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My question is that if we have a triangle, and we divide each of the side by 2 to get a new triangle, what will be the area of the new triangle in context to the original triangle?

Please provide a detailed proof.

Thanks.

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The ratio between the areas of similar figures is equal to the... –  sas Feb 28 at 8:01
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"Please provide a detailed proof." So that you can hand it in as your own work? –  Gerry Myerson Feb 28 at 8:02
    
I have no such intention Gerry, I am a secondary school student trying to understand this concept. –  Aditya Chanana Feb 28 at 8:24
    
Glad to hear it. There are better ways to phrase your request. –  Gerry Myerson Mar 1 at 12:56
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6 Answers

up vote 6 down vote accepted

A demonstration without words might look like this (complicated by SE requiring 30 characters)

4triangles

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Ahh. Mid point theorem! Thank you for the straight forward explanation. –  Aditya Chanana Feb 28 at 10:43
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Hint: Use Heron's formula for the area of a triangle.

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The area of a triangle is given by $$A = \frac{1}{2} \text{base} \cdot \text{height} $$

It should be obvious that halving each side of the triangle halves the base and the height.

The new area will be $$A_0 = \frac{1}{2} \frac{\text{base}}{2} \cdot \frac{\text{height}}{2} $$ $$ A_0 = \frac{1}{2} \frac{1}{2} (\frac{1}{2} \text{ base} \cdot \text{ height} ) $$ $$A_0 = \frac{1}{4} A$$

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I don't think that it is necessary that the height will also be halved. Can you prove it? –  Aditya Chanana Feb 28 at 8:26
    
Any side used as a base is halved by the given of the problem so any corresponding height used must change accordingly , once a particular base is chosen. :) –  neofoxmulder Feb 28 at 8:34
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Another answer without any nasty mathematical symbols or equations (and no pictures, either, except the ones in your mind) :-)

Imagine that the triangle is drawn on a square piece of rubber. Now stretch the rubber square so that it's length and width get doubled. Then the sides of the triangle will be doubled, too. But, what happens to the area of the square? It's now four times as large, right? But the triangle still occupies the same percentage of the square, so the area of the triangle must be four times as big, too.

I'll let you figure out what happens when you shrink the rubber square, instead of stretching it. Shrinking is harder to visualize and describe, but the mathematical principles are the same.

One nice thing about this approach -- it works for any 2D figure, not just triangles.

For extra credit: go through the same sort of reasoning, but now with a 3D object embedded in a cube of rubber or jello.

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Since all sides are halved, the angles between the corresponding sides remain equal. You can show this using similarity of triangles.
$\dfrac{A}{A'} = \dfrac{0.5 s_1 s_2\sin \theta}{0.5 s_1' s_2' \sin \theta}$ where $A, A'$ are areas of older and new triangle respectively, $s_1, s_2$ are any sides of earlier triangle, $s_1', s_2'$ are corresponding sides of new triangle and $\sin \theta$ is the angle between those sides.
Now, you can simplify the ratio of areas using the information $s_1' = s_1/2$ and $s_2' = s_2/2$.

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The area of a triangle can be expressed in terms of two sides and the included angle as $\frac 12 ab \sin C$ - this is proved by noting that if $a$ is the base, then the height is $b\sin C$.

The angle stays the same, while the two sides are halved, giving a factor of $\frac 14$ for the area.

It will not always be possible to find an exact formula for the area or volume of a figure, or to draw a neat diagram for a proof. In $n$ dimensions (for example $n=3$ for solids) changing all dimensions by a linear factor $a$ changes the "volume" (length, area, volume, hyper volume ... ) by a factor $a^n$. This scaling property is a feature of the dimension of an object.

At least this is generally true for well-behaved shapes, which can, for example, be approximated by filling them with cubes or cuboids.

There are shapes which are not so well behaved - you might want to explore fractals, for example, or space-filling curves - as I did when I was at secondary school.

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