lmagine an equilateral triangle and a square. The triangle has sides of 10 m and the square has sides of 15 m. One ant walks completely round the triangle twice and a beetle which walks two and a half times as fast starts walking at the same time around the square. How many laps of the square would the beetle have made in the same time ? .....

perimters:

triangle = 30
square: 60

ant walks 60 m
in same time, beetle walks 2.5*60 = 150 m
150/60 = 2.5 laps

or, just notice that the distance walked by the ant is one lap of the square, so the beetle, walking 2.5 times as fast, walks 2.5 laps on the square.

To solve this problem, let's start by finding the distance covered by the ant when walking around the triangle twice.

An equilateral triangle has equal sides, so each side has a length of 10 m. Since the ant walks completely around the triangle twice, it covers a total distance of 10 m x 3 = 30 m.

Next, let's find the distance covered by the beetle when walking around the square.

A square has equal sides, so each side has a length of 15 m. The beetle walks around the square two and a half times faster than the ant. This means that the beetle covers a total distance of 15 m x 4 x 2.5 = 150 m.

Now, to determine how many laps of the square the beetle would have made in the same time, we can divide the total distance covered by the beetle (150 m) by the distance covered in one lap of the square (15 m).

150 m ÷ 15 m = 10 laps

Therefore, the beetle would have made 10 laps of the square in the same time it took the ant to walk around the triangle twice.