Problem of the Week Problem C and Solution Locate the Fourth Vertex

Problem

Quadrilateral $$BDFH$$ is constructed so that each vertex is on a different side of square $$ACEG$$. Vertex $$B$$ is on side $$AC$$ so that $$AB=4\text{ cm}$$ and $$BC=6\text{ cm}$$. Vertex $$F$$ is on $$EG$$ so that $$EF=3\text{ cm}$$ and $$FG=7\text{ cm}$$. Vertex $$H$$ is on $$GA$$ so that $$GH=4\text{ cm}$$ and $$HA=6\text{ cm}$$. The area of quadrilateral $$BDFH$$ is $$47\text{ cm}^2$$.

The fourth vertex of quadrilateral $$BDFH$$, labelled $$D$$, is located on side $$CE$$ so that the lengths of $$CD$$ and $$DE$$ are both positive integers.

Determine the lengths of $$CD$$ and $$DE$$.

Solution

Solution 1

Since $$ACEG$$ is a square and the length of
$$AG=AH+HG=6+4=10$$ cm, then the length of each side of the square is $$10\text{ cm}$$ and the area is $$10\times 10 = 100\text{ cm}^2$$.

We can determine the area of triangles $$BAH$$ and $$FGH$$ using the formula $$\text{area} = \frac{\text{base}\times\text{height}}{2}$$.

In $$\triangle BAH$$, since $$BA$$ is perpendicular to $$AH$$, we can use $$BA$$ as the height and $$AH$$ as the base. The area of $$\triangle BAH$$ is $$\frac{6\times 4}{2}=12\text{ cm}^2$$.

In $$\triangle FGH$$, since $$FG$$ is perpendicular to $$GH$$, we can use $$FG$$ as the height and $$GH$$ as the base. The area of $$\triangle FGH$$ is $$\frac{4\times 7}{2}=14\text{ cm}^2$$.

The area of $$\triangle BCD$$ plus the area of $$\triangle FED$$ must be the total area minus the three known areas. That is, $$\text{Area }\triangle BCD +\text{Area }\triangle FED=100-12-47-14=27\text{ cm}^2$$.

$$CD$$ and $$DE$$ are both positive integers and $$CD+DE=10$$. We will systematically check all possible values for $$CD$$ and $$DE$$ to determine the values which produce the correct area.

$$CD$$ $$DE$$ Area$$\ \triangle BCD$$ Area$$\ \triangle FED$$ Area $$\triangle BCD\ +$$ Area $$\triangle FED$$
1 9 $$1\times 6 \div 2=3$$ $$9\times 3\div 2=13.5$$ $$3+13.5=16.5\not=27$$
2 8 $$2\times 6 \div 2=6$$ $$8\times 3\div 2=12$$ $$6+12=18\not=27$$
3 7 $$3\times 6 \div 2=9$$ $$7\times 3\div 2=10.5$$ $$9+10.5=19.5\not=27$$
4 6 $$4\times 6 \div 2=12$$ $$6\times 3\div 2=9$$ $$12+9=21\not=27$$
5 5 $$5\times 6 \div 2=15$$ $$5\times 3\div 2=7.5$$ $$15+7.5=22.5\not=27$$
6 4 $$6\times 6 \div 2=18$$ $$4\times 3\div 2=6$$ $$18+6=24\not=27$$
7 3 $$7\times 6 \div 2=21$$ $$3\times 3\div 2=4.5$$ $$21+4.5=25.5\not=27$$
8 2 $$8\times 6 \div 2=24$$ $$2\times 3\div 2=3$$ $$24+3=27$$
9 1 $$9\times 6 \div 2=27$$ $$1\times 3\div 2=1.5$$ $$27+1.5=28.5\not=27$$

Therefore, when $$CD=8$$ cm and $$DE=2$$ cm, the area of quadrilateral $$BDFH$$ is $$47\text{ cm}^2$$.

The second solution is more algebraic and will produce a solution for any lengths of $$CD$$ and $$DE$$ between $$0$$ and $$10$$ cm.

Solution 2

This solution begins the same as Solution 1. Algebra is introduced to complete the solution.

Since $$ACEG$$ is a square and the length of $$AG=AH+HG=6+4=10$$ cm, then the length of each side of the square is $$10\text{ cm}$$ and the area is $$10\times 10=100\text{ cm}^2$$.

We can determine the area of the triangles $$BAH$$ and $$FGH$$ using the formula $$\frac{\text{base}\times\text{height}}{2}$$.

In $$\triangle BAH$$, since $$BA$$ is perpendicular to $$AH$$, we can use $$BA$$ as the height and $$AH$$ as the base. The area of $$\triangle BAH$$ is $$\frac{6\times 4}{2}=12\text{ cm}^2$$.

In $$\triangle FGH$$, since $$FG$$ is perpendicular to $$GH$$, we can use $$FG$$ as the height and $$GH$$ as the base. The area of $$\triangle FGH$$ is $$\frac{4\times 7}{2}=14\text{ cm}^2$$.

The area of $$\triangle BCD$$ plus the area of $$\triangle FED$$ must be the total area minus the three known areas. That is, $$\text{Area }\triangle BCD +\text{Area }\triangle FED=100-12-47-14=27\text{ cm}^2$$.

Let the length of $$CD$$ be $$n$$ cm. Then the length of $$DE$$ is $$(10-n)$$ cm.

The area of $$\triangle BCD$$ is $$\dfrac{BC\times CD}{2}=\dfrac{6\times n}{2}=3n$$.

The area of $$\triangle FED$$ is $$\dfrac{FE\times DE}{2}=\dfrac{3\times (10-n)}{2}=\dfrac{10-n+10-n+10-n}{2}=\dfrac{30-3n}{2}$$.

Therefore, \begin{aligned} \text{Area }\triangle BCD+\text{Area }\triangle FED&=27\\ 3n+\frac{30-3n}{2}&=27\\ 6n+30-3n&=54 & \text{(multiplying both sides by 2)} \\ 3n+30&=54\\ 3n&=24\\ n&=8\\\end{aligned}

Therefore, the length of $$CD$$ is $$8$$ cm and the length of $$DE$$ is $$2$$ cm.

The algebra presented in Solution 2 may not be familiar to all students at this level.