3. Determine the area of the shaded part of rectangle $A B C D$ whose area is $144 \mathrm{~cm}^{2}$, and whose width is $\frac{1}{4}$ of its length. The shaded part consists of three congruent quadrilaterals, and points $E, F$ and $G$ lie on the perpendicular bisector of side $\overline{B C}$.

![](https://cdn.mathpix.com/cropped/2024_06_03_65ab0e546d966e8ed829g-1.jpg?height=446&width=1650&top_left_y=1476&top_left_x=243)
3. In the rectangle $A B C D$, let's denote $|A B|=a$ and $|B C|=b$.

Since the width of the rectangle is $\frac{1}{4}$ of the length of the rectangle, we have $b=\frac{1}{4} a$, or $a=4 b$.

Then the area of the rectangle is $P_{A B C D}=a b=4 b \cdot b$.

1 POINT

Then we have:

$4 b \cdot b=144$,

$b \cdot b=36$,

$b=6 \text{ cm}$, and $a=24 \text{ cm}$.

1 POINT

Since the shaded quadrilaterals are congruent, let's determine the area of one of them. Denote this quadrilateral as $\mathrm{s}$ MNED.

![](https://cdn.mathpix.com/cropped/2024_05_30_a0ea3976463634d3abe5g-2.jpg?height=463&width=619&top_left_y=1079&top_left_x=290)

The area of this quadrilateral is equal to the difference in the areas of two triangles, i.e.,

$P_{M N E D}=P_{M N D}-P_{E N D}$.

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D

![](https://cdn.mathpix.com/cropped/2024_05_30_a0ea3976463634d3abe5g-2.jpg?height=359&width=452&top_left_y=1668&top_left_x=308)

Denote $\overline{M K}$ as the height of triangle $\triangle M N D$.

Since the length of side $\overline{N D}$ is one-third of the length of side $a$ of the rectangle, and the length of height $\overline{M K}$ is the width of rectangle $A B C D$, we have:

$\begin{array}{ll}|N D|=\frac{1}{3} \cdot 24=8 \text{ cm}, & 1 \text{ POINT } \\ |M K|=6 \text{ cm} & 1 \text{ POINT } \\ P_{M N D}=\frac{|N D| \cdot|M K|}{2}=\frac{8 \cdot 6}{2}=24 \text{ cm}^2 . & 1 \text{ POINT}\end{array}$

![](https://cdn.mathpix.com/cropped/2024_05_30_a0ea3976463634d3abe5g-3.jpg?height=290&width=627&top_left_y=246&top_left_x=289)

Denote $\overline{E H}$ as the height of triangle $\triangle E N D$.

Since point $E$ lies on the perpendicular bisector of side $\overline{B C}$ or $\overline{A D}$, the length of segment $\overline{E H}$ is half the width of rectangle $A B C D$, i.e.,

$|E H|=\frac{6}{2}=3 \text{ cm}$.

This means that $P_{E N D}=\frac{|N D| \cdot|E H|}{2}=\frac{8 \cdot 3}{2}=12 \text{ cm}^2$.

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Now, $P_{M N E D}=P_{M N D}-P_{E N D}=24-12=12 \text{ cm}^2$.

Note: To determine the area of $\triangle E N D$, the fact that triangles $\triangle E N D$ and $\triangle M N D$ share the same base $\overline{N D}$, and the lengths of the corresponding heights satisfy

$|M K|=|A D|,|E H|=\frac{1}{2}|A D|$ so $|E H|=\frac{1}{2}|M K|$.

Therefore, we have: $P_{E N D}=\frac{1}{2} P_{M N D}=\frac{24}{2}=12 \text{ cm}^2$.