A Very Nice Geometry Problem | You should be able to solve this! | 2 Methods

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[@Antony_V]

Pythagoras' theorem on triangle OQP: x^2=(x-5)^2+10^2 --> x=12,5

[@sorourhashemi3249]

I love it. Challenging. AB= rad.2x. SQ= rad 2/2 SB. SB= 7.04. So AS×SB=NS×SP===>(rad2x - 7.04)×7.04=15×5===>X=12.5

[@ناصريناصر-س4ب]

We construct the complete circle and get PS*SE=SB*SA, where E is the point of intersection of PQ with the circle, and from it we get the equation 5√2(x√2-5√2)=75, and from it x=25/2

[@prossvay8744]

Connect O to P
OA=OB=OP=x (radius of sermicle)
Angle OAB=OBA=45°
Angle QBS=BSQ=45°
So BQ=QS=5
In ∆ OPQ
OQ^2+PQ^2=OP^2
OQ=x-5
PQ=10
OP=x
(x-x)^2+10^2=x^2
So x=25/2.❤

[@devondevon4366]

12.5 or 25/2
Notice that triangle BSQ is similar to AB0
Hence, OB=x and QB = 5
Draw a line from O to P to form a right triangle, OPQ
OP =x = radius
OQ= x-5 since QB =5
PQ = 10 (5+5)
x^2 = (x-5)^2 + 10^2
x^2 = x^2 - 10x + 25 + 100
10x= 125
x = 125/10
x= 25/2 or 12.5

[@marioalb9726]

Intersecting chords theorem:
5.(2R-5) = 10²
10R -25 = 100
R = 12,5 cm ( Solved √ )

[@gelbkehlchen]

Solution:
Triangle AOB and triangle SQB are right-angled isosceles triangles with angles 90° - 45° - 45°. Therefore QB = SQ = 5.
Pythagoras for triangle OPQ:
x² = OQ²+PQ² = (x-5)²+10² = x²-10x+25+100 = x²-10x+125 |-x²+10x ⟹
10x = 125 |/10 ⟹
x = 12.5

[@alexniklas8777]

Method 3.
AB= x√2; BS= 5√2; AS= x√2-5√2; NS= 15.
AS×BS=PS×NS; (x√2-5√2)5√2=15×5; 10x-50=75; 10x=125; x=12,5
Thanks sir

[@himo3485]

QB=5 PQ=5+5=10
10*10=5*(2x-5) 100=10x-25 10x=125 x=12.5

[@SGuerra]

Parabéns pela escolha da questão. Eu encontrei a mesma solução que você deu! Show!

[@michaeldoerr5810]

The answer is x = 25/2 and is more helpful than the comments. Maybe that is because this video is similar to past videos. I hope that this means that I am studying well!!!

[@santiagoarosam430]

AS*SB=PS(SQ+PQ)→ AS*5√2=5*(5+5+5)→ AS=15√2/2→ AB=(15√2/2)+5√2=25√2/2 =X√2→ X=25/2.
Gracias y un saludo.

[@imetroangola4943]

Pensei da mesma forma! 🫂

[@RealQinnMalloryu4]

(5)^2=25 (5)^2=25 {25+26}=50 90°AOQBSPX/50=1.40AOQBSPX 1.2^20 1.2^2^10 1.1^1^2^5 1^2^1 (AOQBSPX ➖ 2AOQSPX+1).

[@professorrogeriocesar]

Semelhança, depois, Teorema das Cordas.

[@giuseppemalaguti435]

√(x^2-10^2)+5=x...x=12,5

[@sumankundu93]

(x-5)^2+10^2=x^2.

[@johnbrennan3372]

as mult by sb= 5 mult by 15=75 (product of intersecting chord parts equal). Angle oba= 45 degrees(isosceles triangle ) so sb=sqroot 50=5 root2.Therefore as=75/ (5root2)=15/root 2. So ab= 15/root2 + 5 root2=25/root2. This implies that 2(x^2)= 625/2 that is x^2=625/4.So x= 25/ 2

[@devondevon4366]

25/2

[@himadrikhanra7463]

12 unit ?

[@yakupbuyankara5903]

X=12,5

[@DhdhBdhx-m4z]

Sancs