If the area of a triangle is 20cm squared how do you work out the lengths of the sides?
its been over 20 years since i had to do this and im trying to help my son work this out ,no im not giving him tyhe answers to his home work ,i need the formula of how to do this
Answers:
The Area of a triangle is half the area of a square. So take one verticy and take a 90degree line from that up to the other corner.
Oh cr@p I'm starting to forget it all now. Too long ago.
It's something like that.
The area of a triangle is half as much as the area of a rectangle. So if the area of the triangle is 20cm squared, the area of the rectangle is 40 cm squared. Its sides could be 10cm x 4cm or 8cm x 5cm etc. If the triangle is a right angled one, that could mean that its sides are also 10cm, 4cm and then you just need to work out the 3rd side. There is a way to do this using trigonometry, or you could just draw it to scale as 10cm x 4cm and measure the 3rd one!
Try going to http://www.bbc.co.uk/schools/gcsebitesiz.
It might be able to answer you more clearly!
Hope this helps .
There are many formulas and if youve got the area ,you need to go and find the height using pythagoras theorem a nd then find the rest .. Are you sure the area is 20cm??
use a calculator
Ok, u cant work that out unless assume that the triangle is equallateral. SO assuming it is
Area a of triangle = 1/2*length of base*perpendicular height
Let length of base = x
Using pythagoras to find the height
x^2 = (x/2)^2 + (ph)^2
(x^2 - (x/2)^2)^.5 = ph
So,
20 = 1/2*x*(x^2 - (x/2)^2)^.5
400 = 1/4*x^2*(x^2 - x/2)
400 = (x^4)/4 - x^2/8
3200 = 2x^4 - x^2
you can solve this using a quadratic and letting y = x^2
Or, using herons formula Where a,b and c are the lengths of the sides and s= (a+b+c)/2
If we again assume its equalateral
A = (s(s-a)(s-b)(s-c))^.5
20 = ((3x/2)(x/2)(x/2)(x/2))^.5
400=(3x^4)/16
6400/3 = x^4
x= (6400/3)^.25
x=6.7962
You need a bit of extra data to define the shape of the triangle such as the length of one side and an angle
There are an infinite number of triangles with an area of 20 cm^2, so I assume you mean an equilateral triangle with all sides equal. In that case, there is only one with that area.
Let x = the side lengths.
Now draw a line from one vertex to the midpoint of the opposite side and you will create 2 equal right-angled triangles.
Looking at one of those triangles, the hypotenuse = x, and one side = x / 2. Now use Pythagoras to find the height, h.
h^2 = x^2 - (x / 2)^2
= x^2 - x^2 / 4
= 3x^2 / 4
Taking the square root of both sides gives :
h = x * sqrt(3) / 2
Getting back to your equilateral triangle,
the area = (1/2) * base * height
The base will be one whole side equal to x.
So, area = (1/2) * (x) * [x * sqrt(3) / 2]
= x^2 * sqrt(3) / 4
Now equate this general area with your specific area.
Thus, x^2 * sqrt(3) / 4 = 20 cm^2
or, x^2 = 80 / sqrt(3)
Therefore, x = sqrt [ 80 / sqrt(3) ] or approx. 6.796 cm.
you need to know more about the triangle
there are a infinite amount of triangles
possible around an area of 400 cm^2
assume a triangle with all sides equal
let length of base =L=length of each side
area = 1/2 vertical height*base
= 1/2 sqrt(3)L*L
L^2= 2 area/sqrt(3)
=2*20^2/sqrt3
L^2 = (800/sqrt3)
L= 21.49139864 cm
therefore, length of each side of triangle
is 21.49139864cm
i hope that this helps
20cm squared = 1/2 x a x t
2 x 20cm sq = a x t
40cm sq = a x t
(1) a = 1 ; t = 40
(2) a = 2 ; t = 20
(3) a = 3 ; t = 40/3 = 13 1/3
(4) a = 4 ; t = 10
..
(a.1) side 1 = side 2 = side 3
angle 1 = angle 2 = angle 3 = 60
20cm sq = 1/2 x s x s
40cm sq = s x s
s x s = 40cm sq
s = root of 40
s = 6.32455532
(It's completly wrong!)
40cm sq = s x t
40cm sq = s x (root of (s x s - 1/2 s x s))
6400cm sq = s x s x (s x s - 1/2 s x s)
The information you got is insufficient to reach the solution. U should at least have the property of the triangle - equilateral, rectangular etc., else its not just possible.
the area = 1/2 x hieght x base
20 = 1/2 x h x b
b = 40 / h . so if you have the length of any hieght you can get the side which the hieght on.
Area of a triangle is half the base x the perpendicular height.
The question told you more than you told us. For example is the triangle right angled?
If so, the sides each side of the triangle provide the base and perpendicular height for this formula above.
If the triangle is not right angled, have you any extra information to indicate the perpendicular height of the triangle?
The formula is
Area of triangle = 1/2 base x perpendicular height
Good luck with doing homework second time around - perhaps that's when we really learn it!
You don't, unless you have more information.
in an triangle (if you mean equilateral), the perpendicular bisector is the height and half the side is the base. The bisector cuts the triangle into two equal 30 by 60 by 90 triangles. In a 30 by 60 by 90 triangle, the height is square root of 3 * the base.Therefore
A=1/2 (bh)
20cm =1/2 (bh)
bh = 40cm^2
b(square root of 2 * b) = 40cm
b^2*square root of 3 = 40cm
b^2 = 40cm^2/square root of 3
b^2 = 23.1cm
b= 4.8cm
length of one side is 2b
length of the side is 9.6 cm
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Answers:
The Area of a triangle is half the area of a square. So take one verticy and take a 90degree line from that up to the other corner.
