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junior college 2 | H2 Maths
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Hope for a concise solution (not brute force counting).
Date Posted: 4 years ago
Views: 352
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answer ...
(i) 1620 rectangles
(ii) 608 triangles
(i) 1620 rectangles
(ii) 608 triangles
is there any possibility of double counting the triangles? because the ans given to me is 544 (or my given ans could be wrong).
Edit for clarification: Rectangles are correct.
Edit for clarification: Rectangles are correct.
you are right! some of the triangles have been double-counted. smart of you to have reasoned that the difference is due to double counting. sorry that i missed it earlier.
to account for the overlaps, we add as follows ...
case 5. triangles with base 8, height 4, base vertical, vertex above base, and 1 of the other 2 side is vertical
no of ways = 6 x 1 x 1 x 2 = 12
[case 5 accounts for overlap of cases 1 & 4]
case 6. triangles with base 4, height 8, base vertical, vertex above base, and 1 of the other 2 side is vertical
no of triangles = 2 x 5 x 1 x 2 = 20
[case 6 accounts for overlap of cases 2 & 3]
total ways = (54+100+90+60-12-20) x 2
= 544
thanks for highlighting the mistake ...
to account for the overlaps, we add as follows ...
case 5. triangles with base 8, height 4, base vertical, vertex above base, and 1 of the other 2 side is vertical
no of ways = 6 x 1 x 1 x 2 = 12
[case 5 accounts for overlap of cases 1 & 4]
case 6. triangles with base 4, height 8, base vertical, vertex above base, and 1 of the other 2 side is vertical
no of triangles = 2 x 5 x 1 x 2 = 20
[case 6 accounts for overlap of cases 2 & 3]
total ways = (54+100+90+60-12-20) x 2
= 544
thanks for highlighting the mistake ...
Sorry haven’t been checking ans. how does it look like to have base vertical and vertex above base? or is vertex left or right of base?
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