$A$, $B$, $C$ and also $D$ lie in alphabet order ~ above a one so that ABCD forms a kite. $AB = DA = 8 cm$ and $BC = CD = 13 cm$. Discover the area of the kite $ABCD$.
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I believed that i might have the ability to use $AC$ together the diameter that the circle, therefore, $\angle ADC$ would be a right angle. Utilizing that I assumed that I have the right to use the formula for finding the area of a triangle utilizing sine ($\frac12ab \times \sin C$) to discover the area of the two congruent triangles. However, the answer that I acquire does not match with the exactly answer. What am ns doing incorrectly? Any assist will be substantially appreciated!
An overkill solution:
We recognize that the area of dragon is $ef/2$ where $e,f$ are diagonals. By Ptolomey theorem we have:
$$ 2A =ef = ac+bd = 2ac = 2\cdot 104$$ for this reason $A= 104$.
I leaving it together an practice to find why ($M$ is the point where the perpendicular diagonals of the dragon cut)
$$ \frac1BM^2= \frac18^2 + \frac113^2 $$
is correct ( to compare proportionate political parties of similar triangles )
Now as you have done for total kite area
$$ (AC \cdot BD/2 \cdot \frac12) \cdot 2 $$
$AC$ is hypotenuse etc.
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$A,B,C$ and also $D$ space concyclic.$AC$ is the diameter the the circle and also $AD=DC$.The area of square $ABCD$ is $20cm^2$.
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