How to find the allowed range of k in a sector area problem

Key concept: This problem mixes sector area and a small triangle inside the sector. The useful move is to write both areas in terms of $a$, $k$, and $\theta$, then compare them directly.

Step by step

Let the sector have radius $a$ and angle $\theta$, where

$0<\theta<\frac{\pi}{2}$

Since $AD=AE=ka$, triangle $ADE$ has area

$\text{Area of }\triangle ADE=\frac12(ka)(ka)\sin\theta=\frac12k^2a^2\sin\theta$

The whole sector has area

$\frac12a^2\theta$

The line $DE$ divides the sector into two equal areas, so triangle $ADE$ must be half the sector:

$\frac12k^2a^2\sin\theta=\frac14a^2\theta$

Cancel $a^2$:

$k^2\sin\theta=\frac\theta2$

So

$k^2=\frac{\theta}{2\sin\theta}$

We are told that

$\frac{\theta}{\sin\theta}>1$

Hence

$k^2>\frac12 \quad\Rightarrow\quad k>\frac1{\sqrt2}$

Also, because $0<\theta<\frac{\pi}{2}$, we have $\sin\theta<1$ and therefore

$k^2=\frac{\theta}{2\sin\theta}<\frac{\pi/2}{2}=\frac\pi4$

So

$k<\frac{\sqrt\pi}{2}$

Set of possible values of $k$

$\frac1{\sqrt2}<k<\frac{\sqrt\pi}{2}$

Pitfall alert

Don't use the sector formula with the wrong region. The equal-area condition applies to the triangle $ADE$ versus the remaining part of the sector, so the triangle must be exactly half the sector, not the whole thing. Another easy slip is dropping the square on $k$ when substituting the triangle area.

Try different conditions

If the sector angle were fixed to a specific value, then $k$ would be found directly from

$k=\sqrt{\frac{\theta}{2\sin\theta}}$

For angles closer to 0, $\theta/\sin\theta$ is close to 1, so $k$ approaches $1/\sqrt2$. As $\theta$ moves toward $\pi/2$, the value of $k$ increases.

Further reading

sector area, triangle area, equal areas