Find the length X | A Nice Geometry Problem | Important Geometry and Algebra Skills Explained

Find the length X | A Nice Geometry Problem | Important Geometry and Algebra Skills Explained

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[jimlocke]

Solution using double angle formula for tangent, tan(2Θ) = (2tan(Θ))/(1 - tan²(Θ)): Extend AD and DC upward until they meet. Call the intersection point F. Note that tan(Θ) = 10/20 = 1/2. Therefore, tan(2Θ) = (2(1/2))/(1 - (1/2)²) = 1/(1-(1/4)) = 4/3. Furthermore, length AF = 20(tan(2Θ)) = 20(4/3) = 80/3. Length DF = length AF - length AD = 80/3 - 20 = 80/3 - 60/3 = 20/3. Consider ΔFDC and ΔFAB. They are similar by angle-angle (<AFB is common and <FDC and FAB are both 90°). By ratios of sides of similar triangles, (length DC)/(length AB) =(length DF)/(length AF). Thus, substituting X for (length DC), X/20 = (20/3)/(80/3) = 20/80 = 1/4. Multiplying both sides by 20, X = (20)(1/4) = 5. as Math Booster also found.

[manojitmaity]

We can solve it in the following easiest way:
tan(theata)= 1/2
tan(2 theata) =20/(20- x)
Or, 2tan (
solving further
We conclude, x=5

[harikatragadda]

A geometric solution would be to reflect ∆EAB about the EB so that A falls on BC at G. Joining EC makes two congruent right triangles EDC and EGC, since they have the same Hypotenuse.
∠DEC = ½(π-∠AEG) = 90-∠AEB = θ
∆EDC is Similar to ∆EAB by AAA.
Hence X/10 = 10/20
X = 5

[ROCCOANDROXY]

Draw a perpendicular from C to AB intersecting at P implies tan(theta) = 1/2 and tan(2 * theta) = 20/(20 - x) = 4/3 implies x = 5.

[اممدنحمظ]

تمرين جميل جيد . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين .

[spacer]

Let F be midpoint of BC, then ∆EFC and ∆EFB are isosceles ∆ with EF=CF=FB, and ∠FEB=θ, ∠EFC=2θ, ∠CEF=(180-2θ)/2=90-θ
∠CEB=∠CEF+∠FEB=90

Thus ∆CDE ~ ∆EAB
x/10 = 10/20 => x = 5

[spafon]

Since I am trying to learn angle bisector theorem, every time I see the same angle on adjacent triangles I wonder if that theorem may apply. It does apply here. If you extend line segments AD and BC upward until they intersect, define the point where they intersect as F. Let us define c= the length of segment BF and a= the length of segment DF. Then we have by the angle bisector theorem c/(a+10)=20/10. Thus c=2a+20. Then we apply the pythagorean theorem on triangle ABF to get (20+a)^2+ 20^2= c^2=(2a+20)^2. This reduces to 3a^2+40a-400=0, a quadratic. The solution is a=20/3. To get x we then use similar triangles DCF and ABF to give x/a=20/(20+a) or x=20a/20+a= 133.3/26.77=5.

[mariopopesco]

Prolong the BC and AD in a point P. Then in triangle ABP with bisector theorem ant Pitagora theorem you find DP to be 20/3. PDC is similar with PAB, so PD/PA equal with DC/AB. 20/3 / 20+20/3 = x / 20. x=5

[bpark]

There is simpler way using trig. Tan Θ = 10/20 = 1/2. Tan 2Θ = (1/2 + 1/2)/1 - (1/2)(1/2) = 4/3. Extend BC & AD upward until they intersect at point G. AG/20 = tan (2Θ) = 4/3. AG = 80/3. GD = 80/3 - 20 = 20/3. (20/3)/X = tan (2Θ) = 4/3. 4X = 20. X = 5.

[marioalb]

If AE = ED then
angle DEC = θ

Similarity of triangles
DEC ~ ABE
x/10 = 10/20
x = 5 cm ( Solved √ )

[honestadministrator]

A non trigonometric solution :
BF = AB be cut from CB
∆ EAB, ∆ FBE are congruent
EA = EF, angle EFB = 90°
∆ EDC is congruent to ∆ EFC
angle DCE
= (π - angle CBA)/2
= π/2 - angle EBA
= angle AEB
Hereby,
∆ EAB is similar to ∆ CDE
EA/ CD = AB /DE
CD. AB = EA^2

[giuseppemalaguti]

A me risulta col teorema dei sembra che ci sia un errore!!!.. Ecco, ho trovato l'errore.. Risulta 30+4x=50..x=5

[klementhajrullaj]

The perpendicular from E to BC in point F is 10. BF=20 and FC=x. BC=20+x and the perpendicular from C to AB in point G show that GB=20-x and CG=20, so, 😀😉

[رایان-لح]

اگر از B به موازاتAC رسم می کردیم تا امتدادX را قطع کند مسأله آسان تر حل میشد