos2α
x = 7 cm ( Solved √ )
Good brain , high level calculation .
Extend AC up by length 3. Label the end point E, so CE = 3. Construct BC and BE. Note that ΔACB and ΔECB are congruent by side - angle - side. Therefore, BE = AB = 9. Let the point of intersection of BE with the half circle be F. Note that ∠ABC and ∠CBE are equal because ΔACB and ΔECB are congruent. Construct CF. ∠CBF = ∠CBE = ∠ABC . Because inscribed angles ∠CBF and ∠ABC have the same measure, the arcs they subtend, AC and CF, are congruent. The chords that subtend those arcs must be equal, so length CF = AC = 3. Arc BF has the same measure as arc CD, so the chords CD and BF have the same length, x. BE is a secant with length 9 and external segment EF = BE - BF = 9 - x. AE is a secant with length AC + CE = 3 + 3 = 6 and external segment CE = 3. Apply the secant - secant theorem: (CE)(AE) = (EF)(BF), (3)(6) = (9 - x)(9), 18 = 81 - 9x, 9x = 63 and x = 7, as PreMath also found.
Excellent!
Thanks for sharing ❤️🙏
Thank you!
Thanks for watching! 🙏
Decided like you.
Thanks sir!
Glad I could help! 😊
You are very welcome!
Thanks for the feedback ❤️🙏
triangle OAC-4.5, 3, 4.5 ~ triangle ACA` - 3, y, 3 -> 3/4.5 = y/3 -> y=2 ; x=9-y=7
Thanks for sharing ❤️🙏
Similarity of right triangles:
a/3 = 3/9 --> a = 1 cm
x = 9 - 2a = 7 cm (Solved √)
Ptolemy's theorem for inscribed quadrilateral:
d² = 9x + c²
Pytagorean theorem:
d² = (2R)² - c²
Equalling:
9x + c² = (2R)² - c²
9x= (2R)²- 2c²= 9²- 2*3² = 63
x = 7 cm ( Solved √)
√{3^2 - [(9/2)^2 - (x/2)^2]} + x/2 = 9/2
9 - 81/4 + (x^2)/4 = 81/4 - 9x/2 + (x^2)/4
9x/2 = 81/2 - 9
x/2 = 9/2 - 1 = 7/2
x = 7
Excellent!
Thanks for sharing ❤️🙏
@ Thank you 😊
Altura del trapezoide =h : Radio del semicírculo =r=9/2 ---> AD=√(9²-3²)=6√2 ---> Área ABD=3*6√2/2 = 9h/2---> h=2√2 ---> (x/2)²=(r+h)*(r-h) ---> x=7.
Gracias y un saludo cordial.
Excellent!
Thanks for sharing ❤️🙏
Joining O to C.
OC = 4.5, AO = 4.5.
Triangle AOC cosine formula.
3^2 = 4.5^2 + 4.5^2 - 2 * 4.5 * 4.5 * cos AOC.
9 = 40.5 - 40.5 cos AOC.
Cos AOC = 31.5 / 40.5.
Raise perpendicular from point O to ctr. of CD at point P.
Then angle OCP = angle AOC.
Therefore cos OCP = 31.5 /40.5.
Cos OCP = (x / 2) / 4.5.
31.5 / 40.5 = x / 9.
x = 31.5 / 40.5 * 9.
x = 7.
Excellent!
Thanks for sharing ❤️🙏
I did the same, but over complicated it by:
180-(AOC+BOD)=102
X=40.5-40.5*cos102=7
I did the same, but over complicated it by:
180-(AOC+BOD)=102
X=40.5-40.5*cos102=7
Let's find x:
.
..
...
....
.....
