# Math Help - Area of a triangle Question

1. ## Area of a triangle Question

Hi,

I found this question and I could do with some help regarding the answer. It is:

Nine lines are drawn parallel to the base of a triangle to divide each of the other sides into 10 equal segments. If the area of the largest region is 38, find the area of the original triangle.

I believe that it must be ratios and something like:

Ratio of sides = $10x:9x$
Therefore, ratio of areas = $100x^2 : 81x^2$
$100x^2-81x^2=38$
$19x^2=38$
$x^2=2$

Therefore area = $100x^2=200$

Am I right? I feel that this is rather inadequate and too simple.

Thanks very much for any help!

2. ## Re: Area of a triangle Question

To "see" how this works, start with a right triangle: 10 by 24 by 26 will make it easier...

3. ## Re: Area of a triangle Question

Alternatively you could use the standard formula for the area of a triangle with sides $a,b$ and included angle $\theta.$

Area = $\frac{1}{2}ab\sin\theta.$

Call the sloping sides $10m$ and $10n$ say, then the area of the last strip will be

$(\frac{1}{2})100mn\sin\theta - (\frac{1}{2})81mn\sin\theta = (\frac{1}{2})19mn\sin\theta = 36,$ in which case $mn\sin\theta=4,$ so that the area of the whole triangle will be
$(\frac{1}{2}).100.4 = 200.$

4. ## Re: Area of a triangle Question

A model triangle has an altitude of 10 units and a base of10 units. Its area = 50 sq units
Area of largest trapezoid section = 50 - 9*9/2 =9.5 sq units
Fractional area of above 9.5/50 = 0.19
Area of actual triangle* 0.19 =38
AAT = 200 sq units

5. ## Re: Area of a triangle Question

Hello, BobtheBob!

Here's another approach . . .

Nine lines are drawn parallel to the base of a triangle to divide each of the other sides into 10 equal segments.
If the area of the largest region is 38, find the area of the original triangle.

Code:
    -         *
:        *  *
:       *     *
:      *        *
h     *           *
:    *   (9/10)b    *
:   * * * * * * * * * * - - -
:  *                    *  (1/10)h
- *  *  *  *  *  *  *  *  * -
: - - - - - b - - - - - :
The base of the triangle is $b$; its height is $h.$

The bottom trapezoid has height $\tfrac{1}{10}h$ and bases $b$ and $\tfrac{9}{10}b.$

Its area is 38: . $\tfrac{1}{2}\left(\tfrac{1}{10}h\right)\left(b + \tfrac{9}{10}b\right) \:=\:38 \quad\Rightarrow\quad \tfrac{19}{200}bh \:=\:38 \quad\Rightarrow\quad bh \:=\:400$

Therefore: . $\text{Area}_{\Delta} \:=\:\tfrac{1}{2}bh \:=\:200$

6. ## Re: Area of a triangle Question

Thanks! So I was right with 200. However, all the solutions here are far better than mine in my opinion!

Thanks again for the help!

7. ## Re: Area of a triangle Question

Originally Posted by BobtheBob
Nine lines are drawn parallel to the base of a triangle to divide each of the other sides into
10 equal segments. If area of the largest region is 38, find the area of the original triangle.
Givens are:
j = area largest region (38)
h = height of triangle (10)
n = number of regions (10)
Find:
k = area of triangle (?)

k = bh / 2
where b = 2jn(n - 2) / [2(h/n)(n^2 - 3n + 1) + h]
(b = base of triangle)

Substitute the givens and you'll get b = 40, hence k = 40(10) / 2 = 200

Now Bob, you can use that as a "general case" formula;
weird ones like j = 137, h = 23, n = 7 handled quickly!
That one will give you b = 44.903.... and k = 516.384....