Oh cr@p I'm starting to forget it all now. Too long ago.
It's something like that.
The area of a triangle is half as much as the area of a rectangle. So if the area of the triangle is 20cm squared, the area of the rectangle is 40 cm squared. Its sides could be 10cm x 4cm or 8cm x 5cm etc. If the triangle is a right angled one, that could mean that its sides are also 10cm, 4cm and then you just need to work out the 3rd side. There is a way to do this using trigonometry, or you could just draw it to scale as 10cm x 4cm and measure the 3rd one!
Try going to http://www.bbc.co.uk/schools/gcsebitesiz.
It might be able to answer you more clearly!
Hope this helps .
There are many formulas and if youve got the area ,you need to go and find the height using pythagoras theorem a nd then find the rest .. Are you sure the area is 20cm??
use a calculator
Ok, u cant work that out unless assume that the triangle is equallateral. SO assuming it is
Area a of triangle = 1/2*length of base*perpendicular height
Let length of base = x
Using pythagoras to find the height
x^2 = (x/2)^2 + (ph)^2
(x^2 - (x/2)^2)^.5 = ph
So,
20 = 1/2*x*(x^2 - (x/2)^2)^.5
400 = 1/4*x^2*(x^2 - x/2)
400 = (x^4)/4 - x^2/8
3200 = 2x^4 - x^2
you can solve this using a quadratic and letting y = x^2
Or, using herons formula Where a,b and c are the lengths of the sides and s= (a+b+c)/2
If we again assume its equalateral
A = (s(s-a)(s-b)(s-c))^.5
20 = ((3x/2)(x/2)(x/2)(x/2))^.5
400=(3x^4)/16
6400/3 = x^4
x= (6400/3)^.25
x=6.7962
You need a bit of extra data to define the shape of the triangle such as the length of one side and an angle
There are an infinite number of triangles with an area of 20 cm^2, so I assume you mean an equilateral triangle with all sides equal. In that case, there is only one with that area.
Let x = the side lengths.
Now draw a line from one vertex to the midpoint of the opposite side and you will create 2 equal right-angled triangles.
Looking at one of those triangles, the hypotenuse = x, and one side = x / 2. Now use Pythagoras to find the height, h.
h^2 = x^2 - (x / 2)^2
= x^2 - x^2 / 4
= 3x^2 / 4
Taking the square root of both sides gives :
h = x * sqrt(3) / 2
Getting back to your equilateral triangle,
the area = (1/2) * base * height
The base will be one whole side equal to x.
So, area = (1/2) * (x) * [x * sqrt(3) / 2]
= x^2 * sqrt(3) / 4
Now equate this general area with your specific area.
Thus, x^2 * sqrt(3) / 4 = 20 cm^2
or, x^2 = 80 / sqrt(3)
Therefore, x = sqrt [ 80 / sqrt(3) ] or approx. 6.796 cm.
you need to know more about the triangle
there are a infinite amount of triangles
possible around an area of 400 cm^2
assume a triangle with all sides equal
let length of base =L=length of each side
area = 1/2 vertical height*base
= 1/2 sqrt(3)L*L
L^2= 2 area/sqrt(3)
=2*20^2/sqrt3
L^2 = (800/sqrt3)
L= 21.49139864 cm
therefore, length of each side of triangle
is 21.49139864cm
i hope that this helps
20cm squared = 1/2 x a x t
2 x 20cm sq = a x t
40cm sq = a x t
(1) a = 1 ; t = 40
(2) a = 2 ; t = 20
(3) a = 3 ; t = 40/3 = 13 1/3
(4) a = 4 ; t = 10
..
(a.1) side 1 = side 2 = side 3
angle 1 = angle 2 = angle 3 = 60
20cm sq = 1/2 x s x s
40cm sq = s x s
s x s = 40cm sq
s = root of 40
s = 6.32455532
(It's completly wrong!)
40cm sq = s x t
40cm sq = s x (root of (s x s - 1/2 s x s))
6400cm sq = s x s x (s x s - 1/2 s x s)
The information you got is insufficient to reach the solution. U should at least have the property of the triangle - equilateral, rectangular etc., else its not just possible.
the area = 1/2 x hieght x base
20 = 1/2 x h x b
b = 40 / h . so if you have the length of any hieght you can get the side which the hieght on.
Area of a triangle is half the base x the perpendicular height.
The question told you more than you told us. For example is the triangle right angled?
If so, the sides each side of the triangle provide the base and perpendicular height for this formula above.
If the triangle is not right angled, have you any extra information to indicate the perpendicular height of the triangle?
The formula is
Area of triangle = 1/2 base x perpendicular height
Good luck with doing homework second time around - perhaps that's when we really learn it!
You don't, unless you have more information.
in an triangle (if you mean equilateral), the perpendicular bisector is the height and half the side is the base. The bisector cuts the triangle into two equal 30 by 60 by 90 triangles. In a 30 by 60 by 90 triangle, the height is square root of 3 * the base.Therefore
A=1/2 (bh)
20cm =1/2 (bh)
bh = 40cm^2
b(square root of 2 * b) = 40cm
b^2*square root of 3 = 40cm
b^2 = 40cm^2/square root of 3
b^2 = 23.1cm
b= 4.8cm
length of one side is 2b
length of the side is 9.6 cm
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