Question

A semicircle with diameter PQ sists on an isosceles triangle PQR to from a region shaped like a two-dimensional ice-cream cone, as shown in the figure. If \(A\left( \theta \right)\) is the area of the semicircle and \(B\left( \theta \right)\) is the area of the triangle, find \(\)\({\lim _{\theta \to {0^ + }}}\frac{{A\left( \theta \right)}}{{B\left( \theta \right)}}\).

Short answer

The final answer is \({\lim _{\theta \to {0^ + }}}\frac{{A\left( \theta \right)}}{{B\left( \theta \right)}} = 0\).

Step-by-step solution

Draw the semicircle

A semicircle with diameter PQ sists on an isosceles triangle PQR to form a region shaped like a two-dimensionice-cream as shown in the below diagram.

Simplify the obtained condition

Let \(A\left( \theta \right)\)be the area of the semicircle and \(B\left( \theta \right)\) be the area of a triangle.

Need to find \({\lim _{\theta \to {0^ + }}}\frac{{A\left( \theta \right)}}{{B\left( \theta \right)}}\)

Let r be the radius of the semicircle PTQ and h be the height of the isosceles triangle.

Then by the property of the isosceles triangle that bisector of an angle of an isosceles triangle bisects the opposite side and is perpendicular to this side.

In the above figure, RS is perpendicular to PQ and PS=SQ=r(radius of semicircle),

\({\lim _{\theta \to {0^ + }}}\frac{{A\left( \theta \right)}}{{B\left( \theta \right)}} = 0\)\(\angle SRQ = \angle SRP = \frac{\theta }{2}\)and RS=h.

Since the triangle, RSQ is a right-angled triangle,

We have,

\(\begin{aligned}\tan \frac{\theta }{2} &= \frac{{QS}}{{SR}}\\\tan \frac{\theta }{2} &= \frac{r}{h}\\r &= h\tan \frac{\theta }{2}\end{aligned}\)

Apply area of triangle formula:

Area of triangle = \(\frac{1}{2}\left( {base} \right)\left( {height} \right)\).

Area of a triangle (PQR) = \(B\left( \theta \right) = \frac{1}{2}PQ \times RS\)

\(\begin{array}{l} = \frac{1}{2}\left( {2r} \right) \times h\\ = hr\\ = {h^2}\tan \frac{\theta }{2}\end{array}\)

Therefore,the Area of the triangle is \(B\left( \theta \right) = {h^2}\tan \frac{\theta }{2}\).

Apply area of Semicircle formula:

Find the area of the semicircle.

\(\begin{aligned}A\left( \theta \right) &= \frac{1}{2}\pi {r^2}\\ &= \frac{1}{2}\pi {\left( {h\tan \frac{\theta }{2}} \right)^2}\\ &= \frac{\pi }{2}{h^2}{\tan ^2}\frac{\theta }{2}\end{aligned}\)

Apply ratio proportion formula:

Apply ratio proportion between area of triangle and semicircle:

\(\begin{aligned}\frac{{A\left( \theta \right)}}{{B\left( \theta \right)}} &= \frac{{\frac{\pi }{2}{h^2}\tan \frac{\theta }{2}}}{{{h^2}\tan \frac{\theta }{2}}}\\ &= \frac{\pi }{2}\tan \frac{\theta }{2}\end{aligned}\)

Apply limit on both sides.

\(\begin{aligned}{\lim _{\theta \to {0^ + }}}\frac{{A\left( \theta \right)}}{{B\left( \theta \right)}} &= {\lim _{\theta \to {0^ + }}}\frac{\pi }{2}\tan \frac{\theta }{2}\\ &= \frac{\pi }{2}{\lim _{\theta \to {0^ + }}}\tan \frac{\theta }{2}\\ &= \frac{\pi }{2}(0)\\ &= 0\end{aligned}\)

Therefore, \({\lim _{\theta \to {0^ + }}}\frac{{A\left( \theta \right)}}{{B\left( \theta \right)}} = 0\)