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A rectangular seminar hall has a length of 20 ft and a breadth of 16 ft.

(a) Find the least common multiple (LCM) of 16 and 20.

(b) What is the side length of the largest square marble tile that can be used to pave the entire hall with tiles of the same size?

Sagot :

GinnyAnswer
Certainly! Let's break down and solve the problem step-by-step.

### Part (a): Find the least common multiple (LCM) of 16 and 20.

Step 1: Recall the definition of the least common multiple. The LCM of two numbers is the smallest positive integer that is divisible by both numbers.

Step 2: To find the LCM of two numbers [tex]\(a\)[/tex] and [tex]\(b\)[/tex], we can use the formula:

[tex]\[ \text{LCM}(a, b) = \frac{a \times b}{\text{GCD}(a, b)} \][/tex]

where GCD stands for the greatest common divisor.

Step 3: For this problem, the length of the hall is 20 ft, and the breadth is 16 ft.

Using the provided answer, we know:
[tex]\[ \text{LCM}(16, 20) = 80 \][/tex]

So, the least common multiple of 16 and 20 is 80 ft.

### Part (b): Find the length of the biggest square marble needed to pave the hall with square marbles of the same size.

Step 1: Recall that to find the size of the largest square tile that can exactly fit both dimensions of the hall, we need to find the greatest common divisor (GCD) of the two dimensions.

Step 2: The GCD of two numbers [tex]\(a\)[/tex] and [tex]\(b\)[/tex] is the largest positive integer that divides both [tex]\(a\)[/tex] and [tex]\(b\)[/tex] without leaving a remainder.

Step 3: The length of the hall is 20 ft, and the breadth is 16 ft.

Using the provided answer, we know:
[tex]\[ \text{GCD}(16, 20) = 4 \][/tex]

So, the length of the biggest square marble needed to pave the hall is 4 ft.

### Summary:
(a) The least common multiple (LCM) of 16 and 20 is 80 ft.
(b) The length of the biggest square marble needed to pave the hall with square marbles of the same size is 4 ft.
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