My Math Forum Height unknown but need area.

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 March 22nd, 2014, 09:56 PM #1 Senior Member   Joined: Nov 2013 Posts: 247 Thanks: 2 Height unknown but need area. I have a triangle with lengths 3.2, 3.8, and 2.4. I need the area but have to figure out the height because the triangle is not a right triangle. I would have 2 sides unknown if I just try to do the Pythagorean theorem. How can I figure out the height without doing the Pythagorean theorem if at all possible? Now since this is scalene would I get 3 different areas if I use the 3 different altitudes and bases?
 March 22nd, 2014, 10:06 PM #2 Senior Member   Joined: Mar 2014 Posts: 112 Thanks: 8 Heron's formula
 March 22nd, 2014, 10:09 PM #3 Senior Member   Joined: Mar 2014 Posts: 112 Thanks: 8 Also, you could use the law of cosines and determine the angle to determine the height. My above post generally gives the area of a triangle.
March 23rd, 2014, 12:01 PM   #4
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Re: Height unknown but need area.

Quote:
 Originally Posted by caters Now since this is scalene would I get 3 different areas...
It's the same triangle, isn't it, so why would the area change? The height and the base are inversely proportional, so if the height increases, the base decreases. The easiest way is to use Heron's formula, as the john has said

March 23rd, 2014, 02:27 PM   #5
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Re: Height unknown but need area.

Quote:
Originally Posted by eddybob123
Quote:
 Originally Posted by caters Now since this is scalene would I get 3 different areas...
It's the same triangle, isn't it, so why would the area change? The height and the base are inversely proportional, so if the height increases, the base decreases. The easiest way is to use Heron's formula, as the john has said
Yeah it is but you could use the law of sines, law of cosines, or law of tangents to figure out the angle and than figure out the height from sides + angles and than use 1/2 b * h to figure out the area.

Now if the Heron's formula works for triangles will it work for other polygons like quadrilaterals for example if you have 3,2,4,5 as sides of a quadrilateral (don't know if you would actually get a polygon but we will do with it) you have this:
$P= 3+2+4+5 = 14$
$S= 7$
$A= \sqrt{(7*4)(7*5)(7*3)(7*2)}$
$\sqrt{28 * 35 * 21 * 14}$
$\sqrt{288120}$
$98 * \sqrt{30}$

Would this be the actual area of a quadrilateral with sides 3, 2, 4, and 5?

 March 23rd, 2014, 03:18 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,978 Thanks: 2229 In general, the area of a quadrilateral can't be determined from just the lengths of its sides (as quadrilaterals with the given sides but different areas would exist).
March 23rd, 2014, 03:45 PM   #7
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Quote:
 Originally Posted by skipjack In general, the area of a quadrilateral can't be determined from just the lengths of its sides (as quadrilaterals with the given sides but different areas would exist).
Here is a list of the types of quadrilaterals:
Concave
Convex
Tangential (externally tangent to a circle)
Cyclic (internally tangent to a circle)
Trapezoid
Kite
Rhomboid
Rhombus
Rectangle
Oblong (used to mean rectangle that is not square)
Square
Parallelogram
Right Trapezoid (1 right angle)
Isosceles Trapezoid (at least 2 sides equal)
3 sides equal trapezoid (special case of Isosceles Trapezoid)
Bicentric (internally and externally tangent to a circle)

How can you figure out which of these types of quadrilaterals are possible with given side lengths? How would you find the area of a concave quadrilateral?

Here is the hierarchy:
Simple
Inside that is convex and concave
Inside convex is Tangential, Trapezoid, and Cyclic
Inside Tangential is Kite and Bicentric
Inside Trapezoid is Parallelogram, Right Trapezoid, and Isosceles Trapezoid
Inside Cyclic is Isosceles Trapezoid and Bicentric
Inside Kite is Rhombus
Inside Parallelogram is Rectangle and Rhombus
Inside Right Trapezoid is Rectangle
Inside Isosceles Trapezoid is Rectangle and 3 Sides Equal Trapezoid
Inside Rhombus, 3 Sides Equal Trapezoid, Bicentric, and Rectangle is Square

 March 23rd, 2014, 07:45 PM #8 Senior Member   Joined: Sep 2012 From: British Columbia, Canada Posts: 764 Thanks: 53 Re: Height unknown but need area. You can use Brahmagupta's Formula for cyclic quadrilaterals, which can be generalized for any non-cyclic quadrilateral.
 March 23rd, 2014, 11:18 PM #9 Newbie   Joined: Mar 2014 Posts: 10 Thanks: 0 Re: Height unknown but need area. Use Heron's formula for any triangle. ~Cheezees Fiverr.com/Cheezees
 March 24th, 2014, 09:00 AM #10 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: Height unknown but need area. Fiverr.com/Cheezees http://www.fiverr.com/cheezees/do-up-to ... -questions And you only charge 5 bucks?

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### I am off parallelogram, rhombus, a rectangle and a trapezium who am I

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