# Find the area of the shaded region

• MHB
• anemone
In summary, the problem involves two circles, with points P and Q as their centers. Chord AB is tangent to the circle with center P. The line PQ is parallel to chord AB and AB=x units. The task is to find the area of the shaded region in yellow.
anemone
Gold Member
MHB
POTW Director
Points $$\displaystyle P$$ and $$\displaystyle Q$$ are centers of the circles as shown below. Chord $$\displaystyle AB$$ is tangent to the circle with center $$\displaystyle P$$. Given that the line $$\displaystyle PQ$$ is parallel to chord $$\displaystyle AB$$ and $$\displaystyle AB=x$$ units, find the area of the shaded region in yellow.

[TIKZ]
\draw [<->] (0.2,1) -- (5.8, 1);
\begin{scope}
\draw (3,0) circle(3);
\end{scope}
\begin{scope}
\draw (1,0) circle(1);
\end{scope}
\coordinate[label=above: P] (P) at (1,0);
\coordinate[label=above: Q] (Q) at (3,0);
\coordinate[label=left: A] (A) at (0.2,1);
\coordinate[label=right: B] (B) at (5.8,1);
\coordinate[label=above: $x$] (x) at (2.8,1);
\filldraw (1,0) circle (2pt);
\filldraw (3,0) circle (2pt);

\draw[fill=red,fill opacity=0.6] (1,0) circle (1);
\draw[fill=yellow,fill opacity=0.5] (3,0) circle (3);
[/TIKZ]

I am not seeking for help, I just want to share an interesting geometry problem at MHB with the hope our members enjoy it. :)

For those who are interested, you are welcome to post your solution here! I initially wanted to post this in the Challenges subforum, but was afraid this might not difficult enough to be claimed as a challenge problem. (Smile)

Solution of MarkFL:
[TIKZ]
\draw [<->] (0.2,1) -- (5.8, 1);
\begin{scope}
\draw (3,0) circle(3);
\end{scope}
\begin{scope}
\draw (1,0) circle(1);
\end{scope}
\coordinate[label=below: P] (P) at (1,0);
\coordinate[label=below: Q] (Q) at (3,0);
\coordinate[label=left: A] (A) at (0.2,1);
\coordinate[label=right: B] (B) at (5.8,1);
\coordinate (T) at (3,1);
\coordinate[label=above: $\frac x2$] (x) at (1.9,1);
\coordinate[label=above: $R$] (R) at (2.2,0.0001);
\coordinate[label=above: $r$] (r) at (3.2,0.1);
\filldraw (1,0) circle (2pt);
\filldraw (3,0) circle (2pt);
\draw (Q)-- (A);
\draw (Q)-- (T);
\draw[fill=red,fill opacity=0.6] (1,0) circle (1);
\draw[fill=yellow,fill opacity=0.5] (3,0) circle (3);
\draw (3,0.9) rectangle +(-0.1, 0.1);
[/TIKZ]
Let the radius of the smaller circle be $$\displaystyle r$$ and the radius of the bigger circle be $$\displaystyle R$$.

Next, build a right-angled triangle where the base of it is half of $$\displaystyle AB$$, i.e. $$\displaystyle \dfrac{x}{2}$$.

Now, by applying the Pythagoras' theorem, we get

\displaystyle \begin{align*}\text{Area of shaded region}&=\pi(R^2-r^2)\\&=\pi\left(\dfrac{x}{2}\right)^2\\&=\dfrac{\pi x^2}{4}\end{align*}

Last edited:

## 1. How do you find the area of the shaded region?

The area of the shaded region can be found by subtracting the area of any non-shaded regions from the total area of the shape.

## 2. What is the formula for finding the area of the shaded region?

The formula for finding the area of the shaded region will depend on the shape of the region. For example, the formula for finding the area of a rectangle is length x width, while the formula for finding the area of a circle is πr².

## 3. Do you include the shaded region in the final area calculation?

No, the shaded region is typically not included in the final area calculation. It is only used as a visual representation of the area that is being subtracted from the total area.

## 4. Can you use the same method to find the area of any shape?

Yes, the method of subtracting non-shaded regions from the total area can be applied to any shape. However, the formula used to find the area may vary depending on the shape.

## 5. How can I check if my answer for the area of the shaded region is correct?

You can check your answer by using a different method of finding the area, such as using a different formula or breaking the shape into smaller, easier to calculate shapes. You can also use an online calculator or ask a peer or teacher to review your calculations.

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