Finding area of rhombus

Question

I’m trying to solve the following question but I’ve stuck.

AFAI find by using a property that opposite sides are congruent in a rhombus, angles of DAC, DCA, BAC, and BCA are equal each other.

Answer 1

Since AMD is a right triangle you have that the square of the length of a cathetus equals the product of the lengths of its orthographic projection on the hypotenuse times the length of this, i.e. if we denote with $K$ the midpoint of $AC$, we have $AD^2=AM \cdot AK$, so $AD^2=41 \cdot \frac{41+9}{2}=41 \cdot 25$. Now we can use Pythagoras' theorem on the triangle $ADK$ (it's a right triangle since $ABCD$ rhombus) and $DK^2=AD^2-AK^2=41 \cdot 25- 25^2=(41-25)\cdot 25=16 \cdot 25$, therefore $DK=4 \cdot 5=20$. Now we have everything since the area will be $$ DK \cdot AC=20 \cdot 50=1000$$

Answer 2

EDIT 1:

$$ AM.MC= MB^2,\, AB^2 =AM.AC $$

BM (red) is perpendicular bisector of AD

$$ AB=L = 5 \sqrt{41} $$

$$ \sin \theta = \dfrac{20}{5 \sqrt{41}}$$

If each side has length L and vertices have angles $( \theta , \pi- \theta ) $ then total area is

$$ L^2 \sin 2 \theta =1000 $$

etc..

Answer 3

Here's a start. Let $L$ be the side length, $T$ the length of $MD$ and $\alpha$ angle $MAD$. Then $$ T = L \tan \alpha = 41 \cos \alpha $$ $$ T^2 = 9^2 + L^2 = 18L \cos \alpha . $$ That should provide sufficient information to find $L$, $T$ and $\alpha$. There's probably an algebraic shortcut that finds the area directly without needing the separate variables.

Answer 4

There are two similar triangles and the rest is just Pythagoras.