Solution will be displayed step by step

Solution

Solution provided by AtoZmath.com

Want to know about AtoZmath.com and me

This is demo example. Please click on Find button and solution will be displayed in Solution tab (step by step)

Area of Square

Diagonal `(d) = sqrt(2) a` Perimeter `(P) = 4a` Area `(A) = a^2 = d^2/2` Example : I know that for a square Side(a) = 10 . From this find out Area of the square. `"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 : 1 / 13 [ Circle ]       Try problem - 1 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 ]       Try problem - 2 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 ]       Try problem - 3 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 ]       Try problem - 4 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 ]       Try problem - 5 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 ]       Try problem - 6 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 ]       Try problem - 7 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 ]       Try problem - 8 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 ]       Try problem - 9 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 ]       Try problem - 10 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 ]       Try problem - 11 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 ]       Try problem - 12 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 ]       Try problem - 13 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`