Can you solve for X? | (Circles) |
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- Опубликовано: 29 сен 2024
- Learn how to solve for X. Important Geometry and Algebra skills are also explained: Pythagorean theorem; Circle formula; circle theorem. Step-by-step tutorial by PreMath.com.
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……another clear & succinct demonstration !
Thanks for sharing Professor !
Glad you liked it!
You are very welcome!
Thanks for the feedback ❤️
Area of the small circle=81πcm^2
πr^2=81π
So r=9cm
Area of the Large circle= 624π cm^2
πR^2+625π
So R=25π
Connect O to P and P to C (C on AB)
in ∆ OPC
OC^+CP^2=OP^2
OC=OA-BP=25-9=16cm
CP=x ; OP=25+9=34
(16)^2+x^2=(34)^2
So x=30cm.❤❤❤
Circle O:
A = πR²
625π = πR²
R² = 625
R = √625 = 25
Circle P:
A = πr²
81π = πr²
r² = 81
r = √81 = 9
Draw OA and PB. As AB is tangent to circle O at A and circle P at B, ∠OAB = ∠ABP = 90°. Draw PM, where M is the point on OA where PM is perpendicular to OA. As PM is parallel to AB, MA is parallel to PB, and ∠MAB = ∠ABP = 90°, MABP is a rectangle, MA = PB = 9, and PM = AB = x.
Let T be the point of tangency between circles O and P. Draw OP. As the point of tangency between two circles is collinear with the centers, OP passes through T. As OT = 25 and PT = 9, OP = 25+9 = 34. As OA = 25 and MA = 9, OM = 25-9 = 16.
Triangle ∆OMP:
OM² + PM² = OP²
16² + x² = 34²
x² = 1156 - 256 = 900
x = √900 = 30 cm
The answer is x = 30 cm. Another example of easier than it looks. I think that this should be an example of how useful the concept of collinearity is!!!
Pytagorean theorem:
x² = (R+r)² - (R-r)²
x = 30 cm ( Solved √ )
Solution:
Big Circle
Radius R = 25
Small Circle
Radius r = 9
Connecting point "O" to point "P" (collinear) we have hypotenuse 34 cm. The others legs are 16 (25 - 9) and "x"
Applying Pythagorean Theorem:
16² + x² = 34²
x² = 1.156 - 256
x² = 900
x = 30 cm ✅
The right triangle has three integer side lengths, the 8-15-17 primitive pythagorean number triple, stretched by the factor 2.
Nice video, thanks Professor, and a happy Sunday 😊 for you and the friends of the channel 😀
Thanks . Easy. But I love it. I extended from B to the right side equal to radius of small circle and point it as "M". Triangle PBM==>9^2+9^2=PM^2. PM =12.428 . Connect "O" to M triangle OAM==>OA^2+(X+9)^2=(12.427+9+25)^2==>X=30
My way of solution ▶
The area of the large circle, A₁
the area of the small circle, A₂
⇒
A₁= π*r₁²
81π= πr₁²
r₁= √81
r₁= 9 cm
A₂= π*r₂²
625π= πr₂²
r₂= √625
r₂= 25 cm
b) if we connect the centers of the circles we have the length [OP] :
[OP]= r₁+r₂
[OP]= 9 +25
[OP]= 34 m
c) Let's consider the trapezoid OABP
[PB]= [KA]
A ∈ [OA]
and AK ⊥ KP
⇒
[KA]= 9 cm
[OK]= 25 - 9
[OK]= 16 cm
[AB]= x
[KP] // [AB]
[KP]= x
d) By applying the Pythagorean theorem for the right triangle ΔOKP :
[OK]²+[KP]²= [PO]²
[OK]= 16 cm
[KP]= x
[PO]= 34 cm
⇒
16²+x²= 34²
x²= 34²-16²
x²= 900
x= 30 cm
STEP-BY-STEP RESOLUTION PROPOSAL :
01) OA = 25 cm
02) PB = 9 cm
03) Draw two Lines; one Horizontal passing through Point P, and another Vertical passing through Point O. Define the interception Point as Point C.
04) OC = (OA - PB) ; OC = (25 - 9) ; OC = 16 cm
05) OC = 16 cm
06) OP = (25 + 9) = 34 cm
07) X^2 = 34^2 - 16^2
08) X^2 = 1.156 - 256
09) X^2 = 900 ; X = sqrt(900)
10) X = 30 cm
Thus,
OUR ANSWER :
The Length of X equal 30 Centimeters.
I do prefer them a little trickier than this:
radii are 25 and 9
Make a point Q such that OPQ is a right triangle.
The two know sides are OP = 34 and OQ = 16.
The remaining side is the same length as x.
34^2 - 16^2 = 1156 - 256 = 900.
sqrt(900) = 30, so x = 30 (cm).
Let's find x:
.
..
...
....
.....
The radii PB and OA of the two circles can be calculated from their known areas:
(625π)cm² = π*OA²
625cm² = OA²
⇒ OA = √(625cm²) = 25cm
(81π)cm² = π*PB²
81cm² = PB²
⇒ PB = √(81cm²) = 9cm
Now let's add point R on OA such that ABPR is a rectangle. In this case OPR is a right triangle and we can apply the Pythagorean theorem. Since the two circles have exactly one point of intersection, we can conclude that the distance of their centers is equal to the sum of their radii. So we obtain:
OP² = OR² + PR²
(OA + PB)² = (OA − AR)² + PR²
(OA + PB)² = (OA − PB)² + PR²
OA² + 2*OA*PB + PB² = OA² − 2*OA*PB + PB² + PR²
4*OA*PB = PR²
⇒ x = AB = PR = √(4*OA*PB) = √[4*(25cm)*(9cm)] = √(900cm²) = 30cm
Best regards from Germany
Very well explained
Thanks Sir for your efforts
That’s very nice
Good luck with glades
❤❤❤❤
Sqx= sq(R+r)-sq(R-r)=(R+r+R-r)(R+r-R+r)
=4.R.r=4. 25 .9
--> x= 2. 5. 3= 30 units😅😅😅
We can use formula x = sqrt(d1×d2) therfore => x = sqrt(50×18) =30
x=30
x^2 = (R +r)^2 - (R -r)^2.
x^2 = R^2 + 2Rr + r^2 - R^2 + 2Rr - r^2.
x^2 = 4Rr.
x^2 = 4 x 25 x 9 = 900.
x = 30.
X=sqrt(4r1*r2)
X=sqrt(4*25*9).
=Sqrt(900)=30
X=30cm
I thought X was the length of the entire red line (64).
∆CPO = pyth.triple = 2(8 - 15 - 17) → x = 30
That’s one I could do in my head. 😂
x*x*π=625π x=25 y*y*π=81π y=9
25+9=34 25-9=16
x²+16²=34² x²=34²-16²=(34+16)(34-16)=50*18=900 x=30cm
Much simple.
30，book work.😮
Thank you!
Very good!
30 cm
30
Is this meant to be a problem?☹
x^2=OP^2-(AO-BP)^2, OP=AO+BP, x^2=34^2-16^2=900, x=\/900, х=30.
This is insultingly simple ... took me about 10 seconds, no pen, paper, or calcuator needed.
Thanks , you helped me to know things about Geometry today , keep doing videos like this videos, they will help many beginners like me in mathematics
too easy...
What do you expect? The group name is titled PREmath