SOLUTION: a rectangle is drawn so the width is 9 inches longer than the height. If the rectangle's diagonal measurement is 45 inches, find the height

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Question 1167042: a rectangle is drawn so the width is 9 inches longer than the height. If the rectangle's diagonal measurement is 45 inches, find the height
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52778)   (Show Source): You can put this solution on YOUR website!
.

H = height

length = H+9


Diagonal 


    H^2 + (H+9)^2 = 45^2


    2H^2 + 18H + 81 = 2025

    2H^2 + 18H - 1944 = 0

     H^2 + 9H - 972 = 0


Factor


    (H-27)*(H+36) = 0


The roots are  H= 27  and  H= -36.


Only positive root works.


ANSWER.  The height is 27 inches.

Solved.

--------------

A post-solution notice 

    This triangle is nothing else as a famous (3,4,5)-right angled triangle with the similarity factor of 9.



Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


The side lengths are x and x+9.

Two sides and the diagonal form a right triangle. Solve using the Pythagorean Theorem.






Solve by factoring, or using the quadratic formula, or with a graphing calculator....

If a formal algebraic solution is not required, the solution can be found quickly knowing that the side lengths are integers -- so the three side lengths are a Pythaogrean Triple.

The smallest Pythagorean Triple is 3-4-5. Since the diagonal of the rectangle (hypotenuse of the right triangle) is 45, try 27-36-45 for the side lengths of the triangle.

Those side lengths satisfy the conditions of the problem....

ANSWER: The height is 27; the width is 36.
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