Question

Four circular cardboard pieces of radii 7 cm are placed on a paper in such a way that each piece touches other two pieces. Find the area of the portion enclosed between these pieces.

• A. $42c{m}^2$
• B. $48c{m}^2$.
• C. $22c{m}^2$.
• D. $38c{m}^2$.

Answer 1

The correct option is A

$42c{m}^2$

Given, that four circular cardboard pieces arc placed on a paper in such a way that each piece touches other two pieces.

Now, we join centre of all four circles to each other by a line segment since, radius of each circles is 7 cm.

So, $AB=2\times Radiusofcircle$

$=2\times 7=14cm$

$\Rightarrow AB=BC=CD=AB=4cm$

Which shows that, quadrilateral ABCD is a square with each of its side is 14 cm.

We know that, each angle between two adjacent sides of a square is ${90}^{\circ }$.

$\therefore$ Area of sector with $\angle A={90}^{\circ }$

$\frac{\angle A}{{360}^{\circ }}\times \pi r^2=\frac{{90}^{\circ }}{{360}^{\circ }}\times \pi \times (7{)}^2$

$=\frac{1}{4}\times \frac{22}{7}\times 49=\frac{154}{4}=\frac{77}{2}$

$=38.5c{m}^2$

$\therefore$ Area of all four sectors $=4\times Areaofsectorwith\angle A$

$=4\times 38.5$

$=154c{m}^2$

And area of square ABCD $=(sideofsquare{)}^2$

$=(14{)}^2=196c{m}^2$ $[\therefore Areaofsquare=(side{)}^2]$

So, area of shaded region enclosed between these pieces = Area of square ABCD - Area of each sector

$=196-154$

$=42c{m}^2$

Hence, required area of the portion enclosed between these pieces is $42c{m}^2$.