Question

A door of width of 6m has an arc above it having a height of 2m as shown. Find the radius of the arc.

Answer 1

Hint: Assume that the radius of the circle is r. Hence determine the length OE in terms of r. Use Pythagoras theorem in triangle AOE and hence form an equation in r. Solve for r and hence determine the radius of the circular arc of the door.

Complete step-by-step answer:
Let the radius of the circular arc of the door be r.
Hence, we have OA = OF = r.
Since FE =2 m, we have
OE = OF-FE = r-2
Also given that AB = 6m.
Since the perpendicular from the centre to the chord bisects the chord, we have AE = EB = 3m
We know that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs of the triangle. This is known as Pythagoras theorem.
Now in triangle AOE by Pythagoras theorem, we have
$A{{O}^{2}}=O{{E}^{2}}+A{{E}^{2}}$
Substituting the values of AO, OE, and AE, we get
${{r}^{2}}={{\left( r-2 \right)}^{2}}+{{3}^{2}}$
We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
Hence, we have
${{r}^{2}}={{r}^{2}}-2\times 2\times r+{{2}^{2}}+{{3}^{2}}$
Adding $4r-{{r}^{2}}$ on both sides, we get
$4r=4+9=13$
Dividing by 4 on both sides, we get
$r=\dfrac{13}{4}$
Hence the radius of the circular arc is $\dfrac{13}{4}m$

Note: Alternative solution- Using sine rule and the distance of circumcentre from a side of the triangle.

We know that the length of the side a of triangle ABC is given by $a=2R\sin A$
Here a = 6cm
Hence, we have
$2R\sin A=6\Rightarrow R\sin A=3\text{ }\left( i \right)$
Also, we know that the distance of the circumcentre from side a is given by $R\cos A$
Hence, we have
$R\cos A=OE=R-2\text{ }\left( ii \right)$
Squaring and adding equation (i) and (ii), we get
${{R}^{2}}\left( {{\sin }^{2}}A+{{\cos }^{2}}A \right)=9+{{\left( R-2 \right)}^{2}}$
We know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1,\forall x\in \mathbb{R}$
Hence, we have
${{R}^{2}}={{\left( R-2 \right)}^{2}}+9$
Subtracting ${{\left( R-2 \right)}^{2}}$ from both sides, we get
${{R}^{2}}-{{\left( R-2 \right)}^{2}}=9$
Using ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$, we get
$\begin{align}
  & \left( R-R+2 \right)\left( R+R-2 \right)=9 \\
 & \Rightarrow 2R-2=\dfrac{9}{2} \\
 & \Rightarrow 2R=\dfrac{13}{2} \\
 & \Rightarrow R=\dfrac{13}{4} \\
\end{align}$
Hence the radius of the circle is $\dfrac{13}{4}$, which is the same as obtained above.