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Area of Equilateral Triangle

Perimeter `(P) = 3 a` Area `(A) = sqrt(3)/4 a^2` Example : I know that for a equilateral triangles Side = 6 . From this find out Area of the equilateral triangles.

`"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 : 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.

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.