A Circle Inside
A quadrilateral above has the same length of sides, has the same opposing angles and inscribes a circle whose circumference is 8 π . Determine the area of the quadrilateral whose smaller angles are equal to 6 0 ∘ .
Give your answer to 1 decimal place.
The answer is 73.9.
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5 solutions
Same solution. :)
Very simple problem. Solved using basic trigonometry and applying the principles of simmetry. The answer is the square root of (5461 + 1/3). Generalised:
A = 2 r 2 / t a n ( 9 0 ° − α ) + 2 r 2 / t a n ( α )
Where
r = d / 2 π
This can, probably, be further simplified
But if you can't write this number in the field
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The question should specify what to write... I'm not the creator of the problem though. I personally put 73.9 and that made it for me
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We use the exact same figure in the 1st post , and the circle is labelled tangent at $Q$
So , $R=\frac{8\pi}{2\pi}=4$
Hence , we intersect the diagonals to get $AP$ as $2R$, then $\triangleQPA$ 's other leg is
$\sqrt{8^2-4^2}=4\sqrt{3}$
Also , for \$triangleQPB$ 's other leg is unknown , we label it as x .
$\therefore \tan ^30\circ=\frac{x}{4}=\frac{1}{\sqrt{3}}$
$\therefore x\sqrt{3}=4$ , after cross-multiplying .
So , $x=\frac{4}{\sqrt{3}}$
Then , $AB=4\sqrt{3}+\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}*\sqrt{3}}{\sqrt{3}}+\frac{4}{\sqrt{3}}=$
$\frac{16}{sqrt{3}}$ Then , the area of the quadrilateral is $8*\frac{16}{\sqrt{3}}$ , when rounded off is $73.9$ cm$^2$, as $8$ cm is the height of the rhombus
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Thanks for a nice formula. Proof can be had from the sketch I have given in my solution.
General formula for area of a rhombus when radius of the incircle r and one angle
α
is known would be,
4
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I actually managed to solve this all in my head. Took me a day on and off whilst walking the dog. Got 128/3 * sqrt(3) but did need a calculator to convert that to decimal. Radius=4 and the rest was similar triangles using 1,2,sqrt(3) as the 3 sides to get the side length of the equilateral as 16/3 * sqrt(3). The vertical was 8. Easy then to get area of equilateral triangle and double for quadrilateral.
Relevant wiki: Incircles and Excircles