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Difficulty: HardArea of Two-Dimensional Shapes

In rectangle ABCDABCD, the length of side ABAB is 1212 and the length of side BCBC is 99. Point PP lies on the diagonal ACAC such that AP=13ACAP = \frac{1}{3} AC. What is the area of triangle BPDBPD?

Answer: 18

Answer

18
The correct answer is 18. The area of the right triangle ABDABD is half of the area of rectangle ABCDABCD, which is 12×92=54\frac{12 \times 9}{2} = 54. Since point PP lies on diagonal ACAC such that AP=13ACAP = \frac{1}{3} AC, triangle ABPABP has a base of APAP along line ACAC and shares vertex BB with triangle ABCABC. Thus, its area is 13\frac{1}{3} of the area of triangle ABCABC, which is 543=18\frac{54}{3} = 18. Similarly, triangle ADPADP shares vertex DD with triangle ADCADC and has base APAP, so its area is 13\frac{1}{3} of the area of triangle ADCADC, which is 543=18\frac{54}{3} = 18. Because AP=13AC<12ACAP = \frac{1}{3} AC < \frac{1}{2} AC, point PP lies inside triangle ABDABD. Therefore, the area of triangle BPDBPD is the area of triangle ABDABD minus the areas of triangles ABPABP and ADPADP, which is 54−18−18=1854 - 18 - 18 = 18.

Step-by-Step Solution

1
Calculate the area of triangle ABDABD.
Area(△ABD)=12×92=54\text{Area}(\triangle ABD) = \frac{12 \times 9}{2} = 54
The diagonal BDBD divides the rectangle ABCDABCD into two congruent right triangles, each with an area equal to half of the rectangle's total area.
2
Find the areas of triangles ABPABP and ADPADP using the ratio of APAP to ACAC.
Area(△ABP)=13×Area(△ABC)=18\text{Area}(\triangle ABP) = \frac{1}{3} \times \text{Area}(\triangle ABC) = 18 and Area(△ADP)=13×Area(△ADC)=18\text{Area}(\triangle ADP) = \frac{1}{3} \times \text{Area}(\triangle ADC) = 18.
Triangles ABPABP and ABCABC share the same altitude from vertex BB to the line containing diagonal ACAC. Therefore, the ratio of their areas is equal to the ratio of their bases, which is APAC=13\frac{AP}{AC} = \frac{1}{3}. The same logic applies to triangles ADPADP and ADCADC with vertex DD.
3
Subtract the areas of triangles ABPABP and ADPADP from the area of triangle ABDABD to find the area of triangle BPDBPD.
Area(△BPD)=54−18−18=18\text{Area}(\triangle BPD) = 54 - 18 - 18 = 18
Since AP=13ACAP = \frac{1}{3} AC, which is less than half the length of the diagonal, point PP lies strictly within the interior of triangle ABDABD. Thus, the area of triangle ABDABD is partitioned into the areas of triangles ABPABP, ADPADP, and BPDBPD.

Key Concept

Partitioning the area of a polygon and using the ratio of bases for triangles sharing a vertex to compute sub-areas.

Alternative Method

Alternatively, place the rectangle in a coordinate system with BB at the origin (0,0)(0,0), CC at (12,0)(12,0), AA at (0,9)(0,9), and DD at (12,9)(12,9). The coordinates of point PP on diagonal ACAC (from (0,9)(0,9) to (12,0)(12,0)) at one-third of the distance from AA to CC are x=0+13(12−0)=4x = 0 + \frac{1}{3}(12 - 0) = 4 and y=9+13(0−9)=6y = 9 + \frac{1}{3}(0 - 9) = 6. The area of triangle BPDBPD with vertices B(0,0)B(0,0), P(4,6)P(4,6), and D(12,9)D(12,9) can be found using the shoelace formula: Area=12∣0(6−9)+4(9−0)+12(0−6)∣=12∣36−72∣=18\text{Area} = \frac{1}{2} |0(6 - 9) + 4(9 - 0) + 12(0 - 6)| = \frac{1}{2} |36 - 72| = 18.
Estimated Time:2m 0s
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Area of Two-Dimensional Shapes

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