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Numerical Methods

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College Algebra

Area -

Area - Rectangle

Formula :
Diagonal `(d) = sqrt(l^2 + b^2)`
Perimeter `(P) = 2(l+b)`
Area `(A) = l b`

5. I know that for a rectangle = and = . From this find out of the rectangle.

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`

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