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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:

Attachment 23964

(Sorry bad drawing!)

Ratio of sides = $\displaystyle 10x:9x$

Therefore, ratio of areas = $\displaystyle 100x^2 : 81x^2$

$\displaystyle 100x^2-81x^2=38$

$\displaystyle 19x^2=38$

$\displaystyle x^2=2$

Therefore area = $\displaystyle 100x^2=200$

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

Thanks very much for any help!

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...

Re: Area of a triangle Question

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

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

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

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

$\displaystyle (\frac{1}{2}).100.4 = 200.$

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

Re: Area of a triangle Question

Hello, BobtheBob!

Here's another approach . . .

Quote:

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 $\displaystyle b$; its height is $\displaystyle h.$

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

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

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

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!

Re: Area of a triangle Question

Quote:

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....