Draw on rope tricks to guess the area: The Wknd puzzle with Dilip D’Souza

There’s a square within a circle, a boundary all around. It’s a few simple rules of mathematics to guess the measures involved, but can you crack the code?

Since it’s cricket season, two quick and relatively easy puzzles about the Brabourne Stadium, the venue of many memorable Test matches. None since early 1975, sadly, but never mind that.

1) You tie one end of a rope at the boundary of the Brabourne, walk across to the diametrically opposite point on the boundary, pull the rope tight and tie it down there. You measure the length of the rope between the two knots: exactly 100m.

Then you walk back to the centre of the field (assume it is a perfect circle). You want to raise the rope enough right there for the visiting basketball star, Kevin Durant, to walk under. Let’s say he’s 2m tall. Clearly you have to loosen the rope, lengthen it and tie it down again.

But by how much do you need to lengthen it?

a) 1cm b) 10cm c) 1m d) 10m e) 100m

2) A prankster has suspended a massive rigid plastic square above the stadium, perpendicular to the ground. Three of its corners are 5m, 8m and 9m above the ground. The fourth corner is the highest. But how far off the ground is it?

Bonus: What’s the area of the square?

Hints:

1) At the centre, you are 50m from the boundary. The rope has to be 2m off the ground. Imagine the right-angled triangle that’s formed. Two of its sides are 2m and 50m, so what’s the third side (the hypotenuse)? Think Pythagoras.

2) Best to draw a picture. You’ll find more imaginary right-angled triangles. Between the lowest corner and the second-lowest, a height difference of 3m. Between the lowest and the second-highest, 4m.

Scroll down for the answers.

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Answers:

1) b) 10cm. Actually 8cm will be enough. By Pythagoras, the third side is about 50.04m — 4cm more than 50m. Multiply by 2 to get 8cm.

2) On the right-angled triangles, the two shorter sides are 3m and 4m. So the highest corner is 3m higher than the second-highest - 9 + 3 = 12m off the ground.

Bonus: By Pythagoras, the third side of the right triangles — the side of the square — is 5m. Thus the square’s area is 5 x 5 = 25 sq m.