Brain-Teaser 20: The kite
From The Sunday Times, 6th August 1961 [link]
Some problems that look simple enough at first can prove to be remarkably tricky. Consider, for instance, the kite pictured above. The shaded areas are squares of equal size, the sides of each square being 15 inches. The width of the kite from A to B is exactly 42 inches.
What is the length of the kite from C to D?
[teaser20]
John Crabtree and 
S2T2 
Jim Randell 9:29 am on 7 February 2021 Permalink |
The figure is a right kite [ @wikipedia ], so is composed of 2 identical right-angled triangles joined along their common hypotenuse CD (which is the length we wish to determine).
Writing: AC = BC = a, and AD = BD = b, then the value we wish to determine, CD, is:
We are told the other diagonal, AB, has length 42, hence:
And the sides of the squares that run from the centre to the edges of the kite are radii of the incircle, so:
Writing: x = ab, y = a + b, we get:
Also:
So:
So the value of CD is:
Solution: CD = 43.75 inches.
And the lengths of the sides of the kite are 26.25 in and 35 in.
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John Crabtree 10:22 pm on 8 February 2021 Permalink |
Let the mid-point of AB be F, and the point where the squares meet be G
By Pythagoras, FG = √(2 * 225 – 441) = 3
Let z be the angle formed by GAF.
Tan(z) = t = 3 / 21 = 1 / 7
CD = 21 [tan(45 + z) + tan(45 – z)] = 21 [(1+t)/(1-t) + (1-t)/(1+t))
= 21 [4/3 + 3/4] = 21 * 25/12 = 43.75 inches.
AC = 21 * 5/4 = 26.25 in. AD = 21 * 5/3 = 35 in.
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