Introduction to arc length and sector area:
Consider a circle whose center is O and radius r. Let A and B be two points on the circle in such a way that arc AB = r.
The length of the arc AB of the circle is equal to the product of the radius of the circle and angle in radians subtended at the center of the circle.
That is, arc length AB = r × ?, here ? is in radians.
The area of the sector AOB is `(1/2) r^2theta` .
We can also write the above two formula as follows:
(i) arc length AB = 2?r × ` theta/360`
(ii) The area of the sector AOB is `(theta/360)` `Pir^2` .
Now let us see few problems on this topic “arc length and area sector”.
Example1 on Arc Length and Area Sector
An arc of a circle subtends 150 at the center. If the radius of the circle is 4cms, find the length of the arc and area of the sector formed.
Sol: Here ? = 15^0, r = 4cms.
Let us convert the degree value of ? to radians.
? = 15^0 = `(( Pi xx 15)/180)`
= `Pi/12` .
The length of the arc is given by r × ? = 4 × `pi/12`
= `pi/3` cm.
The area of the sector formed = `(1/2)`` r^2` ?
= `(1/2)` ` 4^2` `(pi/12)`
= `(2 pi )/3 ` sq cm.I like to share this solve linear system of equations with you all through my article.
Example2 on Arc Length and Area Sector
A wire of length 10cm is bent so as to form an arc of a circle of radius 4cm. What is the angle at the center in radians?
Sol: Given: The length of the arc is given by r × ? = 10
? = `10/r`
? = `10/4` = 2.5 radians.
Example3 on Arc Length and Area Sector
Ex 3: If the distance between the sun and the earth is 1.495 × 1068 km and the angle subtended by the sun at a point on the earth is half a degree, find approximately the diameter of the sun.
Sol: Given: ? = (1/2)^0 = (`Pi/180` ) × `(1/2)`
= `Pi/360` rad.
If d km be the diameter of the sun, then we have: `(s/r)` = ?
` (d/ (1.495 xx 1068))` = `Pi/360`
Therefore, d = (1.495 × 1068) × `Pi /360`
= 1.304637 × 10^6 km.
Consider a circle whose center is O and radius r. Let A and B be two points on the circle in such a way that arc AB = r.
The length of the arc AB of the circle is equal to the product of the radius of the circle and angle in radians subtended at the center of the circle.
That is, arc length AB = r × ?, here ? is in radians.
The area of the sector AOB is `(1/2) r^2theta` .
We can also write the above two formula as follows:
(i) arc length AB = 2?r × ` theta/360`
(ii) The area of the sector AOB is `(theta/360)` `Pir^2` .
Now let us see few problems on this topic “arc length and area sector”.
Example1 on Arc Length and Area Sector
An arc of a circle subtends 150 at the center. If the radius of the circle is 4cms, find the length of the arc and area of the sector formed.
Sol: Here ? = 15^0, r = 4cms.
Let us convert the degree value of ? to radians.
? = 15^0 = `(( Pi xx 15)/180)`
= `Pi/12` .
The length of the arc is given by r × ? = 4 × `pi/12`
= `pi/3` cm.
The area of the sector formed = `(1/2)`` r^2` ?
= `(1/2)` ` 4^2` `(pi/12)`
= `(2 pi )/3 ` sq cm.I like to share this solve linear system of equations with you all through my article.
Example2 on Arc Length and Area Sector
A wire of length 10cm is bent so as to form an arc of a circle of radius 4cm. What is the angle at the center in radians?
Sol: Given: The length of the arc is given by r × ? = 10
? = `10/r`
? = `10/4` = 2.5 radians.
Example3 on Arc Length and Area Sector
Ex 3: If the distance between the sun and the earth is 1.495 × 1068 km and the angle subtended by the sun at a point on the earth is half a degree, find approximately the diameter of the sun.
Sol: Given: ? = (1/2)^0 = (`Pi/180` ) × `(1/2)`
= `Pi/360` rad.
If d km be the diameter of the sun, then we have: `(s/r)` = ?
` (d/ (1.495 xx 1068))` = `Pi/360`
Therefore, d = (1.495 × 1068) × `Pi /360`
= 1.304637 × 10^6 km.
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