Do two right triangles with the same length hypotenuse have the same area?

Question

I watched computer monitors and I asked myself, do two monitors with the same display diagonal have the same display area?

I managed to find out that the answer is yes, if two right triangles with the same length hypotenuse have the same area. The answer is trivial if the two triangles are identical, and according to Thales' theorem I know that there are cases when two right triangles have the same length hypotenuse, but they are not identical (the legs length and the angles are different).

So my final question:

Do two, not identical right triangles with the same length hypotenuse have the same area?

I guess that the answer is yes, but I can't give a proof that I'm 100% sure.

Answer 1

No. Consider the following picture:

$\hspace{90pt}$

All the triangles have the same hypothenuse, but one of them has area of $r^2$ while others might have it arbitrarily small.

I hope this helps $\ddot\smile$

Answer 2

The answer is already given but I couldn't resist this explanation.

Imagine a ladder that is leaned against the wall and slowly sliding down the wall. At every time you will get a right triangle with the same hypotenuse. If we let $f(t)$ be the area at time $t$ then $f(t)$ will be continuous and at some time (when the ladder hits the floor) it will be zero. At previous times it is positive so by the intermediate value theorem it will achieve every value in some interval $[0, x]$ for some positive $x$.

Answer 3

No. With the hypothenuse as base, the area is the same if and only if the height is the same. Using Thales, you can see that there are points on the semicircle that have different distances from the hypothenuse/diameter.

Answer 4

Just take two right triangles with the same hypothenuse:

$$1^2+2^2=(\sqrt 5)^2,\quad S=1$$ or $$ \left(\sqrt{5/2}\right)^2+\left(\sqrt{5/2}\right)^2 = (\sqrt 5)^2, \quad S= \frac 54.$$

Answer 5

A nice counterexample is given by one right triangle with legs $1$ and $7$, and another one with legs $5$ and $5$. Both have the same hypotenuse by Pythagoras, but the areas are $3.5$ and $12.5$, respectively.

(For integer areas, use legs $(1,8)$ and $(4,7)$.)

Answer 6

No, you can have right triangles with hypotenuse $l$ and any area $A$ in the range $0<A\leq l^2/4$.

Answer 7

Compare a half square with a right triangle with the same hypotenuse, but one of the other sides close to zero.

Answer 8

No. One way to see this is just to draw a segment as your hypotenuse, then draw a couple of right triangles on it. Pick your right angle points far enough apart and you will see right off that they can't possibly have the same area. In fact, if you pick a point close enough to one end of the segment, you can get an area as close to zero as you like.

More mathematically, $c^2 = a^2 + b^2$ and area $= ab/2$. So, picking $a$ for example, $a=\sqrt{c^2 - b^2}$, so $A=b\sqrt{c^2 - b^2}/2$. And, now you can see that $b$ can take any value as long as $b<c$, and that will give a whole range of values for $A$, not a single number.

Answer 9

The simple answer is no Because equation for area is base*height/2 where hypotenuse is proposanal to square of it.

take as an example of 3 ,4, 5 side triangle & 3.35553 ,3.35553 & 5 side triangle & you'll get your answer.

Answer 10

Another way to prove that two right triangles with the same hypotenuse length do not necessarily have equal areas, is as follows:

Consider a formula for calculating the area $ A $ of the right triangle with base $ b $ and hypotenuse $ c $: $ A=\frac {1}{2} bh $. Now, consider Pythagorian theorem for such a triangle: $c^2=b^2+h^2 $. What we want to show is that $ A $ is only a function of $ c $. But trying to solve for $ b $ in the second equation yields $ b=\sqrt{c^2-h^2}$ (not caring about the negative solution, obviously), substituting it into the first formula gives $ A=\frac {1}{2}\sqrt{c^2-h^2} h$, so $ A=f (c, h) $. Repeating the same procedure with $ h $ gives $ A=\frac {1}{2}\sqrt {c^2-b^2} b $, so $ A=f (c, b) $, which means that it is not possible to express the area of a right triangle in terms only of its hypotenuse.

