A geometry problem by Chirag Jindal

Observe that the radius is $\dfrac{2}{3}$ of the height. If the height is $\dfrac{l\sqrt{3}}{2} = 7\sqrt{3}$ , $r = \dfrac{14\sqrt{3}}{3}$ .

Now, the shaded area is $S_{circle} - S_{triangle}$ .

$S_{circle} = \pi \left ( \dfrac{14\sqrt{3}}{3} \right )^2 = \dfrac{196}{3}\pi$

$S_{triangle} = \dfrac{14 \cdot 7\sqrt{3}}{2} = 49\sqrt{3}$

$S_{shaded} = \dfrac{196}{3}\pi - 49\sqrt{3} \approx 120.38$

We calculate the height of the triangle by Pythagorean Theorem, so: $14^2-7^2=147$

$Height=\sqrt{147}$

Then we draw the medians: Alt text

We call $C$ the centroid.Since the centroid divides each median in ratio $2:1$ the length of $C$ to any corner it's $\dfrac{2}{3}$ of the height:

$\dfrac{\sqrt{147}}{3}\times2\approx{8.08}$

So, now we calculate the area of the circle:

$\pi(\dfrac{2}{3}\sqrt{147})^2\approx{205.25}$

Now the area of the triangle:

$\dfrac{\sqrt{147}\times14}{2}\approx{84.87}$

Then the shaded area is:

$\pi(\dfrac{2}{3}\sqrt{147})^2-\dfrac{\sqrt{147}\times14}{2}\approx{\boxed{120.38}}$

Since it is an equilateral, draw line from centre of circle connecting to each of the vertices of triangle. Since there are three equal sections, this means that the three triangles formed from the split have a central angle of 120. The angle of the other two sides are 30. Now, apply the law of cosines, and since the "b value and "c" values are equal, rearrange the equation to c = \frac{a}{2cosC}, where a is 14 and C is 30.

Let's label the center of the circle Point D. Since Triangle ABC is an equilateral triangle we can the find the radius by finding the length of BD or CD or AD, to find this lets create Triangle BCD. Using the angles of a triangle we find that angle ABC is 60 degrees and angle DBC is exactly half of that given that Triangle ABC is equilateral so, angle DBC is 30 degrees which leaves angle DCB 30 degrees. Now we can apply The Law of Sines so 14/sin(120) = r/sin(30) so using algebra we then discover that the radius is then 8.08.... so then if we apply the theorem π*r^2 the area of the circle is then 205.25.... and to find the shaded region we just subtract the area of Triangle ABC which using Heron's formula we then find out the area is 84.87... so 205.25-84.87=120.38. The area of the shaded region is 120.38

area of triangle=1/2*altitude/base then we minus the this area from area of circle we get the area of shaded region

3r=3^1/2 side
then area of circle = 616/3 and area of triangle is 254.62 we will subtract both and obtain the answer 120.38

Let the center of the circle be M, Draw two radii from M to B & C <<Now we have an isosceles triangle>>, Then draw a perpendicular from M to BC to cut it at L and bisects it, since the angle of the equilateral equals 60 and the radius we drawn bisected it, Therefore the angle MCL=30` and CML=60 sin(60)=CL/radius= 7/radius, Then the radius equals 7/sin(60) = 14sqrt(3)/3 cm and with the law of the area of a circle (pi)(r^2) we can get the area of the circle. Now we'll get the area of the triangle with Heron's Formula: sqrt[21(21-14)(21-14)(21-14)] = 49sqrt(3) cm. To get the Area of the shaded part we'll subtract the area of the triangle from the area of the circle: [(Pi)(14sqrt(3)/3)^2] - [49sqrt(3)] =120.3802305 cm^2 (cm square)