Question

Find the area of the shaded region in the following figures.
(a).

(b).

(c).

Answer 1

Hint: In this problem, we have to find the area of the shaded portion from the given figures. In the first figure we have a square with four circles in it, where we can find the area of the four quadrants inside the square and subtract it with the area of the square to get the shaded region. In problem 2, we can subtract the area of the rectangle from the area of the circle. In the third problem, we can find the area of each semicircle and subtract the unshaded part from the shaded part to get the shaded region.

Complete step by step answer:
Here we have to find the area of the shaded region in the given figure.

We can now find the area of the shaded region (blue region).
Here we have the side of the square as 7 + 7 = 14cm
We know that the area of the square formula is \[{{a}^{2}}\] and area of quadrant is \[\dfrac{\pi {{r}^{2}}}{4}\].
We can now subtract the area of square form the area of 4 quadrants, we get
Area of shaded region = Area of square – Area of 4 quadrants.
\[\Rightarrow {{\left( 7 \right)}^{2}}-4\times \dfrac{\pi {{\left( 7 \right)}^{2}}}{4}\]
We can now solve the above steps, we get
\[\begin{align}
  & \Rightarrow 14\times 14-\dfrac{22}{7}\times 7\times 7 \\
 & \Rightarrow 196-154=42c{{m}^{2}} \\
\end{align}\]
Therefore, the area of the shaded region is 42sq.cm.
(b).

Here we can find the area of the shaded region by subtracting the area of the rectangle from the area of the circle.
We can now find the area of the circle.
Here we have to find the radius by finding the diagonal of the rectangle(which is the diameter of the circle) using the Pythagoras theorem.
\[\begin{align}
  & \Rightarrow Diagonal=\sqrt{{{\left( 5 \right)}^{2}}+{{\left( 12 \right)}^{2}}}=\sqrt{25+144} \\
 & \Rightarrow Diagonal=\sqrt{169}=13cm \\
\end{align}\]
We know that the radius is half the diameter, so we can write as,
\[\Rightarrow Radius=\dfrac{13}{2}\]
We can now write the area of circle \[\pi {{r}^{2}}\], we get
\[\Rightarrow Area=\dfrac{22}{7}\times \dfrac{13}{2}\times \dfrac{13}{2}=132.78c{{m}^{2}}\]
We can now subtract the Area of rectangle \[l\times b\] from the area of the circle, we get
Area of shaded region = Area of circle – Area of rectangle.
\[\Rightarrow 132.78 - 12\times 5=72.78c{{m}^{2}}\]
Therefore, the area of the shaded region is 72.78sq.cm.
(c).

Here we can find the area of each semicircle and subtract the unshaded region from the shaded ones to get the area of the shaded region.
We know that the area of a semicircle is \[\dfrac{\pi {{r}^{2}}}{2}\].
Here we have four semicircles, where the largest one has 14cm radius, and the below semicircle has 7cm as radius and the remaining 2 semicircles have 3.5 as the radius.
We can now subtract the unshaded region form the shaded ones,
Area of shaded region = Area of shaded semicircles – Area of unshaded semicircles
\[\Rightarrow \dfrac{\pi {{\left( 14 \right)}^{2}}}{2}+\dfrac{\pi {{\left( 7 \right)}^{2}}}{2}-2\times \dfrac{\pi {{\left( 3.5 \right)}^{2}}}{2}\]
We can now simplify the above step, we get
\[\begin{align}
  & \Rightarrow \dfrac{22}{7}\times \dfrac{{{\left( 14 \right)}^{2}}}{2}+\dfrac{22}{7}\times \dfrac{{{\left( 7 \right)}^{2}}}{2}-2\times \dfrac{22}{7}\times \dfrac{{{\left( 3.5 \right)}^{2}}}{2} \\
 & \Rightarrow 308+77-38.5=346.5c{{m}^{2}} \\
\end{align}\]
Therefore, the area of the shaded region is 346.5sq.cm.

Note: We should always remember that the area of the square formula is \[{{a}^{2}}\] and area of quadrant is \[\dfrac{\pi {{r}^{2}}}{4}\], the Area of rectangle \[l\times b\] and the area of circle \[\pi {{r}^{2}}\]. The area of the semicircle is \[\dfrac{\pi {{r}^{2}}}{2}\]. We should also remember that the diameter is twice the radius of the circle.