Asked by Aleena
One corner of a square is folded to its centre to form an irregular pentagon.the areas of the pentagon and of the square are consecutive integers.what is the area of the square?
Answers
Answered by
Reiny
Make a sketch.
It is easy to see that the height of the triangle that is folded over is 1/4 of the diagonal of the square, so ....
let the side of the square be x
then its diagonal is √2x
and the height of the triangle bent over is √2x/4
and the fold at which the bending takes place would be √2x/2 (we have 45° angles)
area of triangle bent over = (1/2)(√2x/2)(√2x/4)
= 2x^2/16 = x^2 /8
area of pentagon = x^2 - x^2/8 = 7x^2/8
x^2 - 7x^2/8 = 1 (difference between consecutive integers is 1)
times 8
8x^2 - 7x^2 = 8
x^2 = 8
the area of the triangle is x^2/8 = 8/8 = 1 square unit
It is easy to see that the height of the triangle that is folded over is 1/4 of the diagonal of the square, so ....
let the side of the square be x
then its diagonal is √2x
and the height of the triangle bent over is √2x/4
and the fold at which the bending takes place would be √2x/2 (we have 45° angles)
area of triangle bent over = (1/2)(√2x/2)(√2x/4)
= 2x^2/16 = x^2 /8
area of pentagon = x^2 - x^2/8 = 7x^2/8
x^2 - 7x^2/8 = 1 (difference between consecutive integers is 1)
times 8
8x^2 - 7x^2 = 8
x^2 = 8
the area of the triangle is x^2/8 = 8/8 = 1 square unit