Let M be the midpoint of CD. Since OCD is an isosceles triangle, the two triangles OCM and ODM are congruent right triangles and we can apply the Pythagorean theorem. With r being the radius of the semicircle we obtain:
OC² = OM² + CM² = OM² + (CD/2)² = OM² + (x/2)² = OM² + x²/4 ⇒ OM² = OC² − x²/4 = r² − x²/4
Now let's add point E on AB such that OECM is a rectangle. In this case ACE is a right triangle and we can apply the Pythagorean theorem again:
AC² = AE² + CE² = AE² + OM² = (OA − OE)² + OM² = (OA − CM)² + OM² = (r − x/2)² + OM² = r² − rx + x²/4 + OM²
⇒ OM² = AC² − (r² − rx + x²/4) = 3² − r² + rx − x²/4 = 9 − r² + rx − x²/4
Now we are able to calculate the value of x:
OM² = r² − x²/4 = 9 − r² + rx − x²/4
2r² − 9 = rx
⇒ x = (2r² − 9)/r = [2(AB/2)² − 9]/(AB/2) = (2*AB²/4 − 9)/(AB/2) = (AB² − 2*9)/AB = (9² − 18)/9 = (81 − 18)/9 = 63/9 = 7
Best regards from Germany
Excellent!
Thanks for sharing ❤️🙏
別解法 ｏＥ＝ｘ/2。三角形ＡＯＥで三平方の定理を使い、（Ｘ／２）二乗＋ｈｈ＝9*9 １式。三角形ＣＡＯで(9-X/2)二乗＋ｈｈ＝3*3 2式。1式-2式より、９ｘ＝6*12。Ｘ＝8ともとまる。
Thanks for the feedback ❤️🙏
We use an orthonormal center O and first axis (OB). We name t =angleBOD. Then D((9/2).cos(t); (9/2).sin(t)) and B((9/2); 0)
VectorBD((9/2).(cos(t) -1): (9/2).sin(t) and BD^2 = (81/4).[(cos(t)^2 -2.cos(t) +1 +(sin(t)^2] = (81/4). [2 - 2.cos(t)] = (81/4).[4.(sin(t/2)^2]
Then BD = (9/2).(2.sin(t/2)) = 9.sin(t/2). As we know that BD = 3, we then have sin(t/2) = 1/3 and then cos(t) = 1 - 2.(sin(t/2))^2 = 1 - (2/9) = 7/9
The absissa of D is (9/2).cos(t) = (9/2).(7/9) = 7/2, and x is twice this value, so x = 7
Excellent!
Thanks for sharing ❤️🙏
ABCD is a cyclic quadrilateral.
Used product of diagonals equals sum of products of opposite sides.
In this example the diagonals are congruent.
Substituting gives 9 squared minus 3 squared equals 9 times x plus 9.
Get x =7
A trig solution. Connect CO and DO. The angles AOC and BOD equal 2*arcsin(1.5/4.5)=38.9424. Angle COD = 180-(2*38.9424) = 102.1151. Hence, x= 2*4.5*sin(102.1151/2)=7
I've said it before ... and I'm not sure why its always forgotten, but here it is again: for right triangles, the line H is always
H = AB/C where C is hypotenuse
As simple as that looks, it always holds true, and is even fairly easy to prove. Moreover, the part of C on the A side is AA/C and the part of C on the B side is BB/C. Always.
So using Thale's theorem, A = 3, B = sqrt(9^2 - 3^2) = sqrt(72), and C = 9, the given.
X will be C (9) minus two of the "parts of C on the A side", or
X = 9 - (2AA/C) = 9 - (2 * 3 * 3 / 9) = 9 - 18/9 = 9 - 2 = 7
And that's that. Just remember those 3 identities for right triangles and their height
H = AB/C, (A bit of C) = AA/C, and (B bit of C) = BB/C
I didn't use the square symbol because by not using it, the trio is much easier to visually remember.
GoatGuy
x=7
Excellent!
Thanks for sharing ❤️🙏
Bom dia Mestre
Hello dear😀
Thanks ❤️🙏
Draw AD. By Thales' Theorem, as A and B are ends of a diameter and D is a point on the circumference, then ∠BDA = 90°. We can use Pythagoras to find the length of AD, but frankly it's unnecessary. All we need to know is that ∆BDA is a right triangle.
Drop a perpendicular from D to E on AB. If ∠DAB = α and ∠ABD = 90°-α = β, then as ∠DEB = 90°, then ∠BDE = 90°-β = α. ∆BDA and ∆DEB are thus similar by AAA.
