Can you solve for X? | (Semicircle) | #math #maths | #geometry

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Can you solve for X? | (Semicircle) | #math #maths | #geometry

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[marioalb]

A = ½πR² = 50π
R = √100 = 10 cm
Intersecting chords theorem:
(x+4)² = x. (2R-x)
x²+8x+16 =2Rx - x²
2x²+(8-2R)x+16=0
x² + (4-10)x +8 = 0
x = 2 and 4cm (two solutions √)

[Mastero-djuw]

We can also apply the chord power theorem if we complete the circle :

(x) ( 20 - x ) = ( x + 4 )²

20x - x² = x² + 8x + 16

move everything from the left side to the right side

2x² - 12x + 16 = 0

x² - 6x + 8 = 0

( x - 4 )( x - 2 ) = 0

x = 4 or x = 2

[misterenter-izrz]

(10-x)^2+(x+4)^2=100, x^2-6x+8 =0, x=2 or 4. Simple puzzle.😊

[michaeldoerr]

It is nice that there is a geoentry problem with TWO solutions. I shall find more on your channel and elsewhere!!!

[ChuzzleFriends]

A = (πr²)/2
50π = 1/2 * π * r²
50 = r²/2
r² = 100
r = 10
So, BC = CP - BP = 10 - x by the Segment Addition Postulate.
Draw radius AC. We get a right triangle △ABC. Use the Pythagorean Theorem.
a² + b² = c²
(10 - x)² + (x + 4)² = 10²
(100 - 20x + x²) + (x² + 8x + 16) = 100
2x² - 12x + 116 = 100
2x² - 12x + 16 = 0
2x² - 12x + 16 = 0
x² - 6x + 8 = 0 (Factorable)
x² - 2x - 4x + 8
x(x - 2) - 4(x - 2)
(x - 4)(x - 2)
Solve for x.
x - 4 = 0 or x - 2 = 0
x = 4 x = 2
Both values seem reasonable. Substitute.
BC = 10 - 2 or 10 - 4 = 8 or 6
AB = 2 + 4 or 4 + 4 = 6 or 8
Use the Pythagorean Theorem on △ABC again, just to check the solutions.
36 + 64 ? 100
100 = 100
64 + 36 ? 100
100 = 100
Both values ARE correct.
So, the value of x is either 2 units or 4 units.

[roykay]

I was surprised to see 2 possible solutions, and then both work! Cool.

[aljawad]

Apply the Pythagorean theorem to triangle ABC, and the two sides involving X will add up to a quadratic that equals the square of the radius.

[adgfx]

(10-x)^2+(x+4)^2=100=>2x^2-12x+16=0=>x^2-6x+8=0=>(x-2)(x-4)=0=>x={2, 4} ans

[krishnarao]

the Diameter subtends 90 degree anywhere on the circle, rest pythogaras, gives just 1 ccombined equation, x2-6x+8=0; then x=2 or x=4 :)) loved it!!

[prossvay]

Semicircle area= 50π
So 1/2(π)r^2=50π
So r=10
Connect C to A
In ∆ ABC
BC^2+AB^2=AC^2
BC=CP-BP=10-x
AB=x+4 ; AC=10
So (10-x)^2+(x+4)^2=10^2
So x=2 or x=4.❤❤❤

[quigonkenny]

Semicircle C:
Aᴄ = πr²/2
50π = πr²/2
r²/2 = 50
r² = 2(50) = 100
r = √100 = 10

Draw CA. As a radius of semicircle C, CA = r. As CP = CA = r and BP = x, CB = r-x.

Triangle ∆ABC:
AB² + CB² = CA²
(x+4)² + (r-x)² = r²
x² + 8x + 16 + r² + x² - 2rx = r²
2x² + 8x + 16 - 2rx = 0
2x² + 8x + 16 - 20x = 0
2x² - 12x + 16 = 0
x² - 6x + 8 = 0
(x-2)(x-4) = 0
x = 2 | x = 4

If x = 2, AB = 2+4 = 6, CB =10-2 = 8. 6²+8² = 10² ✓

If x = 4, AB = 4+4 = 8, CB = 10-4 = 6. 8²+6² = 10² ✓

Both solutions are valid. x = 2 or x = 4.

[sergioaiex]

Solution:

Semicircle area = 50π

50π = πr²/2 (÷π)
r²/2 = 50
r² = 100
r = √100
r = 10

Connecting C to A, the distance is equal to the radius and is 10. Therefore, the cathetes are (10 - x) and (x + 4). So we will apply the Pythagorean Theorem

(10 - x)² + (x + 4)² = 10²
100 - 20x + x² + x² + 8x + 16 = 100
2x² - 12x + 16 = 0 (÷2)
x² - 6x + 8 = 0

Let's solve the quadratic equation by grouping and factoring

Changing -6x by -2x - 4x, we have:

x² (- 2x - 4x) + 8 = 0
(x² - 2x) (- 4x + 8) = 0
x (x - 2) - 4 (x - 2) = 0
(x - 2) (x - 4) = 0

x = 2 Accepted
x = 4 Accepted

The 2 values of "x" are accepted because they are validated in the Pythagorean Theorem

Therefore:
x = 2 and x = 4

[alexniklas]

πr^2=50π; r^2=100; r=10.
(2r-x)x=(x+4)^2;
x1=2; x2=4.

[nandisaand]

Its funny how Professor Premath always takes the shortcut ("...and now lets recall the famous identity...") when squaring a binomial (as opposed to doing extended FOIL method), yet always goes the long way when factoring a quadratic.

[LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL :

01) Circle Area = 100Pi sq un
02) R^2 = 100 ; R = sqrt(100) ; R = 10
03) AC^2 = BC^2 + AB^2
04) 100 = (10 - X)^2 + (X + 4)^2
05) 100 = 100 - 20X + X^2 + X^2 + 8X + 16
06) 2X^2 - 12X + 16 = 0 ; X^2 - 6X + 8 = 0
07) X^2 - 2X - 4X + 8 = 0
08) X*(X - 2) - 4*(X - 2) = 0
09) (X - 4)*(X - 2) = 0
10) X = 4 or X = 2

ANSWER :

Two Options, wether X = 2 or X = 4. Take a chance!!

[himo]

r*r**π*1/2=50π r=10
BC=10-x
(10-x)^2+(x+4)^2=10^2
100-20x+x^2+x^2+8x+16=100
2x^2-12x+16=0 x^2-6x+8=0
(x-2)(x-4)=0 x=2, 4

[wackojacko]

@ 0:46 🤔 ...the after effects of determining X equal to 2 and 4. 🙂

[nomad]

Я решил с помощью теоремы о пересекающихся хордах

[calvinmasters]

My first two attempts produced j-numbers so I must have been wrong.
I went over my work and found the mistakes.

[triplem]

It cannot be x=2 because otherwise AB=6 and BC=8, but we can see in the drawing that AB is larger than BC.
So the only real solution is x=4