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avatar+85 

Not so hard problems
1.

2.

3.

4.

 Mar 28, 2024
 #1
avatar+394 
+1

1)

It's all ratios, so you can set BC = 3.

Then the radius of the semicircle is 1.

Label the point where the Semicircle is tangent to AB as D.

Label the center of the semicircle as O

Triangle ADO and triangle ACB are similar, and sset DA as x.

Using the properties of similar triangles:

\(\frac{x}{1}=\frac{\sqrt{x^2+1}+1}{3}\).

\(3x=\sqrt{x^2+1}+1 \)

\(x=\frac{3}{4}\).

\(AC = \frac{5}{4}+1=\frac{9}{4}\).

\(\frac{AC}{BC}=\frac{\frac{9}{4}}{3}=\frac{3}{4}\).

 Mar 28, 2024
edited by hairyberry  Mar 28, 2024
 #2
avatar+394 
+1

3)

Draw a tangent to the chord of P, forming a rectangle, labeling that point D.

QD = the length of the external tangent.

PD = 4

PQ = 20

\(QD^2=20^2-4^2\)

\(QD=8\sqrt{6}\).

 Mar 28, 2024
 #3
avatar+394 
+1

4)

Because PT and PU are both tangents, PT = PU (this is easy to prove using HL congruency), and PTO is congruent to PUO.

Because PTO = 90, then POT = 180-90-33=57. These two triangles are congruent. 

So arc TU has a length 57*2=114.

 Mar 28, 2024
edited by hairyberry  Mar 28, 2024
 #4
avatar+394 
0

2)

Is this question complete? I think it might be non-unique.

 Mar 28, 2024
edited by hairyberry  Mar 28, 2024
 #6
avatar+394 
+2

Thanks booboo44, I realized it is possible smiley, thanks to your work.

But I'm getting a different answer, is there anything wrong with my work?

Set the radius of the outer circle as y.

Set the radius of the inner circle as x.

We want:

\(y^2\pi-x^2\pi=\pi(y^2-x^2)\).

 

By the pythagorean theorem, we know

\(y^2-x^2=5^2=25\).

Therefore the area is \(25\pi\)

hairyberry  Mar 29, 2024
 #8
avatar+128399 
+2

Exactly correct, hairyberry  .....

 

The shaded area always  = 25 pi   for  any R, r   such that

R^2  - r^2  =  5^2

 

See my response below

 

cool cool cool

CPhill  Mar 29, 2024
 #5
avatar+458 
0

For problem 2:

 

Let's denote the following:

 

O: Center of both circles

 

A and B: Endpoints of the chord (chord length AB = 10)

 

P: Point of tangency between the chord and the smaller circle

 

r: Radius of the smaller circle

 

R: Radius of the larger circle

 

Since the chord is tangent to the smaller circle at point P, we know that OP is perpendicular to AB. Additionally, since O is the center of both circles, line segment OP bisects chord AB (AP = BP = 5).

 

Finding the Radius of the Smaller Circle (r):

 

Using the Pythagorean Theorem in right triangle APO:

 

AP^2 + OP^2 = r^2 (since AP = BP = 5)

 

5^2 + OP^2 = r^2

 

We don't have the value of OP yet, but we can find it using the information about the larger circle and the chord.

 

Finding the Radius of the Larger Circle (R):

 

In right triangle OBP:

 

OB^2 + OP^2 = R^2 (since OB = half the chord length = 5)

 

5^2 + OP^2 = R^2

 

Relating the Radii (R and r):

 

Since the chord is tangent to the smaller circle, the distance between the center (O) and the point of tangency (P) is the radius of the smaller circle (r).

 

Additionally, this distance (OP) is also the difference between the radii of the larger and smaller circles (R - r).

 

Therefore, we have: OP = r and OP = R - r

 

Setting these two expressions equal to each other:

 

r = R - r

 

Solving for r (the smaller radius):

 

2r = R r = R/2

 

Substitute r in terms of R back into Equation 1:

 

5^2 + (R/2)^2 = R^2

 

Expand and rearrange:

 

R^2 - 5R - 25 = 0

 

Factor the equation:

 

(R - 10)(R + 2.5) = 0

 

Since the radius cannot be negative, we have R = 10.

 

Therefore, the radius of the smaller circle (r) is r = R/2 = 5.

 

Area of the Ring-Shaped Region:

 

The area of the ring is the difference between the area of the larger circle and the area of the smaller circle:

 

Area of Ring = π * R^2 - π * r^2

 

Substitute the values of R and r:

 

Area of Ring = π * (10)^2 - π * (5)^2

 

Area of Ring = π * (100 - 25)

 

Area of Ring = 75π

 

Therefore, the area of the ring-shaped region is 75π square units.

 Mar 29, 2024
 #7
avatar+128399 
+1

2.  This  problem  has no unique solution for R , r

But it does have a definite solution for the shaded area

 

R = radius of  large circle

r = radius of  small circle

R^2  - r^2   = (chord length / 2)^2

R^2 - r^2  =  5^2

 

One obvious  integer solution is  that R  = 13  and r =  12  since R^2 - r^2 =  5^2

Shaded area =  pi ( 13^2 - 12^2)  =  pi [ 169 -144 ]  =  25 pi

 

But

R^2  - 5^2  =  r^2

If we let  R = 20

And r  =  sqrt (375)

The shaded area =  pi (20^2 - (sqrt 375)^2 )  =  pi [ 400 - 375] =  25 pi

 

In general, any  R^2 - r^2   producing 5^2 will work

 

cool cool cool

 Mar 29, 2024
edited by CPhill  Mar 29, 2024
 #9
avatar+128399 
+1

3.  Let  QR   =   8 + x

Let M  be the point of tangency with the radius of the large circle  and N be the same  with the smaller circle

 

We have similar triangles    MPR    and NQR

 

PR / MP =   QR / QN

 

(28 + x)  / 12  =  (8 + x) /  8

 

8(28+x)  = 12 ( 8 + x)

 

224 + 8x  = 96 + 12x

 

224  - 96 =   4x

 

128  = 4x

 

128/4  = x  =  32

 

QR =  8 + 32  =  40

 

cool cool cool

 Mar 29, 2024
 #10
avatar+128399 
+1

1.

B              

       3        E

3            1

 

C    1   D                           A

 

For convenience, let BC = 3  and the radius of the semi-circle =  1

BC = BE

 

Let D be the center of the semi-circle   and let E  be the point where the semi-circle touches AB

 

Triangle EAD  similar to Triangle CAB

 

DE / EA = BC /  CA

 

1 / (BA - 3)  = 3 / CA

 

CA =  3 (BA - 3)

 

AC = 3BA - 9

 

BA^2  = 3^2 + (3BA - 9)^2

 

BA^2  = 9 + 9BA^2 - 54Ba + 81

 

BA^2  = 9BA^2 -54BA + 90

 

8BA^2  - 54BA + 90 =  0

 

Solving this produces .....BA =  15/4

 

AC = 3(15/4)  - 9  =  9/4

 

AC /  BC  =   (9/4) / 3  =  9 / 12  =   3 / 4

 

cool cool cool

 Mar 29, 2024

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