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The longer base of an isosceles trapezoid is equal in length to a diagonal of the trapezoid. The shorter base is equal in length to the altitude of the trapezoid. Show that the ratio of the bases is 5:3.
1 Answer
- falzoonLv 71 decade agoFavorite Answer
Let y = the small base and x = the large base.
Draw the two altitudes from each corner of the small base.
The large base will then be made up of a section of length y
and two end sections each of length (x - y)/2.
Draw a diagonal which is length x. Make this the hypotenuse
of a triangle which has legs of length y and [y + (x - y)/2].
Therefore, by Pythagoras,
x^2 = y^2 + [y + (x - y)/2]^2
= y^2 + [y + x/2 - y/2]^2
= y^2 + {x/2 + y/2)^2
= y^2 + [(x + y)/2]^2
= y^2 + (x + y)^2 / 4
Multiply through by 4 :
4x^2 = 4y^2 + (x + y)^2
= 4y^2 + x^2 + 2xy + y^2
Simplify :
3x^2 = 5y^2 + 2xy
Divide through by xy :
3x/y = 5y/x + 2
3(x/y) = 5(y/x) + 2
3(x/y) = 5 / (x/y) + 2
Let x/y = k
so, 3k = 5/k + 2
Multiply through by k :
3k^2 = 5 + 2k
Rearrange :
3k^2 - 2k - 5 = 0
Factorise :
(k + 1)(3k - 5) = 0
Thus, k = -1 or 5/3
That is, x/y = -1 or 5/3
But x/y cannot be negative, so x/y = 5/3, or x:y = 5:3.