Problem 7 - Entrance Test

The side length of the square ABCD is 10. Area of square ABCD = side * side = 10 * 10 = 100. Point P is on AB such that AP = 4. Since AB = 10, PB = AB - AP = 10 - 4 = 6. Point Q is on BC such that BQ = 3. Since BC = 10, QC = BC - BQ = 10 - 3 = 7. We can find the area of triangle PDQ by subtracting the areas of the three right triangles surrounding it from the area of the square: Area(PDQ) = Area(Square ABCD) - Area(Triangle ADP) - Area(Triangle BPQ) - Area(Triangle CDQ). 1. Area(Triangle ADP): This is a right triangle with legs AD and AP. AD = 10 (side of square). AP = 4 (given). Area(ADP) = (1/2) * base * height = (1/2) * AD * AP = (1/2) * 10 * 4 = 20. 2. Area(Triangle BPQ): This is a right triangle with legs PB and BQ. PB = 6 (calculated). BQ = 3 (given). Area(BPQ) = (1/2) * PB * BQ = (1/2) * 6 * 3 = 9. 3. Area(Triangle CDQ): This is a right triangle with legs CD and QC. CD = 10 (side of square). QC = 7 (calculated). Area(CDQ) = (1/2) * CD * QC = (1/2) * 10 * 7 = 35. Now, calculate Area(PDQ): Area(PDQ) = 100 - 20 - 9 - 35 Area(PDQ) = 100 - (20 + 9 + 35) Area(PDQ) = 100 - 64 Area(PDQ) = 36.