Bob finds a rectangular sheet of paper and measures its diagonal to be 8cm. Bob then cuts out equal squares of side 1cm at two opposite corners of the paper and measures the distance between the two closest vertices of these squares to be 4√2cm. What is the area of the original rectangular paper?
call the sizes a and b. If the diagonal is 8cm then a^2 + b^2 = 8^2 = 64.
After the cutouts, the square of the distance between the closest vertices of the cut-out squares is (a-2)^2 + (b-2)^2, since each side has been reduced by 1cm from each cut-out. If that distance is 4√2cm then its square is 32cm. That gives us two equations in two unknowns.
We can expand the second, and then remove the quadratic terms by substituting the first:
a^2 - 4a + 4 + b^2 - 4b + 4 = 32
64 - 4(a + b) + 8 = 32
-4(a+b) = 32 - 8 - 64 = -40
(a+b) = 10
Now use b = 10-a in the first equation:
a^2 + (10-a)^2 = 64
a^2 + 10^2 - 20a + a^2 = 64
2a^2 - 20a + 36 = 0
a^2 - 10a + 18 = 0
Since the sum of the two roots is 10 from the linear coefficient, and since a + b is 10, then the two roots just exchange the labels a and b. The sides, then are 5 +/- 1/2√(10^ - 4*18) = 5 +/- √7.
The area of the original rectangle is (5 + √7)(5 - √7) = 25 - 7 = 18cm^2
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Posted by Paul
on 2023-08-28 15:10:15 |
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