Question

With the vertices A, B and C of a triangle ABC as centres, arcs are drawn with radii 5 cm each as shown in the given figure. If AB = 14 cm, BC = 48 cm and CA = 50 cm then find the area of the shaded region. [Use π = 3.14]

Answer 1

Given that ABC is a triangle with sides AB = 14 cm, BC = 48 cm, CA = 50 cm.
Clearly it is a right angled triangle.
Area of $△\mathrm{ABC}=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}=\frac{1}{2}\times 48\times 14=336{\mathrm{cm}}^2.$

Now we need to remove the area of 3 arcs each of radius 5 cm from this area of triangle.
Here, $\angle \mathrm{B}=90°,\angle \mathrm{A}+\angle \mathrm{C}=90°.$
The area of these 3 arcs would exactly be equal to a sector of circle with radius 5 cm and angle = $\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180°$ i.e. a semi circle.
Since we know that area of semi circle is $\frac{1}{2}\mathrm{\pi }{r}^2=\frac{1}{2}\times 3.14\times 5\times 5=\frac{1}{2}\times \frac{314}{100}\times 25=\frac{314}{8}=\frac{157}{4}{\mathrm{cm}}^2.$
Hence, the area of shaded region is = $$\begin{gathered} \mathrm{area}\mathrm{of}△\mathrm{ABC}-\mathrm{combined}\mathrm{area}\mathrm{of}3\mathrm{arcs}=336-\frac{157}{4} \\ =336-39.25=296.75{\mathrm{cm}}^2. \end{gathered}$$