Review question

# What's the area inside both the circle $C$ and the triangle $T$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7495

## Question Diagram when $h > 2/\sqrt{5}$ Diagram when $h < \sqrt{3}/2$

The three corners of a triangle $T$ are $(0,0), (1, 2h), (3,0)$, where $h > 0$. The circle $C$ has equation $x^2 + y^2 = 4$. The angle of the triangle at the origin is denoted as $\theta$. The circle and triangle are drawn in the diagram above for different values of $h$.

1. Express $\tan \theta$ in terms of $h$.
2. Show that the point $(1, 2h)$ lies inside $C$ when $h < \sqrt{3}/2$.
3. Find the equation of the line connecting $(3,0)$ and $(1,2h)$. Show that this line is tangential to the circle $C$ when $h = 2/\sqrt{5}$.
4. Suppose now that $h > 2/\sqrt{5}$. Find the area of the region inside both $C$ and $T$ in terms of $\theta$.
5. Now let $h = 6/7$. Show that the point $(8/5,6/5)$ lies on both the line (from part (iii)) and the circle $C$.

Hence show that the area of the region inside both $C$ and $T$ equals

$\dfrac{27}{35}+2\alpha,$

where $\alpha$ is an angle whose tangent, $\tan \alpha$, you should determine.

You may use the fact that the area of a triangle with corners $(0,0), (a,b), (c,d)$ is $\dfrac{1}{2}\vert ad-bc\vert$.