Triangle problem - what should the equation be?

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ss_0001

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May 23, 2011
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Hi,

I am stuck trying to work out where to start. Eg. What the equation should look like. I think I should be using somehow 1/2 base x height.

There is a triangle with the side length a and the height h which is perpendicular to a. If side a would be 6cm longer and height h 3cm longer, the surface of the triangle is 72cm2 larger. If the height h would be 6cm longer and the side a would be 6cm shorter, the surface would be 9cm2 larger. How long are the side a and the height h?
 
(a+6) / 2 x (h+3) = 72

(a-6) / 2 x (h+6) = 9

Or/

((a+6) / 2) x (h+3) = ((a/2) x h) +72 ??

Is this correct? Not sure what's next...
 
Hello, ss_0001!

[1] There is a triangle with the side length \(\displaystyle a\) and the height \(\displaystyle h.\)
[2] If side \(\displaystyle a\) was 6cm longer and height \(\displaystyle h\) 3cm longer, the area of the triangle is 72 cm2 larger.
[3] If the height \(\displaystyle h\) was 6cm longer and the side \(\displaystyle a\) 6cm shorter, the area would be 9 cm2 larger.
How long are the side \(\displaystyle a\) and the height \(\displaystyle h\)?
Click to expand...

\(\displaystyle \text{[1] The original area of the triangle is: }\:\tfrac{1}{2}ah\)

\(\displaystyle \text{[2] If the side is }a+6\text{ and the height is }h+3\text{, the area is 72 cm}^2\;larger.\)
. . . . \(\displaystyle \tfrac{1}{2}(a+6)(h+3) \:=\:\tfrac{1}{2}ah + 72\) *

\(\displaystyle \text{[3] If the height is }h+6\text{ and the side is }a-6\text{, the is 9 cm}^2\;larger.\)
. . . . \(\displaystyle \tfrac{1}{2}(a-6)(h+6) \:=\:\tfrac{1}{2}ah + 9\) *

\(\displaystyle There\text{ are your equations!}\)

 
ss_0001 said:
(a+6) / 2 x (h+3) = 72 + ah/2

(a-6) / 2 x (h+6) = 9 + ah/2

The lefthand sides are correct, but you forgot to add the original triangle area (expressed in red) to the righthand sides.



((a+6) / 2) x (h+3) = ((a/2) x h) +72

Is this correct?

Yes. You correctly added ah/2 to the righthand side.
Click to expand...

NOTE: Beginning with algebra, we stop using the letter x as a multiplication symbol. From now on, the letter x almost always represents a number.

For a multiplication sign, please type an asterisk instead:

((a+6) / 2) * (h+3) = ((a/2) * h) +72

And, with practice, you'll realize that some multiplication signs and extra parentheses are not needed, with a little rearranging.


(h + 3)(a + 6)/2 = ah/2 + 72

(h + 6)(a - 6)/2 = ah/2 + 9

Here we have a system of two equations in two variables (a and h)


What method(s) have you been taught for solving such a system of equations? 8-)

 


It's been awhile, since we gave you some guidance. I hope that you were able to resolve your issues. For future readers of this thread, here is one solution approach to the system of equations.


(h + 3)(a + 6)/2 = ah/2 + 72

(h + 6)(a - 6)/2 = ah/2 + 9


Eliminate fractions by multiplying both sides of each equation by the denominator 2

(h + 3)(a + 6) = ah + 144

(h + 6)(a - 6) = ah + 18


Expand the lefthand sides

ah + 6h + 3a + 18 = ah + 144

ah - 6h + 6a - 36 = ah + 18


Subtract ah from both sides of each equation

6h + 3a + 18 = 144

-6h + 6a - 36 = 18


Solve either of these equations for symbol a, and then substitute the resulting expression for symbol a into the other equation


Solving the first equation for symbol a yields

a = -2h + 42


Subtituting this expression into the second equation yields

-18h + 216 = 18


We now have an equation that contains only one symbol, so we can solve it

h = 11


We already found a formula for evaluating a when we know h

a = -2(11) + 42

a = 20


The base of the original triangle is 20 cm, and the height is 11 cm.

I leave the verification for you! 8-)

 
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