# Area of triangle.

• Nov 22nd 2010, 12:03 AM
LaMandie
Area of triangle.
The question is as follows:
The three sides of a triangle are 10m, 15m, and 20m. What is the area of the triangle?

My diagram:
Attachment 19805
(20 - x)^2 - x^2 = 15^2 - 10^2

(20 -x)(20 - x) - x^2 = 125

400 - 40x -2x^2 = 125

-40x - 2x^2 = -275

x - 2x^2 = 6.875

-2x^2 = 6.875x

x^2 = -3.4375x

x = -3.4375

Obviously it's not the right answer, but I have no idea how to get the right one. Answers are greatly appreciated!
• Nov 22nd 2010, 12:08 AM
mr fantastic
Quote:

Originally Posted by LaMandie
The question is as follows:
The three sides of a triangle are 10m, 15m, and 20m. What is the area of the triangle?

My diagram:
Attachment 19805
(20 - x)^2 - x^2 = 15^2 - 10^2

(20 -x)(20 - x) - x^2 = 125

400 - 40x -2x^2 = 125

-40x - 2x^2 = -275

x - 2x^2 = 6.875

-2x^2 = 6.875x

x^2 = -3.4375x

x = -3.4375

Obviously it's not the right answer, but I have no idea how to get the right one. Answers are greatly appreciated!

Use Heron's formula: Heron's formula - Wikipedia, the free encyclopedia
• Nov 22nd 2010, 12:47 PM
Wilmer
Quote:

Originally Posted by LaMandie
-40x - 2x^2 = -275
x - 2x^2 = 6.875

Why did you not also divide the -2x^2 by -40?
There are other similar errors.
• Nov 22nd 2010, 01:07 PM
Quote:

Originally Posted by LaMandie
The question is as follows:
The three sides of a triangle are 10m, 15m, and 20m. What is the area of the triangle?

My diagram:
Attachment 19805
(20 - x)^2 - x^2 = 15^2 - 10^2

Using Pythagoras' theorem

h^2=15^2-(20-x)^2=10^2-x^2

225-(400-40x+x^2)=100-x^2

-175+40x=100

(20 -x)(20 - x) - x^2 = 125

400 - 40x -2x^2 = 125 ......you should have 400-40x+x^2-x^2 here

-40x - 2x^2 = -275

x - 2x^2 = 6.875

-2x^2 = 6.875x

x^2 = -3.4375x

x = -3.4375

Obviously it's not the right answer, but I have no idea how to get the right one. Answers are greatly appreciated!

You have correctly applied Pythagoras' theorem but should have cancelled the squares of x.
• Nov 22nd 2010, 06:01 PM
Soroban
Hello, LaMandie!

Quote:

The three sides of a triangle are 10m, 15m, and 20m.
What is the area of the triangle?

My diagram:
Attachment 19805

$\displaystyle (20 - x)^2 - x^2 \:=\: 15^2 - 10^2$ . Right1

$\displaystyle (20 -x)(20 - x) - x^2 \:=\: 125$

$\displaystyle 400 - 40x -2x^2 \:=\: 125$ . . . . no

$\displaystyle (20-x)^2 - x^2 \:=\:125 \quad\Rightarrow\quad 400 - 40x + x^2 - x^2 \:=\:125$

. . $\displaystyle 400 - 40x \:=\:125 \quad\Rightarrow\quad 40x \:=\:275 \quad\Rightarrow\quad x \:=\:\frac{275}{40} \:=\:\frac{55}{8}$

Then: .$\displaystyle h^2 + x^2 \:=\:10^2 \quad\Rightarrow\quad h^2 + \left(\frac{55}{8}\right)^2 \:=\:100$

. . . . . $\displaystyle h^2 \:=\:100 - \frac{3025}{64} \:=\:\frac{3375}{64}$

. . . . . $\displaystyle h \:=\:\sqrt{\frac{3375}{64}} \:=\:\frac{15}{8}\sqrt{15}$

Therefore: .$\displaystyle \text{Area} \;=\;\frac{1}{2}bh \;=\;\frac{1}{2}(20)\left(\frac{15}{8}\sqrt{15}\ri ght) \;=\;\dfrac{75\sqrt{15}}{4}$

• Nov 22nd 2010, 08:25 PM
sugarT
hey there :)

i have not used heron's formula before but you could do something else with some simple trig.

1) find either angle using the Cosine Rule
2) then find the area using A = BCsinA/2
where A is the angle you found and B and C are the two sides that make the angle.
• Nov 22nd 2010, 11:05 PM
Wilmer
Well, to each his own.
I like this one, prefer it over Heron's:
Area = cSQRT(a^2 - x^2) / 2 where x = (a^2 - b^2 + c^2) / (2c)

Of course, a,b,c are triangle's sides: a =< b =< c