Answer 11

Let $c$ denote the length of the hypotenuse and $\theta$ denote one of the adjacent angles. Then one leg of the triangle has length $c \sin{\theta}$ and the other has length $c \cos{\theta}$. The area of the triangle is $\frac{1}{2}(c \sin{\theta})(c \cos{\theta}) = \frac{1}{2} c^2 \sin{\theta} \cos{\theta}= \frac{1}{4} c^2 \sin (2 \theta)$.

Clearly, this area is not a constant. It reaches a minimum of zero as $\theta \rightarrow 0$, and a maximum of $\frac{1}{4}c^2$ when $\theta = \frac{\pi}{4} = 45°$.

Answer 12

As others have already correctly said and justified, the answer is no. Although, I thought it would be nice to point out that for monitors (and TVs), there are standard aspect ratios (the ratio of sides of the monitor, or equivalently, the ratio of the sides of the right triangle). For instance, for old monitors and TVs, the aspect ratio was 3:4 and it is 9:16 for the HD TVs and monitors. Therefore, for monitors (and TVs) of the same aspect ratio, the length of the hypotenuse, determines the length of the sides and therefore the area of the triangle.

If we consider the aspect ratio 3:4, we have: $3l = 4w$, where $l$ and $w$ are length and width of the display respectively. Thus, according to pythagorean theorem, we have: $$(1+\frac{9}{16})l^2 = \frac{25}{16}l^2 = h^2$$ thus, we have: $l=\frac{4}{5}h$ and $w=\frac{3}{5}h$. Therefore, the area of the triangle would be:

$$S_{tr} = \frac{1}{2} \cdot \frac{12}{25}h^2$$

and the area of the display is simply twice the area of the triangle; and are both determined only by the length of the hypotenuse and the aspect ratio.

Answer 13

h^2=a^2+b^2 ;Area=A=a*b/2 ; h=hypotenuse ,a, b are the sides of right angled triangle The expressions can be re-written as-- C=constant = h^2 =a^2 +b^2 = (1/2)[(a+b)^2 +(a-b)^2 ] ; A =(1/8) [(a+b)^2 -(a-b)^2] let (a+b)=x ;(a-b)=y Then the question becomes :-- Given x^2 +y^2 =C = constant ; prove that x^2 - Y^2 = constant ( or not) The equation x^2 +y^2 =C represents a circle ; it can have values (x,y) = (0,h) ;(h,0) ; (h/sqrt(2) , h/sqrt(2)) etc. ---this shows that x^2 -y^2 is not a constant , but has varying values.

Answer 14

Computer monitors probably always have the same ratio of width to height (16:9 since several years ago), so imposing this additional constraint, the answer is Yes because the triangles are then identical.

Otherwise the answer is No, as clearly explained above.

Answer 15

A basic algegra approach lets us not only demonstrate that there can be multiple different values of area for a given hypotenuse, it lets us work out the range of possible values for that area and calculate the corresponding lengths of the other two sides.

Define $A$ as area of the triangle, $c$ as the length of the hypotonuse and $a$ and $b$ as the length of the other two sides. $A > 0$, $a > 0$, $b > 0$, $c > 0$

$$a^2 + b^2 = c^2$$ $$A = \frac{ab}{2}$$ $$2A = ab$$ $$b = \frac{2A}{a}$$

Substituting

$$a^2 + 4\frac{A^2}{a^2} = c^2$$

$$a^4 + 4A^2 = a^2c^2$$

$$a^4 - c^2a^2 + 4A^2 = 0$$

Solving the quadratic in $a^2$.

$$a^2 = \frac{c^2\pm\sqrt{c^4-4*4A^2} }{2}$$

$$a = \sqrt{\frac{c^2\pm\sqrt{c^4-16A^2} }{2}}$$

• If $c^4 < 16A^2$ (i.e. if $c^2 < 4A$) there are no soloutions.
• If $c^4 = 16A^2$ (i.e. if $c^2 = 4A$) there is one soloution.
• If $c^4 > 16A^2$ (i.e. if $c^2 > 4A$) there are two soloutions.