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Geometry Difficulty 3.0 AMC 10/12 Prove it Canada

In the grids below, dots are $1$ unit apart horizontally and $1$ unit apart vertically. Shapes are created by connecting dots with straight lines. In Figure 1, the shaded shape is a rectangle with width $8$ and height $6$ that has a $1 \times 1$ square removed. What is the area of the shaded shape in Figure 1? In Figure 2, the shaded shape is a rectangle with width 8 and height $6$ that has a triangle removed. What is the area of the shaded shape in Figure 2? In Figure 3, trapezoid $ABCD$ has $AD=5$ , $BC=3$ and $CD=8$ . Point $G$ is placed vertically below $C$ and point $H$ is placed vertically below $D$ , so that $GH$ is parallel to $CD$ and the area of trapezoid $ABGH$ is twice the area of $ABCD$ . Determine the length of $BG$ .

Figure 1

Figure 2

Figure 3

Solution

The area of the rectangle without the square removed is $8\times6=48$ . The area of the square that has been removed from the rectangle is $1\times1=1$ . The area of the shaded shape in Figure 1 is $48-1=47$ . The removed triangle has base $6$ and height $2$ , and thus has area $\frac12\times6\times2=6$ . Following the work done in (a), the area of the shaded shape in Figure 2 is $48-6=42$ . We begin by determining the area of $ABCD$ . Construct line segment $BE$ perpendicular to $AD$ with $E$ on $AD$ , dividing $ABCD$ into right-angled triangle $ABE$ and rectangle $EBCD$ , as shown. [[IMAGE0]] The area of $\triangle ABE$ is $\frac12\times 8 \times 2=8$ . The area of rectangle $EBCD$ is $8\times3=24$ , and so the area of $ABCD$ is $8+24=32$ . Therefore, the area of trapezoid $ABGH$ is $2\times32=64$ . Trapezoid $ABGH$ can be divided into right-angled triangle $ABE$ and rectangle $EBGH$ . The area of $\triangle ABE$ is 8, and so the area of rectangle $EBGH$ is $64-8=56$ . The area of $EBGH$ is $EB\times BG=8\times BG=56$ , and so $BG=\dfrac{56}{8}=7$ .

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