SOLUTION: A rectangle is inscribed in a square so that each vertex of the rectangle is located on one side of the square, and the sides of the rectangle are parallel to the diagonals of the

Algebra ->  Rectangles -> SOLUTION: A rectangle is inscribed in a square so that each vertex of the rectangle is located on one side of the square, and the sides of the rectangle are parallel to the diagonals of the       Log On


   

Question 1113575: A rectangle is inscribed in a square so that each vertex of the rectangle is located on one side of the square, and the sides of the rectangle are parallel to the diagonals of the square. Suppose that one side of the rectangle is twice the length of the other and that the diagonal of the square is 12 meters long. Find the sides of the rectangle. Do it in statement reason.
Found 3 solutions by rothauserc, math_helper, greenestamps:
Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
1. use Pythagorean Theorem to determine y^2 where y is the length of one side of the square since the interior angles of the square are right angles
:
y^2 + y^2 = 12^2
:
2y^2 = 144
:
y^2 = 72
:
2. use Pythagorean Theorem to determine x since the interior angles of the rectangle are right angles
:
(2x)^2 + x^2 = y^2
:
4x^2 + x^2 = 72
:
5x^2 = 72
:
x^2 = 72/5
:
x = 6square root(2/5) = 3.79 meters
:
*************************************
length of rectangle is 7.58 meters
width of rectangle is 3.79 meters
*************************************
:

Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!


Given information:
(i) The diagonal of the square is given as 12m —> sides are 6sqrt%282%29 m
(ii) L = 2W for the inscribed rectangle.
—
I used areas to find the dimensions of the rectangle:
( area of the rectangle ) + ( area of the 4 triangles ) = ( area of the square )
—
LW + Area(4 triangles) = +%28+6sqrt%282%29+%29%5E2+ = 72 (eq 1)
—
The short legs of the small triangle are of length k, where k%5E2+%2B+k%5E2+=+W%5E2+ —> +k%5E2+=+W%5E2%2F2+
so the area of one small triangle = +%281%2F2%29%28k%5E2%29+=+%281%2F2%29%28W%5E2%2F2%29+=+%281%2F4%29W%5E2+
Area of both small triangles = +%281%2F2%29W%5E2+
—
By a similar approach, the area of both large triangles is:
+%281%2F2%29L%5E2+ = +%281%2F2%29%282W%29%5E2+=++2W%5E2+
—
Substituting the triangle areas (and L=2W), (1) becomes:
+%282W%29%28W%29+%2B+%281%2F2%29W%5E2+%2B+2W%5E2+=+72+
+%289%2F2%29W%5E2++=+72+
++W%5E2+=+144%2F9+=+16+ —> W=4 —> L=8
—
Ans: The length and width of the inscribed rectangle are +highlight%28+L=8m+%29 and +highlight%28+W=4m%29+
—
Check by summing areas:
Small triangles: +%281%2F2%29%284%5E2%29+=+8+
Large triangles: +2%284%5E2%29+=+32+
Rectangle: ++4%2A8+=+32+
8+32+32 = 72 = area of square, looks ok.





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


The solution by tutor rothauserc is not right. It appears his calculations show the length of the diagonal of the rectangle is the same as the length of a side of the square, which is not right....

The solution by tutor math helper is correct.

Here is another solution that is essentially the same as his but in a form that requires much less work.

(1) The diagonal of the square is 12; that means the side of the square is 6*sqrt(2).
(2) The triangles (outside the rectangle and inside the square) are both isosceles right triangles, and we know the hypotenuse of the larger triangles is twice the hypotenuse of the smaller triangles. That means the legs of the larger triangles are twice the length of the legs of the smaller triangles.
(3) The 6*sqrt(2) length of a side of the square is the sum of the lengths of a leg of the larger triangle and a leg of the smaller triangle. So the leg of the larger triangle is 4*sqrt(2) and the leg of the smaller triangle is 2*sqrt(2).
(4) Then the hypotenuses of the triangles (the lengths of the sides of the rectangle) are 8 and 4.