1. ## 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:

(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!

2. 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:

(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

3. 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.

4. 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:

(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.

5. Hello, LaMandie!

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

My diagram:

$\displaystyle (20 - x)^2 - x^2 \:=\: 15^2 - 10^2$ . 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: .$\displaystyle \text{Area} \;=\;\frac{1}{2}bh \;=\;\frac{1}{2}(20)\left(\frac{15}{8}\sqrt{15}\ri ght) \;=\;\dfrac{75\sqrt{15}}{4}$

6. 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.

7. 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