Area of a Sector Segment calculator

We use cookies to improve your experience on our site and to show you relevant advertising. By browsing this website, you agree to our use of cookies. Learn more

AtoZmath.com

College Algebra

Matrix & Vector

Numerical Methods

Statistical Methods

Operation Research

AREA -

Area of a Sector Segment

Formula :
Length of the arc `= l = ( pi r theta)/180`
Area of a minor sector `= ( pi r^2 theta)/360`

13. I know that for a sector & segment = and = . From this find out of the sector & segment.

`pi = `

Decimal Place =

SolutionExampleArea FormulaAll Problems

Problem : 1 / 13 [ Circle ] Enter your problem 1. I know that for a circle Radius = 10 . From this find out Area of the circle.


Solution:	`"Here radius " (r)=10 "(Given)"` `` `"Diameter " (d)= 2 * r` ` = 2 * 10` ` = 20` `"Perimeter" = 2 pi r` ` = 2 * 22/7 * 10` ` = 440/7` `Area = pi r^2` ` = 22/7 * (10)^2` ` = 2200/7`

Problem : 2 / 13 [ Semi-Circle ] Enter your problem 2. I know that for a Semi-Circle Radius = 10 . From this find out Area of the Semi-Circle.


Solution:	`"Here radius "(r)=10" (Given)"` `` `"Diameter "(d)= 2 r` ` = 2 * 10` ` = 20` `` `"Circumference" = pi r` ` = 22/7 * 10` ` = 220/7` `"Perimeter" = pi r + 2 r` ` = 22/7 * 10 + 2 * 10` ` = 220/7 + 20` ` = 360/7` `"Area" = (pi r^2)/2` ` = (22/7 * (10)^2)/2` ` = 1100/7`

Problem : 3 / 13 [ RegularHexagon ] Enter your problem 3. I know that for a Regular Hexagon Side = 10 . From this find out Area of the Regular Hexagon.


Solution:	`"Here Side "(a)=10" (Given)"` `"Perimeter" = 6 a` ` = 6 * 10` ` = 60` `"Area" = sqrt(3)/4 * 6 * a^2` ` = sqrt(3)/4 * 6 * 10^2` ` = 259.8076`

Problem : 4 / 13 [ Square ] Enter your problem 4. I know that for a square Side(a) = 10 . From this find out Area of the square.


Solution:	`"Here a" = 10" (Given)"` `"Diagonal" = sqrt(2) * "a"` ` = sqrt(2) * 10` ` = 14.1421` `"Perimeter" = 4 * "a"` ` = 4 * 10` ` = 40` `"Area" = "a"^2` ` = 10^2` ` = 100`

Problem : 5 / 13 [ Rectangle ] Enter your problem 5. I know that for a rectangle Length = 5 and Breadth = 12 . From this find out Area of the rectangle.


Solution:	`"Here one-Side " (l) = 5" and other-Side " (b) = 12" (Given)"` `"Diagonal" = sqrt(l^2 + b^2)` ` = sqrt(5^2 + 12^2)` ` = sqrt(25 + 144)` ` = sqrt(169)` ` = 13` `"Perimeter" = 2 * "(sum of Sides)"` ` = 2 * (5 + 12)` ` = 34` `"Area = Product of Sides"` ` = 5 * 12` ` = 60`

Problem : 6 / 13 [ Parallelogram ] Enter your problem 6. I know that for a parallelogram a = 9 , b = 22 and h = 14 . From this find out Area of the parallelogram.


Solution:	`"Here " a=9, b=22, h=14" (Given)"` `"Perimeter" = 2 * (a + b)` ` = 2 * (9 + 22) ` ` = 62` `"Area" = a * h` ` = 9 * 14` ` = 126`

Problem : 7 / 13 [ Rhombus ] Enter your problem 7. I know that for a rhombus d1 = 10 and d2 = 24 . From this find out Area of the rhombus.


