# Question

How to do both qns?
Tks

(a) Since 15cm is made up from sum of 5 circular discs’ diameter,

Radius of a disc = 15/10 cm = 1.5cm

Area of a disc = 3.14 ( 1.5)² = 7.065 cm²

(b) After discounting 3cm length of the first layer, there were (19 – 3)/ 2.5= 6.4 layer or 6 layers possible not counting the first layer.

Total of 7 layers means there are total = 5 + (4+5)*3 = 32 discs possible.

So the area left uncovered = 19 * 15 – 32 * 7.065 = 285 – 226.08 = 58.92 cm²

Don’t forget you can alternate 5 discs and 4 discs per row.

Dear Autolytus,

Tks for pointing out the mistake!

We’re all in this together! 🙂

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From diagram, height of disc (see bottom row) is 3 cm, so diameter is 3 cm, radius = 1.5 cm.

(a)

Area of disc = 1/4 x π x d2 (or just πr2)= 2.25π cm2 = 7.065 cm2.

(b)

We’ve already taken up 3+2.5+2.5 = 8 cm of height, so 11 cm height left.

That’s another 4 rows max, so total discs = 5 + 4 + 5 + 4 + 5 + 4 + 5 = 32 discs.

Total disc area = 32 x 7.065 = 226.08 cm2.

Area left uncovered = (15 x 19) – 226.08 = 58.92 cm2.

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