AB/BD = BD/EB
9/3 = 3/EB
9EB = 9
EB = 1
Drop another perpendicular from C to F on AB. As FE and BC are parallel, and as DE and CF are both perpendicular to FE and BC and thus parallel to each other, and as ∠FED = ∠EDC = ∠DCF = ∠CFE = 90°, then FEDC is a rectangle, CF = DE and FE = DC = x.
As BD = CA, DE = EF, and ∠CAF = ∠EBD, then ∆DEB and ∆AFC are congruent. AF = EB = 1.
AB = AF + FE + EB
9 = 1 + x + 1 = x + 2
[ x = 9 - 2 = 7 units ]
Excellent!
Thanks for sharing ❤️🙏
Another way to do is, since CE is perpendicular to AB, we can say that CE**2 = y * (9-y) or 9y-y**2. And CE**2 is also equal to 9 - y**2 (from the right angle triangle ACE). Thus 9y - y**2 = 9 -y**2. So 9y = 9. And thus y = 1
φ = 30° → sin(φ) = 1; ∆ ABC → ABC = δ; sin(BCA) = 1; AC = BD = 3; CD = x = ?
AB = 9 = AE + EO + BO = 2r; sin(AEC) = 1; CE = h
AB = 9 → BC = 6√2 → h = 3(6√2)/9 = 2√2; sin(δ) = 1/3 = AE/3 → AE = 1 → x = 9 - 2 = 7
Ptolemy's theorem for such a simple problem is an overkill.
1. Find the diagonal using the Pythagorean theorem, l = sqrt (9 ^ 2 - 3 ^ 2) = sqrt (72).
2. Find the height of the trapezoid from the right triangle. 3 * sqrt (72) = 9h -->
h = sqrt (8).
3. Extend the height to the intersection with the circle - we get a right triangle with sides
2h, x , 9.
4. Using the Pythagorean theorem, x ^ 2 + 4h ^ 2 = 81 --> x = 7.
Ah, very good. I didn't know Thale's Theorem.
I dropped a vertical from C and then used Law of cosines to find A.
Same result.
I used ptolemy's theorem for this
[Sum of 1st opposite sides]×[Sum of other two opposite sides]=[Multiplication of diagnols]
Another way to go about it a*2=b*2+c*2-2xbxc cos theta. Therefore, 3*2=4.5*2+4.5*2-2x4.5x4.5xcos theta. This implies that theta= 39 degrees. Then 180-(2x39)=102 degree. Now that we know the angles we can use heron's formulas to calculate the area of triangle ODB. So S=(4.5+4.5+3)/2=6. Now Area of Triangle ODB=sqt(6(6-4.5)x(6-4.5)x(6-3)=1/2x4.5xh. From there we know that the Height =2.83. This will be the same Heights for triangle COD. Therefore 1/2x(2.83)=4.5x4.5xsin102. THUS x will equal 6.99 which is 7
Another method:
When we have CB = 6.sqrt(2) just as you did (with the Pytagorean theorem in triangle ACB, as you did), we also have AD = 6.sqrt(2) by symetry.
Now we use the Ptolemy theorem in ABCD which is inscripted in a circle: AC.BD + AB.CD = BC.AD, so 3.3 + 9.x = (6.sqrt(2)).(6.sqrt(2)),
so 9 + 9.x = 72 and then 9x = 33 and x = 7
Excellent!
Thanks for sharing ❤️🙏
it is a system of 2 nonlinear equations:
10 print "premath-can you calculate the value of x trapezoid feb2025":dim x(3),y(3)
20 l1=9:l2=3:r=l1/2:la=l2:lb=r:lc=r:lh=(la^2-lb^2+lc^2)/2/lc
30 h=sqr(la^2-lh^2):lx=2*(r-lh):print "x=";lx
40 x(0)=0:y(0)=0:x(1)=l1:y(1)=0:x(2)=2*r-lh:y(2)=h:x(3)=lh:y(3)=h
50 masx=1200/2/r:masy=850/r:if masx
run in bbc basic sdl and hit ctrl tab to copy from the results window
Suppose that P on AB such that CP//DB
then we have CD=BP=AB-AP
We know that AB=9
Goal: Find AP
Notice that ∆OAC~∆CAP(AA)
then (9/2):3=3:AP => AP=2
Thus CD=7
Clever!
