Question

If in a isosceles triangle pqr pq = pr=5cm, is inscribed in a circle of radius 15 cm, find the area of the triangle

Answer 1

formula of 2 pie r is used

Answer 2

Answer:

The area is $= 27.99 cm^{2}$.

Step-by-step explanation:

Since we have a triangle inscribed in a circle, what we first do is divide 360 by 3, in order to get the angles of the triangle, so:

⇒ $\frac{360º}{3}$ $= 120º$

Then we get a triangle, which is one third of the main triangle, with an angle of 30 degrees. So by trigonometry properties we have:

$sen(30) = \frac{\frac{b}{2} }{15}$

⇒ $\frac{b}{2} = 15sen(30)$

⇒ $b = 15$ cm

Thus the basis of our triangle is equal to 15 cm.

Now we only need to compute our high so we can aply the formula of the area of the triangle.

So:

$cos(30) = \frac{h - 15}{15}$

⇔ $h = 15cos(30) + 15$

⇒ $h = 15 \frac{\sqrt{3} }{2} +15$ $= 27.99 cm^{2}$

Hence, the area is $= 27.99 cm^{2}$.