This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

#### Burning Ropes

Light one rope at both ends and the other at one end. When the first rope finishes burning, light the other end of the second rope. The second rope will finish burning 45 minutes after the start.

##### Extension

What are all the possible times you can measure with two ropes? How about three ropes? Four ropes?

*n*ropes?#### Pole Position

The two poles can be any distance apart; the distance does not affect the heights.

##### Extension

If the heights of the two poles were 12m and 24m tall, what height would the intersection of the lines be?

If the heights of the two poles were

If the heights of the two poles were

*a*m and*b*m tall, what height would the intersection of the lines be?#### Circles

Call the blue area

*B*and the red area*R*.*R*= Area of quarter circle - Area of semicircle - (Area of semicircle -*B*)*R*= ¼*π*(2*r*)^{2}- ½*π**r*^{2}- (½*π**r*^{2}-*B*)*R*=*π**r*^{2}- ½*π**r*^{2}- ½*π**r*^{2}+*B**R*= 0 +*B**R*=*B*##### Extension

Prove that the red and blue areas are equal.

#### Pyramid and Tetrahedron

Let the length of a side of a triangle be 2

By Pythagoras' Theorem, the height of a triangle is

Using Pythagoras' Theorem again, the height of the square-based pyramid is

Therefore, the volume of the square-based pyramid is ⅓×(2

This simplifies to

*L*(I used 2*L*and not L to get rid of fractions in the calculations).By Pythagoras' Theorem, the height of a triangle is

*L*√3.Using Pythagoras' Theorem again, the height of the square-based pyramid is

*L*√2.Therefore, the volume of the square-based pyramid is ⅓×(2

*L*)^{2}×*L*√2.This simplifies to

**.**^{4}/_{3}√2*L*^{3}
Next, we find the area of the tetrahedron.

Call the point on the base of the tetrahedron, directly below the vertex at the top

Using cosine in the triangle made by

Using Pythagoras' Theorem yet again, we find that the height of the tetrahedron is

Therefore, the volume of the tetrahedron is ⅓×½×2

This simplifies to

Call the point on the base of the tetrahedron, directly below the vertex at the top

*A*.Using cosine in the triangle made by

*A*, the corner of the base and the midpoint of a side of the base, the distance from the corner to*A*is^{2}/_{√3}*L*.Using Pythagoras' Theorem yet again, we find that the height of the tetrahedron is

*L*^{2√2}/_{√3}Therefore, the volume of the tetrahedron is ⅓×½×2

*L*×*L*√3×*L*^{2√2}/_{√3}.This simplifies to

**.**^{2}/_{3}√2*L*^{3}
Finally, the ratio of volume of the square based pyramid to the tetrahedron is:

2 : 1

^{4}/_{3}√2*L*^{3}:^{2}/_{3}√2*L*^{3}2 : 1

##### Extension

What would the ratio be if they were isosceles triangles?

#### 8! minutes

8×7×6×5×4×3×2×1 minutes

= 8×7×4×3×1 hours (dividing by 60)

= 7×4×1 days (dividing by 24)

= 4×1 weeks (dividing by 7)

= 4 weeks

= 8×7×4×3×1 hours (dividing by 60)

= 7×4×1 days (dividing by 24)

= 4×1 weeks (dividing by 7)

= 4 weeks

##### Extension

8 is the smallest number

What is the smallest number

*n*such that*n*! minutes is a whole number of weeks.What is the smallest number

*m*such that*m*! seconds is a whole number of weeks?
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