CO=OD=9/2 y²+(x/2)²=(9/2)² y²+(9/2-x/2)²=3²
81/4-x²/4=9-(81/4-9x/2+x²/4) 81/4-x²/4=9-81/4+9x/2-x²/4
162/4-9=9x/2 63/2=9x/2 x=7
Teorema de Ptolomeo. Producto de las diagonales igual a 9x+9
Thanks for the feedback ❤️🙏
CE² = AC² - AE²
CE² = OC² - OE²
AC² - AE² = OC² - OE²
3² - [(9 - x)/2]² = (9/2)² - (x/2)²
9 - (81 - 18x + x²)/4 = (81/4) - (x²/4)
(36 - 81 + 18x - x²)/4 = (81 - x²)/4
36 - 81 + 18x - x² = 81 - x²
18x = 126
x = 7
Just another solution (with excess of trigonometry :)))) :
1. CD = 6*sqrt(2) according to the Thales and Pythagorean theorems;
2. Let a = alpha = angle CAB, then ABC = DCB = D90-a and CDB = D180-a, CBD = D180-(D90-a) -(D180-a) = 2a-D90;
3. Let us apply cosine theorem to triangle ABC:
CB^2 = AC^2 + AB^2 - 2*AB*BC*cos(a) => cos(a) = ( CB^2 - (AC^2 + AB^2))/( 2*AB*BC) = (72 - 9 -81)/(-54) = 1/3;
4. Let us apply sine theorem to triangle CBD:
CB/sin(CDB) = x/sin(CBD) ;
x = CB*sin(CBD)/sin(CDB) =
=6*sqrt(2) * sin(2a-D90)/sin(D180-a) =
= 6*sqrt(2) *(sin(2a)*cos(D90)-cos(2a)*sin(D90))/sin(a)=
= 6*sqrt(2)* (-cos(2a)/sin(a)) =
= 6*sqrt(2)*(1-2*(cos(a))^2)/sqrt(1-(cos(a))^2) =
=6*sqrt(2)*(1-2*(1/3)^2)/sqrt(1-(1/3)^2) =
= 6*sqrt(2)*(1-2/9)/sqrt(1-1/9) =
=6*sqrt(2)*(7/9)/sqrt(8/9) =
= 7 linear units.
7 , using basic similarity
AE= (9-x)/2. AC=3 therefore CE= sqroot [ 9-( 9-x )/2)^2)]. Then using pythag. X=7.
Excellent!
Thanks for sharing ❤️🙏
I first found bc to be sqrt 72. Then calculated triangle area to be 3 times that by half which is equivalent to 9 by half by height that gave me a height of sqrt 8.then calculated half of x being radius squared which is 4.5 minus height squared and square rooting getting 3.5 then finally multiplying by 2 to get 7 for x.
Please keep ❤
Thanks dear❤️🙏
2:00-8:00 ΔABC: AC²=AE•AB =>
=> AE=AC²/AB=3²/9=9/9=1 😁
Trazando OD y DF, obtenemos dos triángulos rectángulos (OFD y BFD) que comparten el cateto (DF) al que llamaremos "h".
Aplicamos Pitágoras en ambos triángulos:
(9/2)²=(x/2)²+h²
3²=[(9-x)/2]²+h²
Despejando h² en ambas ecuaciones tenemos:
h²=(81-x²)/4
h²=(36-81+18x-x²)/4
e Igualando los resultados:
81-x²=36-81+18x-x²
18x=81+81-36
18x=126
X=126/18
X=7
Saludos
r=9/2, cos(
let A be the center of a coordinate system (x-axis along AB)
P(x,y) x^2 + y^2 =3^2 = 9
p(x,y) is also on the circle- (x-4.5)^2+y^2=4,5^2
x^2+2*4.5*x+4.5^2+y^2=4.5^2
x^2+y^2-9x=0
9-9x=0
x=1 this x is not the questioned x
so CD (the answer) = 9-(2*x)=7
7
Excellent!