Solution:	`"Here, we have " d_1 = 10" and " d_2 = 24" (Given)"` `a^2 = (d_1/2)^2 + (d_2/2)^2` `a^2 = (10/2)^2 + (24/2)^2` `a^2 = (5)^2 + (12)^2` `a^2 = 169` `a = 13` `"Perimeter" = 4 * a` ` = 4 * 13` ` = 52` `"Area" = 1/2 " (Product of diagonals)"` ` = 1/2 * d_1 * d_2` ` = 1/2 * 10 * 24` ` = 120`

Problem : 8 / 13 [ Trapezium ] Enter your problem 8. I know that for a trapezium a = 22 , b = 18 , c = 16 , d = 16 and h = 4 . From this find out Area of the trapezium.


Solution:	`"Here " a=22, b=18, c=16, d=16, h=4" (Given)"` `"Perimeter" = a + b + c + d` ` = 22 + 18 + 16 + 16 ` ` = 72` `"Area" = (a + b) * h/2` ` = (22 + 18) * 4/2` ` = 80`

Problem : 9 / 13 [ Scalene Triangle ] Enter your problem 9. I know that for a scalene triangles a = 3 , b = 4 and c = 5 . From this find out Area of the scalene triangles.


Solution:	`"Here " a=3, b=4, c=5" (Given)"` `"We know that,"` `"Perimeter" = a + b + c` ` = 3 + 4 + 5` ` = 12` ` ` `"Semi-Perimeter" = s = (a + b + c)/2` ` = 12/2` ` = 6` `"Here " a=3, b=4, c=5" and semi-Perimeter" = 6` `"We know that,"` `"Area" = sqrt(s (s - a) (s - b) (s - c))` ` = sqrt(6 (6 - 3) (6 - 4) (6 - 5))` ` = 6`

Problem : 10 / 13 [ Rightangle Triangle ] Enter your problem 10. I know that for a rightangle triangles AB = 5 and BC = 12 . From this find out Area of the rightangle triangles.


Solution:	`"Here one Side" = 5" and other Side" = 12" (Given)"` `"We know that,"` `"In triangle ABC, by Pythagoras' theorem"` `AC^2 = AB^2 + BC^2` `AC^2 = 5^2 + 12^2` `AC^2 = 25 + 144` `AC^2 = 169` `AC = 13` `"Perimeter" = AB + BC + AC` ` = 5 + 12 + 13` ` = 30` ` ` `"Here base" = 5" and height = "12 ` `"We know that,"` `"Area" = 1/2 * AB * BC` ` = 1/2 * 5 * 12` ` = 30`

Problem : 11 / 13 [ Equilateral Triangle ] Enter your problem 11. I know that for a equilateral triangles Side = 6 . From this find out Area of the equilateral triangles.


Solution:	`"Here " a = 6" (Given)"` `"We know that,"` `"Perimeter" = 3 * a` ` = 3 * 6` ` = 18` `"We know that,"` `"Area" = sqrt(3)/4 * a^2` ` = 1.732/4 * 6 * 6` ` = 15.5885`

Problem : 12 / 13 [ Isoceles Triangle ] Enter your problem 12. I know that for a isoceles triangles a = 5 and b = 6 . From this find out Area of the isoceles triangles.


Solution:	`"Here base " (b) = 6" and equal side " (a) = 5" (Given)"` `"We know that,"` `"Perimeter" = (2 * "equal Side") + "third Side"` ` = (2 * a) + b` ` = (2 * 5) + 6` ` = 16` `"We know that,"` `"Area" = 1/2 * "base" * "height"` ` = 1/2 * b * sqrt(a^2 - b^2/4)` ` = 1/2 * 6 * sqrt(5^2 - 6^2/4)` ` = 1/2 * 6 * 4 ` ` = 12`

Problem : 13 / 13 [ SectorSegment ] Enter your problem 13. I know that for a sector & segment Radius = 14 and angle of measure = 90 . From this find out length of arc of the sector & segment.


Solution:	`"Here "r = 14" and " theta = 90" (Given)"` `"Length of the arc " = l = (pi r theta)/180` ` = (3.1416 * 14 * 90)/180` ` = 21.98` `"Area of a minor sector "= (pi r^2 theta)/360` ` = (3.1416 * 14^2 * 90)/360` ` = 153.86`

Share this solution or page with your friends.

College Algebra

Copyright © 2023. All rights reserved. Terms, Privacy