1. ## Triangles and Geometry

Hello,

I've got the following question:

The area of the triangle bounded by the three lines x = 0, y = 0, and x + y = a has an area of 8. What is a?
The answer says it's 4 but I have no clue as to why?

Is it because a rectangle is double the area of the triangle in question which is 16? And the length and width would both 4 on this right angled triangle?

It sounds overly complicated so I'm not sure if there's a simpler explanation?

2. Originally Posted by Kataangel
Hello,

I've got the following question:

The answer says it's 4 but I have no clue as to why?

Is it because a rectangle is double the area of the triangle in question which is 16? And the length and width would both 4 on this right angled triangle?

It sounds overly complicated so I'm not sure if there's a simpler explanation?

Hey you are on the right track, if you simply sketch out the given lines you will see that a right angled isosceles triangle. The two equal sides of the triangle will be "a"

Now as we know area of a triangle = 1/2*(b)*(h), here b=h=a,
hence 1/2*(a)(a) = 8 ( area is given as 8)

a^2 = 16

so a = 4

Also alternately, if you double the triangle you will get a Square of side "a"

whose area will be +> a*a=2*8

a^2=16

a=4

hope this helps you out

hope that helps you out

3. Originally Posted by sidjaya86
Hey you are on the right track, if you simply sketch out the given lines you will see that a right angled isosceles triangle. The two equal sides of the triangle will be "a"

Now as we know area of a triangle = 1/2*(b)*(h), here b=h=a,
hence 1/2*(a)(a) = 8 ( area is given as 8)
I can see how the triangle is right angled (since it is bounded by perpendicular lines) but not how it is isosceles?

Could you please elaborate more on how you get from the triangle area formula to b=h=a?

4. Hello, Kataangel!

The area of the triangle bounded by the three lines: . $\begin{Bmatrix}x \:=\:0 \\ y\:=\:0 \\ x+y\:=\:a\end{Bmatrix}$ has an area of 8.
What is $a$ ?
Did you make a sketch?
Code:
        |
* |
a*
|:*
|:::*
|:::::*
|;;;;;;;*
|:::::::::*
- - + - - - - - * - -
0           a *
$\text{The line }x = 0\text{ is the }y\text{-axis.}$

$\text{The line }y = 0 \text{ is the }x\text{-axis.}$

$\text{The line }x+y\:=\:0\text{ has intercepts: }\,(a,0),\:(0,a)$

We have a right triangle with: . $\text{base} = a,\;\text{height} = a$

The area of a triangle is: . $\text{Area} \:=\:\frac{1}{2}\text{(base)(height)}$

So we have: . $\frac{1}{2}(a)(a) \:=\:8 \quad\Rightarrow\quad a^2 \:=\:16 \quad\Rightarrow\quad \boxed{a \:=\:\pm4}$