Question

The diameter of two circular plates are 10cm and 24 cm respectively. What is the diameter of the plate which has an area equal to the combined area of the two given plates?

Answer 1

Hint:
As diameter is the half the radius of the circle. We will first find the radius and then the area of the plates using the formula, $\pi {r^2}$, where $r$ is the radius of each plate. Add the areas to get the combined area and again apply the same formula to find the required diameter.

Complete step by step solution:
To start with, we will find the radius of both the circular plates by taking the half of the diameter.
For the plate with diameter 10cm, the radius is $\dfrac{{10}}{2} = 5cm$ and the radius of the plate with diameter 24 cm is $\dfrac{{24}}{2} = 12cm$
The area of a circle is given by $\pi {r^2}$, where $r$ is the radius of the circle.
Substitute the value of the $r = 5$ for the area of the plate with radius 5cm.
${A_1} = \pi {\left( 5 \right)^2} = 25\pi c{m^2}$
Similarly, substitute the value of the $r = 12$ for the area of the plate with radius 12cm.
${A_2} = \pi {\left( {12} \right)^2} = 144\pi c{m^2}$
Add both the areas to find the total area of both the plates.
$
\Rightarrow {A_1} + {A_2} = \left( {25\pi + 144\pi } \right)c{m^2} \\
\Rightarrow A = \pi \left( {25 + 144} \right)c{m^2} \\
\Rightarrow A = \pi 169c{m^2} \\
$
Now, the area of the plate bigger plate is $\pi 169c{m^2}$
Let the radius of the bigger plate is $R$
Then, $\pi {R^2} = \pi 169$
Divide both sides by $\pi $
$
\Rightarrow {R^2} = 169 \\
   \Rightarrow R = \pm \sqrt {169} \\
   \Rightarrow R = \pm 13 \\
$
But, radius can never be negative. Therefore, the radius is 13cm and the diameter of the bigger plate is given as $13 \times 2 = 26cm$

Hence, the diameter of the plate which has an area equal to the combined area of the two given plates is 26cm.

Note:
We have to not substitute the value of $\pi $ as it will get cancelled when we will substitute the values in the formula of area of circle. Substituting the value of $\pi $ will make the calculations unnecessarily longer.