
Find the area of a cardboard wasted if a sector of maximum possible size is cut out from a square cardboard of size 24 cm. $\left[ \pi =3.14 \right]$.
Answer
511.5k+ views
Hint: Here, first draw the diagram and then subtract the area of the sector which is one-quarter of the circle from the area of the square. Find the area of the cardboard wasted after it is cut off maximum possible size or sector.
Complete step-by-step solution:
Let us first draw the figure to understand the question where we have square cardboard of size 24cm. Here, we need to draw a figure of a square and inscribed in it is a sector of a circle, where $\pi =3.14$. So, the two opposite corners become the ends of an arc such that the radius is 24 cm.
Here the unshaded region is a sector of the circle with a radius of 24 cm.
We need to find the area of the cardboard wasted when the maximum possible sector is cut off or according to the figure, find the area of the shaded region.
Therefore, area required = area of the square – area of the sector ........... (i)
We know,
Area of square = side $\times $ side
= 24 $\times $ 24
= 576 sq. cm.
The area of the sector will be the quarter of the area of the circle. If you can observe, that the sector which will be cut off is a quarter of a complete circle of radius 24 cm.
Therefore, Area of the sector = $\dfrac{1}{4}$$\left( \pi {{r}^{2}} \right)$
$\begin{align}
& =\dfrac{1}{4}\times \pi \times {{\left( 24 \right)}^{2}} \\
& =\dfrac{1}{4}\times \pi \times 576 \\
& =\dfrac{1}{4}\times 1809.56 \\
& =452.38 \\
\end{align}$
Hence, the area of the sector is 452.38 sq. cm.
Now, we have the area of the sector and the area of the square, let us substitute in equation (i) and find the value of the area of the required cut off-board.
Area required = 576 – 452.38
= 123.62 sq. cm.
Hence, the area of the cardboard wasted if a sector of maximum possible size is cut out from a square cardboard is 123.62 sq. cm.
Note: Here, the description of the diagram is important to understand the question. Also, remember to input the units to avoid any deduction in marks. You can either input sq. cm. or $c{{m}^{2}}$. Here from the maximum possible size means we cut the arc by taking the maximum size of the side of given square as radius which is 24. we can use the derivative for maxima and minima by assuming the radius of arc x but that would be a long way to get the solution.
Complete step-by-step solution:
Let us first draw the figure to understand the question where we have square cardboard of size 24cm. Here, we need to draw a figure of a square and inscribed in it is a sector of a circle, where $\pi =3.14$. So, the two opposite corners become the ends of an arc such that the radius is 24 cm.

Here the unshaded region is a sector of the circle with a radius of 24 cm.
We need to find the area of the cardboard wasted when the maximum possible sector is cut off or according to the figure, find the area of the shaded region.
Therefore, area required = area of the square – area of the sector ........... (i)
We know,
Area of square = side $\times $ side
= 24 $\times $ 24
= 576 sq. cm.
The area of the sector will be the quarter of the area of the circle. If you can observe, that the sector which will be cut off is a quarter of a complete circle of radius 24 cm.
Therefore, Area of the sector = $\dfrac{1}{4}$$\left( \pi {{r}^{2}} \right)$
$\begin{align}
& =\dfrac{1}{4}\times \pi \times {{\left( 24 \right)}^{2}} \\
& =\dfrac{1}{4}\times \pi \times 576 \\
& =\dfrac{1}{4}\times 1809.56 \\
& =452.38 \\
\end{align}$
Hence, the area of the sector is 452.38 sq. cm.
Now, we have the area of the sector and the area of the square, let us substitute in equation (i) and find the value of the area of the required cut off-board.
Area required = 576 – 452.38
= 123.62 sq. cm.
Hence, the area of the cardboard wasted if a sector of maximum possible size is cut out from a square cardboard is 123.62 sq. cm.
Note: Here, the description of the diagram is important to understand the question. Also, remember to input the units to avoid any deduction in marks. You can either input sq. cm. or $c{{m}^{2}}$. Here from the maximum possible size means we cut the arc by taking the maximum size of the side of given square as radius which is 24. we can use the derivative for maxima and minima by assuming the radius of arc x but that would be a long way to get the solution.
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