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 am and bm tall, what height would the intersection of the lines be?
If the heights of the two poles were am and bm 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 = ¼π(2r)2 - ½πr2 - (½πr2 - B)
R = πr2 - ½πr2 - ½πr2 + B
R = 0 + B
R = B
R = Area of quarter circle - Area of semicircle - (Area of semicircle - B)
R = ¼π(2r)2 - ½πr2 - (½πr2 - B)
R = πr2 - ½πr2 - ½πr2 + 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 2L (I used 2L 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×(2L)2×L√2.
This simplifies to 4/3√2L3.
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×(2L)2×L√2.
This simplifies to 4/3√2L3.
Next, we find the area of the tetrahedron.
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/√3L.
Using Pythagoras' Theorem yet again, we find that the height of the tetrahedron is L2√2/√3
Therefore, the volume of the tetrahedron is ⅓×½×2L×L√3×L2√2/√3.
This simplifies to 2/3√2L3.
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/√3L.
Using Pythagoras' Theorem yet again, we find that the height of the tetrahedron is L2√2/√3
Therefore, the volume of the tetrahedron is ⅓×½×2L×L√3×L2√2/√3.
This simplifies to 2/3√2L3.
Finally, the ratio of volume of the square based pyramid to the tetrahedron is:
4/3√2L3 : 2/3√2L3
2 : 1
4/3√2L3 : 2/3√2L3
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 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?
What is the smallest number m such that m! seconds is a whole number of weeks?
No comments:
Post a Comment