Area of a Isoceles Triangle calculator

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Area of a Isoceles Triangle

Height $(h) = sqrt(a^2 - b^2/4)$
Perimeter $(P) = 2 a + b$
Area $(A) = (b h)/2$

Example :
I know that for a isoceles triangles a = 5 and b = 6 . From this find out Area of the isoceles triangles.

$"Here base " (b) = 6" and equal side " (a) = 5" (Given)"$

$"We know that,"$

$"Perimeter" = (2 * "equal Side") + "third Side"$

$= (2 * a) + b$

$= (2 * 5) + 6$

$= 16$

$"We know that,"$

$"Area" = 1/2 * "base" * "height"$

$= 1/2 * b * sqrt(a^2 - b^2/4)$

$= 1/2 * 6 * sqrt(5^2 - 6^2/4)$

$= 1/2 * 6 * 4$

$= 12$

Area

1. Circle

Area

$(A) = pi r^2$
Circumference

$(C) = 2 pi r = pi d$
Diameter

$(d) = 2 r$

2. Semicircle

Area

$(A) = 1/2 pi r^2$
Circumference

$(C) = pi r = (pi d)/2$
Perimeter

$(P) = pi r + 2 r$
Diameter

$(d) = 2 r$

3. Regular Hexagon

Perimeter

$(P) = 6 a$
Area

$(A) = sqrt(3)/4 xx 6 xx a^2$

4. Square

Diagonal

$(d) = sqrt(2) a$
Perimeter

$(P) = 4a$
Area

$(A) = a^2 = d^2/2$

5. Rectangle

Diagonal

$(d) = sqrt(l^2 + b^2)$
Perimeter

$(P) = 2(l+b)$
Area

$(A) = l b$

6. Parallelogram

Area

$(A) = ah$
Perimeter

$(P) = 2a + 2b$

7. Rhombus

Radius

$(r_1) = (d_1)/2$
Radius

$(r_2) = (d_2)/2$
Side

$(a) = sqrt(r_1^2 + r_2^2)$
Perimeter

$(P) = 4 a$
Area

$(SA) = (d_1 d_2)/2$

8. Trapezium

Area

$(A) = h/2 (a + b)$
Perimeter

$(P) = a + b + c + d$

Area >> Triangle

9. Scalene Triangle

Perimeter

$(P) = a+b+c$

$S = P/2 = (a+b+c)/2$
Area

$(A) = sqrt(S (S - a) (S - b) (S - c))$

10. Right angle Triangle

Diagonal

$(d) = sqrt(a^2 + b^2)$
Perimeter

$(P) = a+b+c$
Area

$(A) = 1/2(a b)$

11. Equilateral Triangle

Perimeter

$(P) = 3 a$
Area

$(A) = sqrt(3)/4 a^2$

12. Isosceles Triangle

Height

$(h) = sqrt(a^2 - b^2/4)$
Perimeter

$(P) = 2 a + b$
Area

$(A) = (b h)/2$

13. Sector Segment

Length of the arc

$= l = (pi r theta)/180$
Area of a minor sector

$= (pi r^2 theta)/360$

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 [ Semicircle ] Enter your problem 2. I know that for a Semicircle Radius = 10 . From this find out Area of the Semicircle.


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