A quarter-circle with radius $5$ is drawn. A circle is drawn inside the sector, which is tangent to the sides of the sector, as shown. Find the radius of the inscribed circle.
The rest is in asymptote
size( 100 ) ;
// Not to scale. Don't try anything here, kids
pair O = origin ;
draw( circle((-1,1),1) ) ;
pair I = intersectionpoint( arc((-1,1),1,90,180) , O--4dir(135) ) ;
real radius = length( I-O ) ;
pair A = radius*dir(180) ;
pair B = radius*dir(90) ;
draw( O--arc(O,radius,90,180)--cycle ) ;
label( "$O$" , O , SE ) ;
label( "$A$" , A , SW ) ;
label( "$B$" , B , NE ) ;
If the small circle is inscribed in the sector, I assume it is tangent to the larger circle, as well as the radii of the sector. In that case, let
O = center of large circle
P = center of inscribed circle
r = radius of inscribed circle
OP = 5-r
r^2+r^2 = (5-r)^2
r = 5(√2 - 1)
Why did the circle go to school?
Because it wanted to be "well-rounded"!
Now, let's focus on solving this problem. We need to find the radius of the inscribed circle.
To do this, we can use the fact that the line connecting the center of the quarter-circle to the center of the inscribed circle is also a radius of the quarter-circle.
Using the diagram, we can see that the line connecting point O (the center of the quarter-circle) and point A (the center of the inscribed circle) is part of the quarter-circle's radius.
Therefore, the length OA is equal to the radius of the quarter-circle, which is 5.
So, the radius of the inscribed circle is also 5.
Did you hear about the circus fire? It was in tents!
To find the radius of the inscribed circle, we can use the property that the radius of the inscribed circle is perpendicular to the tangent at the point of tangency.
Let's call the center of the quarter-circle O, and the point of tangency of the inscribed circle with the quarter-circle AB. We want to find the length of AB, which is the radius of the inscribed circle.
Since AB is tangent to the quarter-circle, the line segment OA is perpendicular to AB. This means that triangle OAB is a right triangle.
Let's use the Pythagorean theorem to find the length of AB.
In triangle OAB, the length of OA is 5 (the radius of the quarter-circle) and the length of OB is the radius of the inscribed circle, which we want to find.
Using the Pythagorean theorem, we have:
OA^2 = OB^2 + AB^2
Substituting the values we know:
5^2 = OB^2 + AB^2
Simplifying:
25 = OB^2 + AB^2
But we know that AB = 2 * OB, because AB is the diameter of the circle and OB is the radius.
Substituting this into the equation:
25 = OB^2 + (2 * OB)^2
25 = OB^2 + 4OB^2
25 = 5OB^2
Dividing both sides by 5:
5 = OB^2
Taking the square root of both sides:
√5 = OB
Therefore, the radius of the inscribed circle is √5.
To find the radius of the inscribed circle, we need to use the properties of circles and sectors.
First, let's observe the given diagram. We have a quarter-circle with radius 5. We also have a circle inscribed inside the sector formed by the quarter-circle. The inscribed circle is tangent to the sides of the sector at points A and B.
To find the radius of the inscribed circle, let's focus on the right triangle formed by the centers of the quarter-circle (O), the inscribed circle (I), and point B.
Since the inscribed circle is tangent to the side of the sector, we know that IB is perpendicular to OB. This means we have a right triangle OIB.
The hypotenuse OB is the radius of the quarter-circle, which is given as 5.
To find the radius of the inscribed circle, we need to find the length of IB.
We can apply the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides.
In this case, we have:
OB^2 = OI^2 + IB^2
Substituting the known values:
5^2 = OI^2 + IB^2
25 = OI^2 + IB^2
Now, we know that OI is the radius of the inscribed circle. Let's denote it as r.
The length of OB is the sum of the length of OI and the length of IB:
OB = OI + IB
Substituting the known values:
5 = r + IB
We can rearrange this equation to solve for IB:
IB = 5 - r
Now, we can substitute this value back into the Pythagorean theorem equation:
25 = r^2 + (5 - r)^2
Expanding and simplifying:
25 = r^2 + 25 - 10r + r^2
0 = 2r^2 - 10r
Dividing both sides by 2:
0 = r^2 - 5r
Now, we have a quadratic equation. We can solve it by factoring or using the quadratic formula.
Factoring this equation gives us:
0 = r(r - 5)
This equation has two possible solutions: r = 0 or r = 5.
Since the radius of a circle cannot be zero (as it would just be a point), we can conclude that the radius of the inscribed circle is 5.
Therefore, the radius of the inscribed circle is 5 units.