Question

From a cuboidal solid metallic block, of dimensions 15 cm × 10 cm × 5 cm, a cylindrical hole of diameter 7 cm is drilled out. find the surface area of the remaining block

Answer 1

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Answer 2

The surface area of the remaining block 583 $\text{cm}^{2}$.

Step-by-step explanation:

We are given that from a cuboidal solid metallic block, of dimensions 15 cm × 10 cm × 5 cm, a cylindrical hole of diameter 7 cm is drilled out.

Let the length of a cuboidal solid metallic block = L = 15 cm

the breadth of a cuboidal solid metallic block = B = 10 cm

the height of a cuboidal solid metallic block = H = 5 cm

As we know that the total surface of the cuboid is given by;

The Total surface area of cuboid = $2(\text{LB} + \text{BH}+\text{HL})$

                                  = $2[(15 \times 10) + (10 \times 5)+(5 \times 15)]$

                                  = $2\times (150 + 50 +75)$

                                  = $2 \times 275$ = 550 $\text{cm}^{2}$

Now, it is stated that a cylindrical hole of diameter 7 cm is drilled out, so;

The curved surface area of a cylindrical hole = $2 \pi rh$

Here, h = height of cylinder = 5 cm

          r = radius of the hole = $\frac{\text{Diameter}}{2}$  = $\frac{7}{2}$  = 3.5 cm

So, the curved surface area of a cylindrical hole = $2 \times \frac{22}{7}\times 3.5 \times 5$

                                                                                = 110 $\text{cm}^{2}$

Also, the area of the two bases of a cylindrical hole which is in the form of the circle = $\pi r^{2} + \pi r^{2}$

                = $\frac{22}{7}\times 3.5^{2} +\frac{22}{7}\times 3.5^{2}$

                = 77 $\text{cm}^{2}$

Now, the surface area of the remaining block = Total surface area of the cuboidal metallic block + The curved surface area of a cylindrical hole - Area of two circles

= 550 $\text{cm}^{2}$ + 110 $\text{cm}^{2}$ - 77 $\text{cm}^{2}$  = 583 $\text{cm}^{2}$.