Question

Solve the value of the area present between points A, E, and F in the given figure.

Answer 1

We start solving the problem by dividing the given figure into different regions to simply the following. We then use the definitions of the area of a triangle, area of rectangle, area of semi-circle, and area of the sector to find the required area of the region. We then make subsequent calculations to get the required result.

Complete step-by-step solution:
According to the problem we need to find the area of the region formed between points A, E, and F.

We find the area of the region bounded by the points A, E, and F as follows:
Area of region AEF = Area of triangle ABD – Area of region BFD – Area of the region DFE ---(1).
We can see that the triangle ABD is a right-angled triangle with base 8 units and height 4 units.
We know that the area of the triangle is defined as $\dfrac{1}{2}\times \text{base}\times \text{height}$.
Area of the triangle ABD = $\dfrac{1}{2}\times 8\times 4$.
Area of the triangle ABD = 16 sq. units ---(2).
From the figure, we can see that area of region ACF = area of region BDF.
We can also see that area of ACF + area of BDF = Area of rectangle ABCD – Area of semicircle CFD.
We know that the area of the rectangle with length ‘l’ and breadth ‘b’ is $lb$ q. units and the area of the semi-circle with diameter ‘d’ is $\dfrac{\pi {{d}^{2}}}{8}$sq. units.
Area of BDF + Area of BDF = $\left( 8\times 4 \right)-\left( \dfrac{\pi \times {{8}^{2}}}{8} \right)$.
2(Area of BDF) = $32-8\pi$ sq. units.
Area of BDF = $\dfrac{32-8\pi }{2}$.
Area of BDF = $16-4\pi$ sq. units ---(3).
Now, we find the area of EFD as follows:

Here G is the center of the semi-circle. Since the points, E and D lie on the circumference of the semi-circle, the lengths GD and GE are the same. We know that angles opposite to the equal sides in a triangle are equal. We know that the sum of the angles in a triangle is $\pi$ radians.
So, we get Area of region EDF = Area of sector EGDF – Area of the triangle DGE ---(4).
So, we get the angle at G in the sector GDFE as $\pi -2\theta$.
We know that the area of the sector with radius r and angle $\alpha$ is $\dfrac{1}{2}{{r}^{2}}\alpha$, where $\alpha$ is in radians. And we know that the area of the triangle with a length of the sides a, b, and $\beta$ being the angle between those sides is $\dfrac{1}{2}ab\sin \beta$.
From equation (4), we have Area of region EDF = $\left( \dfrac{1}{2}\times {{4}^{2}}\times \left( \pi -2\theta \right) \right)-\left( \dfrac{1}{2}\times 4\times 4\times \sin \left( \pi -2\theta \right) \right)$.
Area of region EDF = $8\pi -16\theta -8\sin \left( 2\theta \right)$. As $\sin \left( \pi -\alpha \right)=\sin \alpha$.
We know that $\sin \left( 2\theta \right)=\dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta }$.
Area of region EDF = $8\pi -16\theta -8\left( \dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right)$.
Area of region EDF = $8\pi -16\theta -\left( \dfrac{16\tan \theta }{1+{{\tan }^{2}}\theta } \right)$ ---(5).
From the triangle ACD, we have $\tan \theta =\dfrac{4}{8}$.
$\Rightarrow \tan \theta =\dfrac{1}{2}$.
$\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{1}{2} \right)$. We substitute these two values in equation (5).
Area of region EDF = $8\pi -16{{\tan }^{-1}}\left( \dfrac{1}{2} \right)-\left( \dfrac{16\times \left( \dfrac{1}{2} \right)}{1+{{\left( \dfrac{1}{2} \right)}^{2}}} \right)$.
Area of region EDF = $8\pi -16{{\tan }^{-1}}\left( \dfrac{1}{2} \right)-\left( \dfrac{8}{1+\dfrac{1}{4}} \right)$.
Area of region EDF = $8\pi -16{{\tan }^{-1}}\left( \dfrac{1}{2} \right)-\left( \dfrac{8}{\dfrac{5}{4}} \right)$.
Area of region EDF = $8\pi -16{{\tan }^{-1}}\left( \dfrac{1}{2} \right)-\dfrac{32}{5}$ ---(6).
We substitute equations (2), (3), and (6) in equation (1).
So, we get Area of region AEF = $16-\left( 16-4\pi \right)-\left( 8\pi -16{{\tan }^{-1}}\left( \dfrac{1}{2} \right)-\dfrac{32}{5} \right)$.
Area of region AEF = $16{{\tan }^{-1}}\left( \dfrac{1}{2} \right)+6.4-4\pi$.
We have found the required area as $16{{\tan }^{-1}}\left( \dfrac{1}{2} \right)+6.4-4\pi$.
$\therefore$ The area of the region AEF is $16{{\tan }^{-1}}\left( \dfrac{1}{2} \right)+6.4-4\pi$.

Note: We should not confuse the diameter of the semicircle with the radius of the semi-circle. While calculating the area of the sector, we should make sure that the angle is taken in radians, not in degrees. We should not make calculation mistakes while solving this problem. We have taken CFD as a semi-circle as two ends of it clearly pass through the vertices of the rectangle. Otherwise, it would not be a semicircle.