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Length of three perpendiculars from three heads on the three arms of a triangle is given. How to construct that triangle?
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hi MD Muzibur Rahman
Welcome to the forum.
At the moment I cannot see a solution to this, but I'll work on it. Here's my thinking so far:
Let's say the sides are a, b and c; and the perpendiculars are h, i and j.
Because the area of the triangle is fixed we know that ah = bi = cj so dividing by hij gives
So it looks like you can find a way to get a, b and c from the known h, i and j. {Maybe the sine rule?)
More when I've got it 'sorted'.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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I've come up with this method, which is rather clumsy but does work.
I'll show with an example.
Say h = 6, I = 8 and j = 5.
Then
Choose a = 20 (say) => b = 15 and c = 24
Construct that triangle and find the 'h' for that triangle. I got 14.95.
G is the intersection of the altitudes. Make G the centre of a dilation (enlargement) with scale factor 6/14.95
Construct the new triangle. It has the required properties. Here it is:
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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