Why are the diagonals of a parallelogram not equal?
Imagine a parallelogram and draw its diagonals. Now the areas of the two triangles on one of the bases is equal. But by Heron's formula, the areas are not equal. So what is the explanation for it.
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$\begingroup$Your reasoning is a little murky, but it looks like you're thinking:
Here are two triangles with the same $a$ and $b$ but different $c$. Since the area can be computed from $a$, $b$ and $c$ using Heron's formula, different $c$s lead to different areas, and so the triangles can't have the same area.
But it is not true that different $c$ necessarily lead to different areas. Heron's formula is $$ \sqrt{s(s-a)(s-b)(s-c)} \qquad\text{where }s=\frac{a+b+c}2$$ Increasing $c$ while holding $a$ and $b$ constant will make $s$ larger, and therefore the $s(s-a)(s-b)$ factors become larger. But the $s-c$ factor becomes smaller by an amount that can dominate the increase in the three other factors -- or, in appropriate cases, can cancel it out exactly!
Indeed, setting $c$ to either $a+b$ or $|a-b|$ will make the area $0$, and for some $c$ between those extremes the triangle will have maximal area, varying continuously with $c$. Every possible area less than the maximum is created by two different $c$, namely one above and one below the $c$ that gives maximal area.
$\endgroup$ $\begingroup$Notice, although the diagonals of a parallelogram are unequal in length. you should consider one diagonal at a time which is common hypotenuse to both the triangles of a parallelogram.
Thus, by drawing any of two unequal diagonals of a parallelogram at a time, we obtain two congruent triangles having all three corresponding sides equal in length & obviously the areas of these triangles are same by using Hero's formula. The sum of areas of these two congruent triangles is equal to the area of parallelogram.
Suppose that $AB=CD=a$ & $BC=AD=b$ are two unequal sides & $AC=d_1$ & $BD=d_2$ are two unequal diagonals of a parallelogram $ABCD$ then
$\bullet$ drawing the diagonal $AC=d_1$, we obtain two congruent triangles $\triangle ABC$ & $\triangle ACD$ each having area $$=\sqrt{s(s-a)(s-b)(s-d_1)}$$ where, $s=\frac{a+b+d_1}{2}$
$\bullet$ drawing the diagonal $BD=d_2$, we obtain two congruent triangles $\triangle ABD$ & $\triangle BCD$ each having area $$=\sqrt{s(s-a)(s-b)(s-d_2)}$$ where, $s=\frac{a+b+d_2}{2}$
$\endgroup$ $\begingroup$I just encountered this question myself and after looking through some answers online I wasn't satisfied with any since I found them all to be over complicated and didn't really get to the point. So I'll post my conclusion here :
Imagine a parallelogram with horizontal lines AB and CD and vertical lines AC and BD. Now imagine dropping a perpendicular line from C onto AB and another perpendicular line from B onto CD. You'll see that these two perpendicular lines and the horizontal lines of the parallelogram form a rectangle.
The diagonal CB of the parallelogram is also one of the diagonals of this rectangle as C and B are 2 of 4 vertices that make up the rectangle. A rectangle's diagonals are equal, so a diagonal equal to CB would be defined by the points where the perpendicular lines dropped from B and C intersect the horizontal lines of the parallelogram and define the remaining 2 vertices of the rectangle.
Try drawing this and you'll see that the 2nd diagonal of the parallelogram, AD, is clearly not equal to this and therefore not equal to the diagonal CD either.
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