Examkin

Question

Difficulty: MediumArea of Two-Dimensional Shapes

In square ABCDABCD, the side length is 1212. Point EE lies on side ABAB such that AE=3BEAE = 3BE, and point FF lies on side ADAD such that AF=FDAF = FD. What is the area of triangle CEFCEF?

Answer: 63

Answer

63
To find the area of triangle CEFCEF, we subtract the areas of the three right triangles surrounding it from the total area of square ABCDABCD. The area of square ABCDABCD is 122=14412^2 = 144. Point EE on side ABAB splits the side of length 1212 into segments AE=9AE = 9 and BE=3BE = 3. Point FF on side ADAD splits the side of length 1212 into equal segments AF=6AF = 6 and FD=6FD = 6. The areas of the three surrounding right triangles are: Area(△AEF)=12×9×6=27\text{Area}(\triangle AEF) = \frac{1}{2} \times 9 \times 6 = 27, Area(△EBC)=12×3×12=18\text{Area}(\triangle EBC) = \frac{1}{2} \times 3 \times 12 = 18, and Area(△FDC)=12×6×12=36\text{Area}(\triangle FDC) = \frac{1}{2} \times 6 \times 12 = 36. Subtracting these from the total area of the square yields Area(△CEF)=144−(27+18+36)=144−81=63\text{Area}(\triangle CEF) = 144 - (27 + 18 + 36) = 144 - 81 = 63.

Step-by-Step Solution

1
Determine the lengths of the segments created by points EE and FF on the sides of the square.
Since the square has a side length of 1212, the length of side ABAB is 1212. Given that AE=3BEAE = 3BE and AE+BE=12AE + BE = 12, we can write 3BE+BE=12  ⟹  4BE=12  ⟹  BE=33BE + BE = 12 \implies 4BE = 12 \implies BE = 3. This gives AE=9AE = 9. Since FF is the midpoint of ADAD (AF=FDAF = FD), we have AF=FD=122=6AF = FD = \frac{12}{2} = 6.
Finding these segment lengths is necessary to compute the base and height of the right triangles at the corners of the square.
2
Calculate the areas of the three right triangles surrounding triangle CEFCEF.
The area of right triangle AEFAEF is 12×AE×AF=12×9×6=27\frac{1}{2} \times AE \times AF = \frac{1}{2} \times 9 \times 6 = 27. The area of right triangle EBCEBC is 12×BE×BC=12×3×12=18\frac{1}{2} \times BE \times BC = \frac{1}{2} \times 3 \times 12 = 18. The area of right triangle FDCFDC is 12×FD×CD=12×6×12=36\frac{1}{2} \times FD \times CD = \frac{1}{2} \times 6 \times 12 = 36.
These three triangles occupy the entire area of the square except for the region defined by triangle CEFCEF.
3
Subtract the sum of the areas of the three right triangles from the total area of square ABCDABCD.
The total area of square ABCDABCD is 122=14412^2 = 144. The area of triangle CEFCEF is 144−(27+18+36)=144−81=63144 - (27 + 18 + 36) = 144 - 81 = 63.
This subtraction removes the corner regions, leaving only the area of the central triangle.

Key Concept

Calculating the area of an inscribed polygon by subtracting the areas of simpler surrounding geometric shapes from a larger bounding shape.

Alternative Method

Alternatively, coordinate geometry can be used. Place the vertex DD at the origin (0,0)(0,0) on the coordinate plane. Then the coordinates of the vertices of the square are D(0,0)D(0,0), C(12,0)C(12,0), B(12,12)B(12,12), and A(0,12)A(0,12). Point EE lies on segment ABAB and is located at (9,12)(9,12). Point FF lies on segment ADAD and is located at (0,6)(0,6). The area of the triangle with vertices C(12,0)C(12,0), E(9,12)E(9,12), and F(0,6)F(0,6) can be found using the Shoelace Formula: Area=12∣12(12−6)+9(6−0)+0(0−12)∣=12∣72+54+0∣=12(126)=63\text{Area} = \frac{1}{2} |12(12 - 6) + 9(6 - 0) + 0(0 - 12)| = \frac{1}{2} |72 + 54 + 0| = \frac{1}{2} (126) = 63.
Estimated Time:1m 30s
Rate this question

Topics

Area of Two-Dimensional Shapes

More questions from this topic