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Problem of the Week
Problem D and Solutions
Made in the Shade
B
A
Problem
Rectangle ABCD has length AB = 9 and width AD = 5. Diagonal BD
is divided into 5 equal parts at W , X, Y , and Z. Determine the area of
the shaded region.
D
X
W
Y
Z
C
Solution 1
Using the formula for area of a triangle =
base×height
,
2
we have area 4ABD =
9×5
2
=
45
2
units2 .
The triangles 4DAW, 4W AX, 4XAY, 4Y AZ and 4ZAB have the same height. Since
DW = W X = XY = Y Z = ZB, the triangles also have equal bases. Therefore,
area 4DAW = area 4W AX
= area 4Y AZ = area 4ZAB
=9 area 4XAY
2
1
1 45
= 5 (area 4ABD) = 5 2 = 2 units .
Similarily, the area of 4BCD is
9×5
2
=
45
2
units2 .
The triangles 4DCW, 4W CX, 4XCY, 4Y CZ and 4ZCB have the same height and equal
bases. ∴ area 4DCW = area 4W CX = area 4XCY
9 = area2 4Y CZ = area 4ZCB
1 45
1
= 5 (area 4BCD) = 5 2 = 2 units .
Therefore, the area of the shaded region is 4 29 = 18 units2 .
Solution 2
Since ABCD is a rectangle, the angle at A is 90◦ . We can then use the Pythagorean
Theorem
√
2
2
2
2
2
to calculate BD = AB + AD = 9 + 5 = 81 + 25 = 106,
√ and so BD = 106, since BD > 0.
Therefore, DW = W X = XY = Y Z = ZB = 51 (BD) = 51 106.
Using the formula area of a triangle = base×height
, base AB = 9 and height AD = 5, we can
2
2
45
calculate area 4ABD = 9×5
=
units
.
2
2
√
Let’s treat BD = 106 as the base √
of 4ABD and let h√be the corresponding height. Since the
45
area of 4ABD is 2 , then we have 106×h
= 45
and so 106 × h = 45, thus h = √45
.
2
2
106
√
4W AX and 4Y AZ both have height h = √45
and base 106
, so
5
√ 106
106
√45
area 4W AX = area 4Y AZ = 12
= 92 units2 .
5
106
Similarily, 4W CX and 4Y CZ both have height h = √45
and base
106
√ 106
√45
area 4W CX = area 4Y CZ = 21
= 92 units2 .
5
106
Therefore, the area of the shaded region is 4 29 = 18 units2 .
√
106
,
5
so