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.
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.
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.
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
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.
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.
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
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
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
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
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
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 ]
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
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
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
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
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.
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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.
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excellant very quick @@scottdort7197
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.
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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.
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Epic as always
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Decided like you.
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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
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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.
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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.
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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。
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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
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
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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
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
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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
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triangle OAC-4.5, 3, 4.5 ~ triangle ACA` - 3, y, 3 -> 3/4.5 = y/3 -> y=2 ; x=9-y=7
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√{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
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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 ]
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AE= (9-x)/2. AC=3 therefore CE= sqroot [ 9-( 9-x )/2)^2)]. Then using pythag. X=7.
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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
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
Bom dia Mestre
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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.
Teorema de Ptolomeo. Producto de las diagonales igual a 9x+9
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x=7
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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
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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
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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 ✅
r=9/2, cos(
X = 9 - 2 × (3^2 / 9) = 7
Yep, as simple as that !
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.
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