Thanks for sharing ❤️🙏
X = 9 - 2 × (3^2 / 9) = 7
Yep, as simple as that !
شكرا لكم على المجهودات
يمكن استعمال
ABC=a
CBD=90-2a
x^2 = CB^2 + CD^2-2CB CD cosCBD
=7
Solution:
Since we are dealing with an isosceles trapezoid, let's label the lengths
AE = a
AO = r = 9/2
CE = h
Applying Pythagorean Theorem in ∆ ACE, it will be:
AE² + CE² = AC²
(a)² + (h)² = (3)²
a² + h² = 9 ... ¹
Once again, applying Pythagorean Theorem in ∆ CEO, it will be:
EO = r - a = 9/2 - a
EO² + CE² = CO²
(9/2 - a)² + (h)² = (9/2)²
81/4 - 9a + a² + h² = 81/4
a² + h² - 9a = 0 ... ²
Replacing Equation ¹ in Equation ², it will be:
9 - 9a = 0
9a = 9
a = 1
Replacing "a = 1" in Equation ¹, to calculate "h", it will be:
(1)² + (h)² = (3)²
1 + h² = 9
h² = 8
h = 2√2
Finally, applying Pythagorean Theorem in ∆ CGO, such that "G" is CD midpoint, it will be:
CG² + GO² = CO²
(x/2)² + (2√2)² = (9/2)²
x²/4 + 8 = 81/4 (×4)
x² + 32 = 81
x² = 49
x = 7
Thus x = 7 units ✅
MY RESOLUTION PROPOSAL :
01) Drop a Vertical Line from D and find Point E between Point O and Point B. Line Segment BE = a lin un
02) OA = OB = OB = OD = 9 / 2 = 4,5 lin un.
03) AD^2 + BD^2 = AB^2
04) AD^2 = AB^2 - BD^2
05) AD^2 = 81 + 9
06) AD^2 = 90
07) AD = sqrt(90)
08) Right Triangles (BDE) and (ABD) are Similar. We can use "The Principle of Proportionality".
09) a / 3 = 3 / 9
08) a = 9 / 9
09) a = 1
10) X = 9 - 2a
11) X = 9 - 2
12) X = 7
MY BEST ANSWER :
As far as my best knowledge, CD = X is equal to 7 Linear Units.
Excellent!
Thanks for sharing ❤️🙏
Per risolvere il problema considero un sistema di assi cartesiani passante per il centro del semicerchio O e traccio una circonferenza di raggio 3 con centro in B (9,0) poi determino la equazione delle due due circonferenze e le metto a sistema per trovare la distanza x che è la semibase minore del trapezio ABCD, quindi CD=2*x
equazione della circonferenza generica dato centro (xc,yc) e raggio r
(x-xc)^2+(y-yc)^2=r
(xc,yc)=(0,0)
metto a sistema le due circonferenze
(x-0)^2+(y-0)^2=(9/2)^2
(x-(9/2))^2+(y-0)^2=3^2
x^2+y^2=81/4
x^2+81/4-2*(9/2)*x+y^2=9
y^2=(81/4)-x^2
x^2+81/4-9*x+81/4-x^2-9=0
81/4+81/4-9=9*x
162/4-9=9*x
(162/4-9)/9=x
x=(81/2-9)/9
CD=2*x
CD=2*(81/2-9)/9
CD=(81-18)/9
CD=63/9
CD=7
Não precisa saber o valor de CB
Da pra fazer por similaridade
Tracando a linha perpendicular partindo de C até AB, temos o ponto É
O triângulo ABC é congruente ao triângulo ACE, de forma que podemos achar a razão das hipotenusas e cateto menor
3/9=AE/3
AE = 1
X = 7
Dá para usar tbm h^2=mn e pitagoras no triângulo ACE para achar:
m=AE
n=EB
h=CE
h^2=m(9-m)
h^2+m^2=3^2
Comparando h^2, temos
m(9-m)=9-m^2
m=1
Excellent!
Thanks for sharing ❤️🙏
7