How far is O, where the diagonals cross, from AD?
A hard-level "Length" problem.The "hourglass" made by crossing diagonals is always similar, as long as there are parallel sides. The ratio of the two parallel sides fixes where the crossing point sits.
GivenAD is parallel to BC. AD = 6 cm, BC = 10 cm, height of the trapezium = 24 cm
Hints
- The one and only way in is that AD and BC are parallel. Let's thicken those two sides first.
- Look at the two triangles facing each other across the crossing point: red is AOD, blue is COB. Together they look like an hourglass.
- Because AD and BC are parallel, the alternate angles are equal: the two red-marked angles match, and so do the two blue-marked ones.
- If two angles match, the third must too — so the red and blue triangles are the same shape (similar). Only the size differs, in the ratio AD : CB = 6 : 10 = 3 : 5.
- Same shape means the heights are in the same ratio, 3 : 5. So O cuts the 24 cm height in exactly that ratio.
- Now just split 24 cm in the ratio 3 : 5. That is 3+5 = 8 parts, so one part is 24 ÷ 8 = 3 cm, and the top is 3 of them.
Answer 9cm
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